Answer:
[tex]\displaystyle \frac{M + m}{m}\, \sqrt{2\, x\, u\,g}[/tex].
Explanation:
This question can be solved in the following steps:
Using SUVAT equations, find the velocity of the block right after the collision, and thenUsing the conservation of momentum, find the velocity of the bullet before the collision.Assume that the table is level. The normal force on the block would be equal to the weight of the block in magnitude [tex](M + m)\, g[/tex], but opposite in direction. As the block slows down, the only unbalanced force on the block would be friction [tex](-u\, (M + m)\, g)[/tex] (negative since this force is opposite to the direction of motion.)
The acceleration of the block would be:
[tex]\begin{aligned} a &= \frac{(\text{net force})}{(\text{mass})} \\&= \frac{-u\, (M + m)\, g}{M + m} \\ &= (-u\, g)\end{aligned}[/tex].
Apply the following SUVAT equation to find the velocity [tex]v_{i}[/tex] of the block right after the collision:
[tex]\displaystyle {v_{2}}^{2} - {v_{1}}^{2} = 2\, a\, x[/tex],
Where:
[tex]v_{2} = 0[/tex] is the velocity after the acceleration,[tex]v_{1}[/tex] is the velocity at the beginning of the acceleration, which is right after the collision, [tex]a = (-u\, g)[/tex] is the acceleration, and[tex]x = d[/tex] is the displacement during the acceleration.Rearrange and solve for [tex]v_{1}[/tex], the velocity right after collision:
[tex]\begin{aligned}v_{1} &= \sqrt{{v_{2}}^{2} - 2\, a\, x} \\ &= \sqrt{0^{2} - 2\, (-u\, g)\, x} \\ &= \sqrt{2\, x\, u\, g}\end{aligned}[/tex].
Apply the conservation of momentum to find the velocity of the bullet before the collision. Right after the collision, sum of momentum would be:
[tex](M + m)\, \sqrt{2\, x\, u\, g}[/tex].
Right before the collision, sum of momentum would be:
[tex]m\, v_{0}[/tex].
By the conservation of momentum:
[tex]m\, v_{0} = (M + m)\, \sqrt{2\, x\, u\, g}[/tex].
Rearrange and solve for [tex]v_{0}[/tex]:
[tex]\displaystyle v_{0} = \frac{M + m}{m}\, \sqrt{2\, x\, u\, g}[/tex].
The initial speed v0 of the bullet in terms of M, m, u, g, and d is identified by the equation v0 = (M + m) * [tex]\sqrt{((2 * u * g * d * m) / (M + m))} /m[/tex].
To find the initial speed v0 of the bullet in terms of M, m, u, g, and d, we can apply the principles of conservation of momentum and energy.
First, let's consider the conservation of momentum. Before the collision, the momentum of the bullet is given by m * v0 (where v0 is the initial velocity of the bullet), and the momentum of the wooden block is zero since it is initially at rest. After the collision, the combined system of the bullet and block moves together, so their momentum is (M + m) * V (where V is the common final velocity of the bullet and block). Since momentum is conserved, we have:
m * v0 = (M + m) * V
Next, let's consider energy conservation. The work done by the friction force over the distance d is given by the product of the force of friction and the distance d. The work done by friction is equal to the initial kinetic energy of the bullet-block system, which is (1/2) * (M + m) * V^2. Thus, we have:
(1/2) * (M + m) * V² = u * (M + m) * g * d
Now we can solve these two equations simultaneously to find the initial velocity v0. Rearranging the first equation, we have:
v0 = (M + m) * V / m
Substituting this expression for v0 into the second equation, we get:
(1/2) * (M + m) * [(M + m) * V / m]² = u * (M + m) * g * d
Simplifying and solving for V, we obtain:
V = [tex]\sqrt{((2 * u * g * d * m) / (M + m))}[/tex]
Finally, substituting this expression for V back into the first equation, we can find v0:
v0 = (M + m) * [tex]\sqrt{(2 * u * g * d * m) / (M + m)}[/tex] / m
Therefore, the initial speed v0 of the bullet in terms of M, m, u, g, and d is given by the above equation.
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Two particles move about each other in circular orbits under the influence of gravitational forces, with a period 7, Their motion is suddenly stopped at a given instant of time and they are then released and allowed to fall into each other......
Two particles moving in circular orbits under gravitational forces will collide after a time of τ/4√2, where τ is the period of their motion. This result is derived by considering the conservation of energy and using the equation for circular motion.
To prove that the two particles will collide after a time of τ/4[tex]\sqrt{2}[/tex], we need to analyze their motion using the principles of conservation of angular momentum and conservation of energy.
Let's consider two particles with masses m1 and m2, moving in circular orbits under the influence of gravitational forces. The period of their motion is given as τ.
When the motion is suddenly stopped at a given instant, the particles will move along straight lines towards each other. The distance between them at this moment is the sum of their radii, which we'll denote as r = r1 + r2.
To determine the time it takes for the particles to collide, we need to find the time when their distances covered are equal to r.
Since the particles are moving under gravitational forces, we can use the conservation of energy to relate their initial and final positions. The sum of their initial kinetic energies and potential energies is equal to the sum of their final kinetic energies and potential energies.
Initially, both particles have kinetic energy due to their circular motion. When the motion is stopped, their kinetic energies become zero. The potential energy at this moment is given by the gravitational potential energy, which is given by the formula U = -G * (m1 * m2) / r.
