Answer:
y = - (½ g / v₀²) x²
Explanation:
This is a projectile launch exercise where there is no acceleration on the x-axis so
x = v₀ₓ t
v₀ₓ = v₀ cos tea
y = [tex]v_{oy}[/tex] t - ½ g t2
v_{oy} = v₀ sin θ
as the sphere is thrown horizontally, the angle is tea = 0º, so the initial velocity remains
v₀ₓ = v₀
v_{oy} = 0
we substitute in our equations
x = v₀ t
y = - ½ g t²
we eliminate the time from these equations, we substitute the first in the second
y = - ½ g (x / v₀)²
y = - (½ g / v₀²) x²
this is the equation of a parabola
A car moving with a constant acceleration covers the distance between two points 80 m apart in 8.0 s. Its velocity as it passes the second point is 15 m/s. What was the velocity at the first point? Show work for brainliest!!
Answer:
Explanation:
3.6
Your classmate’s mass is 63 kg and the table weighs 500 N. Calculate the normal force on the table by the floor. Show your work!
Answer:
[tex]F_N=1234.8N[/tex]
Explanation:
Hello.
In this case, since the normal force is opposite to the total present weight, we can compute it by considering the mass of the classmate with the gravity to compute its weight, and the weight of the table:
[tex]F_N=63kg*9.8m/s^2+500N\\\\F_N=617.4N+500N\\\\F_N=1234.8N[/tex]
Best regards.
25
60
20
1. How many servings can be made with 25 cups of pasta, 30 cups of marinara, and 5
cups of parmesan?
Answer:
60
Explanation:
i think it 60 because u plus 25 +30=55+5=60
Q.9) Consider a plot of the displacement (x) as a function of the applied force (F) for an ideal elastic spring as
shown, see figure. The slope of the curve would be
A) the acceleration due to gravity.
B) the reciprocal of the acceleration of gravity.
C) the mass of the object attached to the spring.
D) the reciprocal of the spring's constant.
Answer:
Option (D).
Explanation:
For an ideal spring, the force is given in terms of displacement as follows :
F=-kx
[tex]x=\dfrac{-F}{k}\\\\\text{or}\\\\x=\dfrac{-1}{k}F[/tex]...(1)
Where,
x is displacement in spring i.e. compression or stretching
k is spring constant of the spring
-ve sign shows that the force exerted by the spring is in opposite direction of the spring's displacement.
The equation of a line is : y=mx+c, m is slope
From equation (1)
Slope = 1/k
or
Slope = reciprocal of the spring's constant.
It would mean that the slope of the curve represents the reciprocal of the spring's constant.
Using the spring constant and displacement relation, the slope of the curve would be the reciprocal of the spring's constant.
Using the Force constant relation :
F = keF = Force ; k = spring constant ; e = extension
F = kxMaking the x the subject of the formula :
x = F/kThe slope, which is the spring constant can be expressed as ;
x = 1/k × FHence, the slope is the reciprocal of the spring constant.
Learn more :https://brainly.com/question/18451126
2. How far can a dog travel if he can travel at 2.5 m/s for 5 seconds?
Answer:
[tex]\boxed {\tt 12.5 \ meters}[/tex]
Explanation:
We want to find how far the dog can travel, or the distance.
Distance can be found by multiplying the speed and the time.
[tex]d=s*t[/tex]
The dog travels 2.5 meters per second and travels for 5 seconds. Therefore,
[tex]s=2.5 \ m/s\\t= 5 \ s[/tex]
Substitute the values into the formula.
[tex]d=2.5 \ m/s \ * 5 \ s[/tex]
Multiply. Note that the seconds, or "s" will cancel each other out when multiplying.
[tex]d= 2.5 \ m * 5[/tex]
[tex]d=12.5 \ m[/tex]
The dog can travel 12.5 meters at 2.5 m/s for 5 seconds.
What does it mean if something is Periodic?
Answer:
In the context of chemistry and the periodic table, periodicity refers to trends or recurring variations in element properties with increasing atomic number. Periodicity is caused by regular and predictable variations in element atomic structure
Explanation:
A truck collides with a sports car in a high-speed head-on collision. . The mass of the truck and its contents is 5 times larger than the mass of the car and its contents. Select greater than, less than or equal to. During the collision, the magnitude of the change in momentum of the truck is _____ the magnitude of the change in momentum of the car. A: Greater than B: Less than C: Equal to
Answer:
A: Greater than
Explanation:
Momentum (P) implies product of the mass (m) and velocity (v) of a body.
P = mv
It is measured in kgm/s.
Thus the greater the mass of a body, the more its momentum even when moving at a slower velocity. The change in momentum of an object is called its impulse.
Let the mass of the car be represented by [tex]m_{c}[/tex] and that of the truck be represented by [tex]m_{T}[/tex].
Thus,
[tex]m_{T}[/tex] = 5[tex]m_{c}[/tex]
P of the truck = 5[tex]m_{c}[/tex]v
P of the car = [tex]m_{c}[/tex]v
Thus, during collision, the change in momentum of the truck is greater than that of the car.
Answer:
The correct option is B
Explanation:
This question seeks to test the knowledge of the change in momentum and deep understanding of the Newton's first law of motion which states that a substance will continue to be in a state of rest or constant motion unless acted upon by an external force.
The formula for change in momentum is mass multiplied by change in velocity. where change in velocity is final velocity minus the initial velocity.
The mass of the truck is already said to be 5 times larger, the change in velocity of the car will definitely be negative because the car will pushed back by the truck. For instance, if the initial velocity of the car is 50m/s and the final velocity backwards is 10m/s, the change in velocity will be (-10-50) which will equal (-60)
However, for the truck, the change in velocity will still remain positive for the final velocity because it is expected that the truck will still move forward despite the heads-on collision. Hence, if the initial velocity is 50m/s and the final velocity is 10m/s, the change in velocity will be (10-50) which will equal (-40).