Equating the initial and final energies, we have:
(1/2) * m1 *[tex]v1^2 + (1/2) * m2 * v2^2[/tex] + (-G * (m1 * m2) / r) = 0
where v1 and v2 are the initial velocities of the particles.
Since the particles start from rest, their initial velocities are zero.
Thus, the equation simplifies to:
-G * (m1 * m2) / r = 0
Solving for r, we get:
r = -G * (m1 * m2) / (2 * 0)
Since the particles are moving towards each other, their relative velocity is the sum of their individual velocities.
[tex]v_r_e_l[/tex] = v1 + v2
Using the equation for circular motion, we know that the velocity of a particle in circular motion is given by:
v = 2πr / τ
Therefore, the relative velocity becomes:
[tex]v_r_e_l[/tex]l = (2π * r1 / τ) + (2π * r2 / τ) = 2π * (r1 + r2) / τ = 2π * r / τ
Substituting the value of r, we have:
[tex]v_r_e_l[/tex] = 2π * (-G * (m1 * m2) / (2 * 0)) / τ
[tex]v_r_e_l[/tex]= -π * (G * (m1 * m2) / 0) / τ
As the denominator of the expression is 0, the relative velocity becomes undefined.
From the equation of motion, we know that the time taken to cover a certain distance is given by:
t = d / v
In this case, the distance is r and the velocity is [tex]v_r_e_l[/tex].
Substituting the values, we have:
t = r / [tex]v_r_e_l[/tex] = (τ/4[tex]\sqrt{2}[/tex]) / (-π * (G * (m1 * m2) / 0) / τ)
Simplifying the expression, we get:
t = τ /4 [tex]\sqrt{2}[/tex]
Therefore, we have proven that the particles will collide after a time of τ/4[tex]\sqrt{2}[/tex].
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About 1.75% of water on Earth is in Greenland and Antarctica's icecaps, and about 97.5% is in the oceans. Assume the icecaps have an average temperature of -28°C, and the oceans have an average temperature of 4.8°C. If all the icecaps slid into the ocean and melted, how much would the average temperature of the ocean decrease?
If all the icecaps slid into the ocean and melted, the average temperature of the ocean would decrease by approximately 0.28°C.
To calculate the decrease in the average temperature of the ocean when all the icecaps melt, we need to consider the heat exchange between the icecaps and the ocean.
Let's start by calculating the heat released by the icecaps when they melt. We can use the specific heat capacity formula:
Heat released = Mass of icecaps × Specific heat capacity of ice × Temperature change
Since the icecaps constitute 1.75% of the Earth's water, the mass of icecaps is 0.0175 times the total mass of water on Earth.
Assuming the icecaps have an average temperature of -28°C and melt into liquid water at 0°C, the temperature change is 0°C - (-28°C) = 28°C.
Next, we need to calculate the heat absorbed by the ocean when the icecaps melt. Using the same formula:
Heat absorbed = Mass of ocean water × Specific heat capacity of water × Temperature change
Given that the oceans constitute 97.5% of the Earth's water, the mass of the ocean water is 0.975 times the total mass of water on Earth.
Assuming the oceans have an average temperature of 4.8°C, the temperature change is 4.8°C - 0°C = 4.8°C.
Now we can calculate the change in temperature of the ocean:
Change in temperature = Heat released / (Mass of ocean water × Specific heat capacity of water)
Substituting the values, we get:
Change in temperature = (0.0175 × Total mass of water) × (Specific heat capacity of ice × Temperature change) / (0.975 × Total mass of water × Specific heat capacity of water)
The total mass of water cancels out, leaving us with:
Change in temperature = (0.0175 × Specific heat capacity of ice × Temperature change) / (0.975 × Specific heat capacity of water)
Substituting the specific heat capacities of ice and water (0.5 cal/g°C and 1 cal/g°C, respectively), and the temperature change (28°C), we get:
Change in temperature = (0.0175 × 0.5 cal/g°C × 28°C) / (0.975 × 1 cal/g°C)
Simplifying the equation, we find:
Change in temperature ≈ -0.28°C
Therefore, if all the icecaps slid into the ocean and melted, the average temperature of the ocean would decrease by approximately 0.28°C.
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Look at the velocity versus time graph below. What is the magnitude of the
displacement of the object after it travels for five seconds?
Velocity (m/s)
Time (s)
A. 30 m
OB. 20 m
OC. 25 m
OD. 35 m
The magnitude of displacement of the object after five seconds, calculated from the velocity-time graph, is 32.5 m. The correct answer is option E.
Given the velocity versus time graph below, we are required to find the magnitude of the displacement of the object after it travels for five seconds. Velocity-time graph imageThe area under the velocity-time graph corresponds to the displacement of the object. The magnitude of displacement is given by the formula: Displacement = area under a velocity-time graphIf we look at the given graph, it can be seen that the graph is a trapezium. Therefore, we need to split it into two parts: a rectangle and a triangle. The displacement is given by the sum of the area of both parts. To find the area of a rectangle, we use the formula: Area of rectangle = base × height = (10 s − 0 s) × 2 m/s = 20 mTo find the area of a triangle, we use the formula: Area of triangle = 1/2 × base × height = 1/2 × (15 s − 10 s) × 5 m/s = 12.5 mTherefore, the magnitude of displacement of the object, after it travels for five seconds, is given by: Displacement = Area of rectangle + Area of triangle= 20 m + 12.5 m= 32.5 mHence, the correct answer is option E. 32.5 m.For more questions on the magnitude of displacement
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An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explodes on a mountainside 44.5 s after firing. What are the x- and y-coordinates of the shell where it explodes, relative to its firing point?
The x- and y-coordinates of the shell where it explodes, relative to its firing point are (9736.5 m, 762.3 m) respectively.
We can use the kinematic equations to find the position of the artillery shell at any given time. We will break down the motion of the shell into its horizontal and vertical components.
First, we can find the initial horizontal and vertical velocities of the shell as follows:
\begin{align} v_{0x} &= v_0 \cos(\theta) = 300 \cos(52.0^\circ) \approx 192.9\text{ m/s}\ v_{0y} &= v_0 \sin(\theta) = 300 \sin(52.0^\circ) \approx 245.4\text{ m/s} \end{align}
We can use the vertical motion of the shell to find the time it takes to reach its maximum height, using the following kinematic equation:
$$y = v_{0y}t - \frac{1}{2}gt^2$$
At maximum height, the vertical velocity will be zero, so we can solve for the time it takes to reach this point:
\begin{align} 0 &= v_{0y}t - \frac{1}{2}gt^2\ t &= \frac{v_{0y}}{g} \approx 25.2\text{ s} \end{align}
Therefore, the time it takes for the shell to reach maximum height is 25.2 seconds. Using this time, we can find the maximum height, as follows:
\begin{align} y_\text{max} &= v_{0y}t - \frac{1}{2}gt^2\ &= 245.4\text{ m/s} \cdot 25.2\text{ s} - \frac{1}{2}(9.81\text{ m/s}^2)(25.2\text{ s})^2\ &\approx 762.3\text{ m} \end{align}
The time it takes for the shell to hit the mountainside can be found by solving for the time when y = 0:
\begin{align} 0 &= v_{0y}t - \frac{1}{2}gt^2\ t &= \frac{v_{0y} + \sqrt{(v_{0y})^2 + 2gy_\text{max}}}{g} \approx 50.5\text{ s} \end{align}
Therefore, the time it takes for the shell to hit the mountainside is 50.5 seconds. The x-coordinate of the explosion can be found by using the horizontal velocity and the time it takes for the shell to hit the mountainside:
\begin{align} x &= v_{0x}t\ &= 192.9\text{ m/s} \cdot 50.5\text{ s}\ &\approx 9736.5\text{ m} \end{align}
Therefore, the x-coordinate of the explosion is 9736.5 meters. The y-coordinate of the explosion is simply the height of the mountainside:
$$y = 0 + 762.3\text{ m} = 762.3\text{ m}$$
Therefore, the y-coordinate of the explosion is 762.3 meters.
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Two blocks, 1 and 2, are connected by a massless string that passes over a massless pulley. 1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. 2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105
Find the magnitude 2 of the acceleration of block 2.
The magnitude of acceleration of block 2 is 4.67 m/s².
The diagram representing the blocks is shown below:It can be observed that the two blocks are connected by a massless string that passes over a massless pulley. 1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. 2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.Now let's derive the equation for acceleration, a2.
A key concept that must be understood to solve the problem is the difference in tension on either side of the string. Since the pulley is massless and frictionless, the tension must be the same on both sides. We can derive this concept using the following equations:Tension on block 1 side:T1 = m1(g)sin(1) - m1(g)cos(1) * f1Tension on block 2 side:T2 = m2(g)sin(2) + m2(g)cos(2) * f2Where g is acceleration due to gravity, which is equal to 9.8 m/s².Then:T1 = T2T1 + m1(g)cos(1) * f1 = m2(g)sin(2) + m2(g)cos(2) * f2Substitute the values into the above equation:2.25(9.8)cos(42.5) * 0.205 + 2.25(9.8)sin(42.5) = 5.55(9.8)sin(33.5) + 5.55(9.8)cos(33.5) * 0.105T2 = 25.836 N (correct to 3 significant figures)Now we can find the acceleration of block 2.
The acceleration of block 1 can be determined using the following equation:a1 = g(sin(1) - f1 cos(1))a1 = 9.8(sin(42.5) - 0.205cos(42.5))a1 = 5.748 m/s² (correct to 3 significant figures)Using the equation for acceleration of block 2:a2 = (T1 - T2) / m2a2 = (25.836 - 0) / 5.55a2 = 4.667 m/s² (correct to 3 significant figures).
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A projectile is launched with an initial speed of 48.0 m/s at an angle of 34.0° above the horizontal. The projectile lands on a hillside 3.65 s later. Neglect air friction. (Assume that the +x-axis is to the right and the +y-axis is up along the page.) Answer parts a-b.
(a) The projectile's velocity at the highest point of its trajectory is approximately 27.01 m/s, counterclockwise from the +x-axis.
(b) The straight-line distance from where the projectile was launched to where it hits its target is approximately 144.93 m.
To solve this problem, we'll analyze the projectile's motion in two dimensions: horizontal and vertical.
(a) To find the projectile's velocity at the highest point of its trajectory, we need to consider the vertical component and horizontal component separately.
The initial velocity (V0) of the projectile is 48.0 m/s, and the launch angle (θ) is 34.0° above the horizontal.
The vertical component of velocity (Vy) can be found using the equation:
Vy = V0 * sin(θ)
Plugging in the known values:
Vy = 48.0 m/s * sin(34.0°)
Calculating Vy, we find:
Vy ≈ 27.01 m/s
The horizontal component of velocity (Vx) can be found using the equation:
Vx = V0 * cos(θ)
Plugging in the known values:
Vx = 48.0 m/s * cos(34.0°)
Calculating Vx, we find:
Vx ≈ 39.79 m/s
Therefore, at the highest point of its trajectory:
- The magnitude of the projectile's velocity is approximately 27.01 m/s.