Assuming the mass of the car is 1g. From the formula for change in momentum
The truck will be assumed to be (5 × 1) × -40 = -200
The car will be assumed to be 1 × -60 = -60
Hence, the magnitude of change in momentum of the truck will be lesser than the magnitude of the change in momentum of the car.
Which are consequences of smoking while pregnant
Answer:
Explanation:
Tobacco. Smoking during pregnancy increases the risk of health problems for developing babies, including preterm birth, low birth weight, and birth defects of the mouth and lip. Smoking during and after pregnancy also increases the risk of sudden infant death syndrome (SIDS).
Answer:
B I - T H deficits, S l D S, S T I L L . B I - T H
Explanation:
2021 edg .
A soccer player heads the ball and sends it flying vertically upwards at a speed of 18.0 m/s . How high above the players ' head does the ball travel ?
Answer:
16.53 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 18.0 m/s.
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =.?
The maximum height reached by the ball can be obtained as follow:
v² = u² – 2gh (since the ball is going against gravity)
0² = 18² – (2 × 9.8 × h)
0 = 324 – 19.6h
Rearrange
19.6h = 324
Divide both side by 19.6
h = 324 / 19.6
h = 16.53 m
Therefore, the maximum height reached by the ball is 16.53 m
A length of 20-gauge copper wire (of diameter 0.8118 mm) is formed into a circular loop with a radius of 25.0 cm. A magnetic field perpendicular to the plane of the loop increases from zero to 10.0 mT in 0.34 s. Find the average electrical power dissipated in the process
Answer:
The average electrical power dissipated in the process is 0.653 mW
Explanation:
Given;
gauge of copper wire, 20 gauge
resistivity from chart, [tex]\rho = 1.68 *10^{-8} \ ohm.m[/tex]
radius of the circular loop, r = 25 cm
magnetic field strength, B = 10 .0 mT
time, t = 0.34 s
Length of the wire, L = 2πr = 2 x π x 0.25 = 1.571 m
Area of the wire, A = πR² ⇒ R = D/2 = 0.8118 mm/ 2 = 0.4059 mm
= π(0.4059 x 10⁻³)² = 0.5177 x 10⁻⁶ m²
The resistance of the wire is given by;
[tex]R = \frac{\rho L}{A}\\\\ R = \frac{1.68*10^{-8}*1.571}{(0.5177*10^{-6})}\\\\ R = 5.098 *10^{-2} \ ohms[/tex]
Now, determine the electric potential;
[tex]E = N\frac{d\phi}{dt}\\\\ E = N(\frac{BA}{dt})\\\\ E = 1(\frac{(10*10^{-3})(\pi*0.25^2)}{0.34} )\\\\E = 0.00577 \ V[/tex]
The average power is given by;
P = V²/R
P = (0.00577²) / (5.098 x 10⁻²)
P = 6.53 x 10⁻⁴ W
P = 0.653 mW
Therefore, the average electrical power dissipated in the process is 0.653 mW
a 100 foot cliff drops vertically to a lake if the angle of elevation from a swimmer to the top of the cliff is 54.6 how far is the swimmer from the clif
Answer:
Explanation:
Please see attached photo for explanation.
In the attached photo, X is the distance of the swimmer from the cliff.
Thus, we can obtain the value of x by using Tan ratio as illustrated below:
θ = 54.6°
Opposite = 100 foot
Adjacent = x =.?
Tan θ = Opposite /Adjacent
Tan 54.6 = 100/x
Cross multiply
x × Tan 54.6 = 100
Divide both side by Tan 54.6
x = 100 / Tan 54.6
x = 71.07 foot
Therefore, the swimmer is 71.07 foot from the cliff.
A projectile is fired with a horizontal velocity of 20 mfs and a vertical velocity of 45
6. What is the magnitude of its original velocity vector?
7. How long will the projectile be in the air?
8. What is its range?
Answer:
6. 49.2 m/s
7. 9.18 s
8. 184 m
Explanation:
6. Use Pythagorean theorem.
v₀² = v₀ₓ² + v₀ᵧ²
v₀² = (20 m/s)² + (45 m/s)²
v₀ = 49.2 m/s
7. Given:
Δy = 0 m
v₀ᵧ = 45 m/s
aᵧ = -9.8 m/s²
Find: t
Δy = v₀ᵧ t + ½ aᵧt²
(0 m) = (45 m/s) t + ½ (-9.8 m/s²) t²
0 = 45t − 4.9t²
0 = t (45 − 4.9t)
t = 9.18 s
8. Given:
v₀ₓ = 20 m/s
aₓ = 0 m/s²
t = 9.18 s
Find: Δx
Δx = v₀ₓ t + ½ aₓt²
Δx = (20 m/s) (9.18 s) + ½ (0 m/s²) (9.18 s)²
Δx = 184 m
What do seismic waves and sound waves have in common? [Seismic and sound waves]
A They're mechanical waves incorrect answer
B They're electromagnetic waves incorrect answer
C They're phonetic waves incorrect answer
D They're permanent waves
Answer:
D
Explanation:
They continously go back and fourth on the ray of movement making the tremendous exponent of the scientificalsource behind it
d is the answer
lkjhgfdeswwwwwwwwwww3e
A rocket is launched from a height of 3 m with an initial velocity of 15 m/s What is the maximum height of the rocket? When will this occur?
If no extra acceleration is added to the rocket, then its velocity at time t is
v = 15 m/s - g t