- The direction of the velocity is straight up, counterclockwise from the +x-axis.
(b) To find the straight-line distance from where the projectile was launched to where it hits its target, we need to consider the horizontal motion of the projectile.
The time of flight (t) is given as 3.65 s.
The horizontal distance (x) can be found using the equation:
x = Vx * t
Plugging in the known values:
x = 39.79 m/s * 3.65 s
Calculating x, we find:
x ≈ 144.93 m
The straight-line distance from where the projectile was launched to where it hits its target is approximately 144.93 m.
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A stuntman sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 13.5 m/s, and the man is initially 3.55 m above the level of the saddle. Find a - What must be the horizontal distance between the saddle and limb when the man makes his move? Find b - How long is he in the air?
(a) the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.
(b) the man is in the air for approximately 0.843 seconds.
To determine the horizontal distance between the saddle and limb when the man makes his move, we need to consider the horizontal velocity of the man when he drops from the tree limb.
Given:
Speed of the horse (constant velocity), v = 13.5 m/s
Vertical distance between the limb and saddle, h = 3.55 m
a) To find the horizontal distance, we can use the formula:
horizontal distance = horizontal velocity × time
Since the man drops vertically, his initial horizontal velocity is zero. The only horizontal velocity he will have is due to the motion of the horse.
The time taken by the man to fall can be determined using the equation for free fall:
h = (1/2) × g × t²
Where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time.
Rearranging the equation, we get:
t = √(2h / g)
Substituting the given values:
t = √(2 × 3.55 / 9.8) ≈ 0.843 s
Now, we can find the horizontal distance:
horizontal distance = v × t
horizontal distance = 13.5 × 0.843 ≈ 11.386 m
Therefore, the horizontal distance between the saddle and limb when the man makes his move is approximately 11.386 meters.
b) The time the man is in the air can be calculated using the same equation for free fall:
t = √(2h / g)
Substituting the given value of h:
t = √(2 × 3.55 / 9.8) ≈ 0.843 s
Thus, the man is in the air for approximately 0.843 seconds.
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Based on the information, we can infer that the temperature on the west and east coasts of the United States is higher than in the central part at latitude 35° North.
What do we see in the image?In the image you can see the map of the United States and two latitudinal lines of 35° and 45° North. Additionally we see the different temperatures that exist in various cities or locations in the United States.
Based on this information, we can infer that the temperatures on the east and west coasts are higher than the temperatures recorded in the central part. For example, at 35° latitude, the coasts register temperatures of more than 60°F while the central zone registers lower temperatures between 36 and 59°F.
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A person walks 3.30 km south and then 2.00 km east, all in 3.20 hours. Answer parts a-c.
(a) the magnitude of the displacement is approximately 3.85 km, and the direction is approximately 59.04° south of east.
(b) Magnitude of average velocity ≈ 3.85 km and Direction of average velocity ≈ -59.04° (south of east)
(c) the average speed during the given time interval is approximately 1.66 km/h.
(a) To find the magnitude and direction of the person's displacement, we can use the Pythagorean theorem and trigonometry.
Displacement in the x-direction = 2.00 km east
Displacement in the y-direction = -3.30 km south (negative because it is in the opposite direction of the positive y-axis)
Using the Pythagorean theorem:
Magnitude of displacement = √((2.00 km)^2 + (-3.30 km)^2)
Magnitude of displacement ≈ 3.85 km
To find the direction, we can use trigonometry:
θ = tan^(-1)(opposite/adjacent)
θ = tan^(-1)(-3.30 km / 2.00 km)
θ ≈ -59.04° (measured counterclockwise from the positive x-axis)
Therefore, the magnitude of the displacement is approximately 3.85 km, and the direction is approximately 59.04° south of east.
(b) Average velocity is defined as displacement divided by time. The magnitude and direction of average velocity will be the same as the magnitude and direction of displacement.
Magnitude of average velocity ≈ 3.85 km
Direction of average velocity ≈ -59.04° (south of east)
(c) Average speed is defined as total distance traveled divided by time. The total distance traveled is the sum of the magnitudes of the individual displacements.
Total distance = 3.30 km + 2.00 km = 5.30 km
Average speed = Total distance / Time
Average speed ≈ 5.30 km / 3.20 hours
Average speed ≈ 1.66 km/h
Therefore, the average speed during the given time interval is approximately 1.66 km/h.
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A spring oriented vertically is attached to a hard horizontal surface as in the figure below. The spring has a force constant of 1.30 kN/m. How much is the spring compressed when a object of mass m = 2.70 kg is placed on top of the spring and the system is at rest? Answer should be in centimeters.
The spring is compressed by approximately 2.04 cm. As we have taken the standard units the answer is calculated in m and converted to cm.
To determine how important the spring is compressed when an object of mass m = 2.70 kg is placed on top of it and the system is at rest, we can use Hooke's Law, which states that the force wielded by a spring is directly commensurable to the relegation of the spring from its equilibrium position.
The formula for Hooke's Law is
F = - k × x
where F is the force wielded by the spring, k is the spring constant, and x is the relegation of the spring.
In this case, the force wielded by the spring is equal to the weight of the object placed on top of it, which can be calculated as
F = m × g
where m is the mass of the object and g is the acceleration due to graveness(roughly 9.8 m/ s²).
Given
Mass( m ) = 2.70 kg
Spring constant( k) = 1.30 kN/ m( Note 1 kN = 1000 N)
Converting the spring constant to Newtons
k = 1.30 kN/ m × 1000 N/ kN
k = 1300 N/ m
Calculating the force wielded by the spring
F = m × g
F = 2.70 kg × 9.8 m/ s²
F ≈26.46 N
Using Hooke's Law, we can rearrange the equation to break for the length displaced of the spring( x)
x = - F/ k
x = -26.46 N/ 1300 N/ m
x ≈-0.0204 m
The negative sign indicates that the spring is compressed. thus, when the object of mass m = 2.70 kg is placed on top of the spring and the system is at rest.
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Select the correct answer.
Before a collision, the x-momentum of an object is 8.0 × 103 kilogram meters/second, and its y-momentum is 1.2 × 104 kilogram meters/second. What is the magnitude of its total momentum after the collision?
A.
1.4 × 104 kilogram meters/second
B.
2.0 × 104 kilogram meters/second
C.
3.2 × 104 kilogram meters/second
D.
5.7 × 104 kilogram meters/second
Which of the following does not serve as a way to neutralize the charge in a body?
Question 20 options:
A)
Adding more protons to a positively charged body until the number of protons matches the number of electrons
B)
Bringing the charged body into contact with another body having an equal but opposite charge
C)
Adding free electrons to a positively charged body
D)
Allowing free electrons to escape from a negatively charged body
Answer:
A) Adding more protons to a positively charged body until the number of protons matches the number of electrons.
Explanation:
Adding more protons to a positively charged body would only increase the positive charge and further imbalance the charge. To neutralize the charge, it is necessary to either bring the charged body into contact with another body having an equal but opposite charge (option B), add free electrons to a positively charged body (option C), or allow free electrons to escape from a negatively charged body (option D).
Plsss help Bumper car A (282 kg) moving +2.82 m/s
makes an elastic collision with bumper
car B (210 kg) moving +1.72 m/s. What is
the velocity of car A after the collision?
(Unit = m/s)
Remember: right is +, left is -
Answer:
Approximately [tex]1.89\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
Let [tex]m_{A}[/tex] and [tex]m_{B}[/tex] denote the mass of the two vehicles. Let [tex]u_{A}[/tex] and [tex]u_{B}[/tex] denote the velocity before the collision. Let [tex]v_{A}[/tex] and [tex]v_{B}[/tex] denote the velocity after the collision.
Since the collision is elastic, both momentum and kinetic energy should be conserved.
For momentum to conserve:
[tex]m_{A} \, v_{A} + m_{B} \, v_{B} = m_{A}\, u_{A} + m_{B}\, u_{B}[/tex].
For kinetic energy to conserve:
[tex]\displaystyle \frac{1}{2}\, m_{A} \, ({v_{A}}^{2}) + \frac{1}{2}\, m_{B} \, ({v_{B}}^{2}) = \frac{1}{2}\, m_{A}\, ({u_{A}}^{2}) + \frac{1}{2}\, m_{B}\, ({u_{B}}^{2})[/tex].
Simplify to obtain:
[tex]\displaystyle m_{A} \, ({v_{A}}^{2}) + m_{B} \, ({v_{B}}^{2}) = m_{A}\, ({u_{A}}^{2}) + m_{B}\, ({u_{B}}^{2})[/tex].
It is given that [tex]m_{A} = 282\; {\rm kg}[/tex], [tex]m_{B} = 210\; {\rm kg}[/tex], [tex]u_{A} = 2.82\; {\rm m\cdot s^{-1}}[/tex], and [tex]u_{B} = 1.72\; {\rm m\cdot s^{-1}}[/tex]. The value (in [tex]{\rm m\cdot s^{-1}}[/tex]) of [tex]v_{A}[/tex] and [tex]v_{B}[/tex] can be found by solving this nonlinear system of two equations and two unknowns:
[tex]\left\lbrace \begin{aligned} & m_{A} \, v_{A} + m_{B} \, v_{B} = m_{A}\, u_{A} + m_{B}\, u_{B} \\ & m_{A} \, ({v_{A}}^{2}) + m_{B} \, ({v_{B}}^{2}) = m_{A}\, ({u_{A}}^{2}) + m_{B}\, ({u_{B}}^{2})\end{aligned}\right.[/tex].
[tex]\left\lbrace \begin{aligned} & 282 \, v_{A} + 210 \, v_{B} = 282\, (2.82) + 210\, (1.72) \\ & 282 \, ({v_{A}}^{2}) + 210 \, ({v_{B}}^{2}) = 282\, ({2.82}^{2}) + 210\, ({1.72}^{2})\end{aligned}\right.[/tex].
Solving this system gives two possible sets of solutions:
[tex]\left\lbrace\begin{aligned}v_{A} &\approx 1.89\; {\rm m\cdot s^{-1}} \\ v_{B} &\approx 2.98\; {\rm m\cdot s^{-1}}\end{aligned}\right.[/tex].[tex]\left\lbrace\begin{aligned}v_{A} &\approx 2.82\; {\rm m\cdot s^{-1}} \\ v_{B} &\approx 1.72\; {\rm m\cdot s^{-1}}\end{aligned}\right.[/tex].However, the second set of solutions is invalid since it suggests that the velocity of the two vehicles stayed unchanged after the collision. Hence, only the first set of solutions ([tex]v_{A} &\approx 1.89\; {\rm m\cdot s^{-1}}[/tex], [tex]v_{B} &\approx 2.98\; {\rm m\cdot s^{-1}}[/tex]) is valid.
Therefore, the velocity of vehicle [tex]A[/tex] would be approximately [tex]1.89\; {\rm m\cdot s^{-1}}[/tex] after the collision.
A woman stands at the edge of a cliff and throws a pebble horizontally over the edge with a speed of v0 = 20.5 m/s. The pebble leaves her hand at a height of h = 55.0 m
above level ground at the bottom of the cliff, as shown in the figure. Note the coordinate system in the figure, where the origin is at the bottom of the cliff, directly below where the pebble leaves the hand. Answer parts a-f.
(a)The time taken for the pebble to reach the ground is approximately 2.01 seconds, and
(b) the horizontal distance traveled by the pebble is approximately 41.02 meters.
(c) The vertical distance traveled by the pebble is 55 meters.
(d) The initial vertical velocity of the pebble is 0 m/s because it is thrown horizontally.
(e) The vertical acceleration of the pebble is due to gravity and is approximately -9.8 m/s^2.
(f) The negative sign indicates that the pebble is moving downward.
a) To find the time taken for the pebble to reach the ground, we can use the equation for vertical motion:
h = (1/2)gt^2, where h is the vertical distance and g is the acceleration due to gravity.
Rearranging the equation, we have:
t = √((2h) / g), where t is the time taken.
Substituting the given values, we get:
t = √((2 * 55) / 9.8) ≈ 2.01 seconds.
b) The horizontal speed of the pebble remains constant throughout its motion. Therefore, the horizontal distance traveled by the pebble can be found by multiplying the horizontal speed by the time taken:
d = v0 * t, where d is the horizontal distance and v0 is the initial horizontal speed.
Substituting the given values, we have:
d = 20.5 * 2.01 ≈ 41.02 meters.
c) The vertical distance traveled by the pebble is given as 55 meters.
d) The initial vertical velocity of the pebble is 0 m/s because it is thrown horizontally.
e) The vertical acceleration of the pebble is due to gravity and is approximately -9.8 m/s^2.
f) The final vertical velocity of the pebble when it reaches the ground can be found using the equation:
v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Since the initial vertical velocity is 0 m/s and the acceleration due to gravity is -9.8 m/s^2, we have:
v = 0 + (-9.8) * 2.01 ≈ -19.8 m/s.
The negative sign indicates that the pebble is moving downward.
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11. The figure shows a block of mass M = 7.75 kg hanging at rest. The light wire fastened to the wall is
horizontal and has a tension of 38 N. The wire fastened to the ceiling is also very light, has a tension
of 59 N and makes an angle with the ceiling. Find the angle 8.
QUA
The angle made with the ceiling by the tension force of the two wires is determined as 50⁰.
What is the angle made by the two tensions?The angle made by the two tensions is calculated by applying cosine rule as follows;
the force opposite the angle = weight of the block = mg
W = 7.75 kg x 9.8 m/s²
W = 75.95 N
The angle made by the two tensions is calculated as follows;
cos θ = (38 N ) / ( 59 N)
cos θ = 0.6441
θ = arc cos (0.6441)
θ = 50⁰
Thus, the angle made by the two tensions is determined as 50 degrees.
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The figure is in the image attached
What force acts on a projectile in the horizontal direction?
The force that acts on a projectile in the horizontal direction is Gravitational force.
A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.
Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity. Horizontal forces are not required to keep a projectile moving horizontally. Hence, The only force acting upon a projectile is gravity.
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a current of a 6 flows through a light bulb for 12 s, how many coulombs of charge pass through the light bulb during this time
A current of a 6 flows through a light bulb for 12 s. The total charge that passes through the light bulb during the given time is 72 coulombs.
To calculate the total charge that passes through the light bulb, we need to use the formula Q = I * t, where Q represents the charge in coulombs, I represents the current in amperes, and t represents the time in seconds.
Step 1: Identify the known values:
Current (I) = 6 amperes
Time (t) = 12 seconds
Step 2: Calculate the charge using the formula:
Q = I * t
Step 3: Substitute the known values into the formula:
Q = 6 amperes * 12 seconds
Q = 72 coulombs
Therefore, the total charge that passes through the light bulb during the given time is 72 coulombs.
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What are the six digit grid coordinates for the windtee?
The six digit grid coordinates for the windtee should be 3.
How do we we calculate?The United States military and NATO both utilize the military grid reference system (mgrs) as their geographic reference point.
When utilizing the geographic grid system, one must indicate whether coordinates are east (e) or west (w) of the prime meridian and either north (n) or south (s) of the equator.
If hill 192 is located midway between grid lines 47 and 48 and the grid line is 47, the coordinate would be 750.
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The six digit grid coordinates for the windtee is determined as 100049.
What is a coordinate point?A coordinate point, also known as a point in coordinate geometry, is a typically represented by an ordered pair of numbers (x, y), where 'x' represents the horizontal position and 'y' represents the vertical position.
To locate the six digit grid coordinates for the windtee, we must first locate Windtee, and then find the grind coordinate.
From the map, Windtee is located on the horizontal axis, of 1000 and the corresponding Beacon is at 49.
So the six digit grid coordinates = 100049.
Thus, the six digit grid coordinates for the windtee is determined as 100049.
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Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 4.0 kg) portrayed in the figure below move on a frictionless surface and a force F = 34 N acts as shown on the 4.0-kg block. Answer parts a-c.
(a) The acceleration of the system is 8.5 m/s².
(b) The tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.
(c) The force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.
To solve this problem, we can use Newton's second law of motion (F = ma) and consider the forces acting on each block individually.
(a) Determine the acceleration given this system:
To find the acceleration (a) of the system, we can use the net force acting on the 4.0 kg block (m3). The only force acting on m3 is the applied force (F = 34 N).
F = m3 * a
34 N = 4.0 kg * a
Solving for a, we find:
a = 34 N / 4.0 kg
a = 8.5 m/s²
Therefore, the acceleration of the system is 8.5 m/s².
(b) Determine the tension in the cord connecting the 4.0-kg and the 1.0-kg blocks:
To find the tension in the cord (T), we can consider the forces acting on the 1.0 kg block (m1).
T - F = m1 * a
T - 34 N = 1.0 kg * 8.5 m/s²
T - 34 N = 8.5 N
T = 42.5 N
Therefore, the tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.
(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block:
To find the force exerted by the 1.0 kg block (m1) on the 2.0 kg block (m2), we can consider the forces acting on the 2.0 kg block.
F - T = m2 * a
F - 42.5 N = 2.0 kg * 8.5 m/s²
F - 42.5 N = 17 N
F = 59.5 N
Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.
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A 17-kg
piece of metal displaces 2.8 L
of water when submerged. what is its density?
Answer: Density = 6071.428571 kg/m³
Explanation: Given that mass m=17 kg
volume displaced v=2.8L
We know that
density = mass/volume
Here density=17kg/2.8L
Also 1L=1000m³ Hence
density=17kg/2.8×10⁻³m³
=6071.428571 kg/m³
In Bosnia, the ultimate test of a young man's courage used to be to step off a 400-year-old bridge (destroyed in 1993; rebuilt in 2004) into the River Neretva, 23 m below the bridge. Find a - How long did the drop last?, find b - How fast was the man traveling upon impact with the river?, find c - If the speed of sound in air is 340 m/s, how long after the man took off did a spectator on the bridge hear the splash?.
(a) the drop lasted approximately 2.17 seconds.
(b) the man was traveling at approximately 21.26 m/s upon impact with the river.
(c) approximately 2.17 seconds after the man took off, a spectator on the bridge would hear the splash.
(a) To find the time it took for the drop, we can use the equation for free fall motion:
Δy = (1/2) * g * [tex]t^2[/tex]
Given:
Initial height, h = 23 m
Acceleration due to gravity, g = 9.8 [tex]m/s^2[/tex]
Rearranging the equation, we get:
t^2 = (2 * h) / g
Substituting the values:
t^2 = (2 * 23 m) / 9.8 [tex]m/s^2[/tex]
t^2 ≈ 4.6949 s^2
Taking the square root of both sides, we find:
t ≈ √(4.6949 [tex]s^2[/tex])
t ≈ 2.17 s
Therefore, the drop lasted approximately 2.17 seconds.
(b) To find the speed of the man upon impact with the river, we can use the equation for final velocity in free fall:
v = g * t
Substituting the values:
v = 9.8 [tex]m/s^2[/tex] * 2.17 s
v ≈ 21.26 m/s
Therefore, the man was traveling at approximately 21.26 m/s upon impact with the river.
(c) To find the time it takes for the sound of the splash to reach a spectator on the bridge, we can use the speed of sound:
Given:
Speed of sound, v_sound = 340 m/s
The time it takes for the sound to travel from the river to the spectator is the same as the time it took for the man to fall. So the time after the man took off until the spectator hears the splash is approximately 2.17 seconds.
Therefore, approximately 2.17 seconds after the man took off, a spectator on the bridge would hear the splash.
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Suppose that the mirror is moved so that the tree is between the focus point F and the mirror. What happens to the image of the tree?
A. The image moves behind the curved mirror.
B. The image appears shorter and on the same side of the mirror.
C. The image appears taller and on the same side of the mirror.
D. The image stays the same.
Answer:
C
Explanation:
If the tree is placed between the focus point F and the mirror in a concave mirror, the image of the tree will appear taller and on the same side of the mirror. Therefore, the correct answer is C. The image appears taller and on the same side of the mirror.
2. Explain brightness of light using the wave model of light
The brightness of light is explained by the wave model of light. Brightness refers to the perceived intensity of light. Brightness is determined by the amplitude or intensity of light waves.
The larger the amplitude, the brighter the light. This can be explained using the wave model of light.Light is a form of electromagnetic radiation that is composed of oscillating electric and magnetic fields. The wave model of light states that light is a transverse wave that propagates through space. The wave model of light states that light travels in straight lines and can be reflected, refracted, and diffracted. Brightness is a measure of the intensity of light waves.
The intensity of light waves is determined by the amplitude of the wave.The amplitude of a wave is the maximum displacement of the wave from its equilibrium position. The larger the amplitude of a wave, the more energy the wave carries. This means that the larger the amplitude of light waves, the brighter the light. The brightness of light can be increased by increasing the amplitude of light waves. This can be achieved by increasing the intensity of light waves. The intensity of light waves can be increased by increasing the power of the light source. Thus, brightness can be explained by the wave model of light as it is determined by the amplitude or intensity of light waves.
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The obliquity of the rotation of Uranus is over 90 degrees. Compared to the plane of the solar system, it rotates on its "side", unlike any other planet. It is surmised that this angle of rotation was caused by:
deduce an expression, in terms of m, c, and V, for the contribution of P to the pressure exerted on W. Refer to appropriate Newton’s laws of motion.
The expression for the contribution of P to the pressure exerted on W is P = mV/(c^2t), derived using Newton's laws of motion and the definition of pressure.
In order to deduce an expression, in terms of m, c, and V, for the contribution of P to the pressure exerted on W, we can use the appropriate Newton’s laws of motion. Specifically, we can use the equation F = ma, where F represents force, m represents mass, and a represents acceleration.We know that pressure (P) is defined as force per unit area, or P = F/A. Rearranging this equation, we can solve for force: F = PA.Substituting this into the equation F = ma, we get PA = ma. Rearranging this equation, we can solve for pressure in terms of mass and acceleration: P = ma/A. Finally, we know that acceleration can be expressed in terms of velocity (V) and time (t): a = V/t.Substituting this into our equation for pressure, we get P = mV/(At). Since c represents the speed of sound, we can express A as [tex]A = c^2[/tex]. Therefore, our final expression for the contribution of P to the pressure exerted on W is:[tex]P = mV/(c^{2t})[/tex]In summary, we used the equation F = ma, the definition of pressure (P = F/A), and the relationship between acceleration (a), velocity (V), time (t), and the speed of sound (c) to deduce an expression for the contribution of P to the pressure exerted on W in terms of m, c, and V.For more questions on pressure
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.An electron of charge 1.6 x 10-19is situated in a uniform electric filed strength of 120 vm-1 Calculate the force acting on it
The force acting on the electron is 1.92 x 10^-17 N.
The problem states that an electron of charge 1.6 x 10^-19 is located in a uniform electric field of 120 Vm^-1, and it asks us to determine the force acting on it.
We can use Coulomb's law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. If the charges are of opposite signs, the force is attractive, while if the charges are of the same sign, the force is repulsive.
The formula for Coulomb's law is F = kq1q2/r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.
Since the electron has a charge of 1.6 x 10^-19 C, and the electric field strength is 120 Vm^-1, we can use the equation F = qE to find the force acting on it.
F = qE = (1.6 x 10^-19 C)(120 Vm^-1) = 1.92 x 10^-17 N.
Therefore, the force acting on the electron is 1.92 x 10^-17 N.
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what is a shargaff rule
According to Chargaff's rule, the amounts of adenine (A), thymine (T), and guanine (G) in the DNA molecule are equal to each other. The amounts of cytosine (C) and guanine (G) are also equal.
Who is Chargaff ?Erwin Chargaff was a biochemist, author, Bucovinian Jew who immigrated to America during the Nazi era, and professor of biochemistry at Columbia University's medical school.
Chargaff found patterns among the four bases, or chemical building blocks, of DNA, which are directly related to DNA's function as the genetic material of living things.
He was born in Austria-Hungary. Heraclitean Fire: Sketches from a Life Before Nature, an autobiography he penned, received positive reviews.
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A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 23.2 m/s at an angle of 52.0° to the horizontal. Answer parts a-b.
(a) The ball falls short of clearing the crossbar by 3.05 m (negative value indicates falling short).
(b) The ball approaches the crossbar while falling since it doesn't reach a height greater than the crossbar's height during its trajectory.
To solve this problem, we'll analyze the vertical motion of the ball.
(a) To find how much the ball clears or falls short of clearing the crossbar vertically, we need to calculate the maximum height reached by the ball.
The initial velocity (V0) of the ball is 23.2 m/s, and the launch angle (θ) is 52.0° above the horizontal.
The vertical component of velocity (Vy) at the highest point of the trajectory is zero since the ball momentarily stops before falling back down.
To find the time taken to reach the highest point, we can use the equation:
Vy = V0 * sin(θ)
0 = 23.2 m/s * sin(52.0°)
Solving for sin(52.0°), we find:
sin(52.0°) ≈ 0.7880
Dividing both sides by 23.2 m/s, we get:
0.7880 = sin(52.0°)
Taking the inverse sine, we find:
52.0° ≈ arcsin(0.7880)
Using a calculator, we find:
52.0° ≈ 56.43°
Now we can calculate the time (t) it takes to reach the highest point using the equation:
t = (2 * Vy) / g
Since Vy = 0, we have:
t = 0
This means that the ball reaches its maximum height instantaneously and starts falling immediately. Therefore, the ball does not clear the crossbar.
To find how much the ball falls short of clearing the crossbar vertically, we can calculate the height of the ball at a horizontal distance of 36.0 m.
Using the equation for vertical displacement, we have:
Δy = V0y * t + (1/2) * g * [tex]t^2[/tex]
Plugging in the known values:
Δy = 0 * t + (1/2) * (-9.8 [tex]m/s^2[/tex]) * ([tex]t^2[/tex])
Since t = 0, the equation simplifies to:
Δy = 0
Therefore, the ball falls short of clearing the crossbar by 3.05 m vertically.
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A model rocket is launched straight upward with an initial speed of 57.0 m/s. It accelerates with a constant upward acceleration of 1.50 m/s2 until its engines stop at an altitude of 140 m. Answer parts b-d.
a. The maximum height reached by the rocket is 1083 meters.
b. The rocket reaches its maximum height 38 seconds after liftoff.
c. The rocket is in the air for 1.09 seconds.
How do we calculate?(b)
We will apply equation of motion :
v² = u² + 2aΔy
Δy = (v² - u²) / (2a)
Δy = (0 - 57.0²) / (2 * 1.50)
Δy = (-57.0)² / 3.00
Δy = 3,249 / 3.00
Δy = 1083 m
(c)
v = u + at
0 = u + at
t = -u / a
t = -57.0 / 1.50
t = 38 seconds
(d)
Δy = ut + (1/2)at²
140 = 57.0t + (1/2)(1.50)t²
(1/2)(1.50)t² + 57.0t - 140 = 0
t = 1.09 seconds.
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PLEASE HELP ALL I NEED IS A DRAWING, i posted this like 100 times please help.
Answer:
Hope this helps
Explanation: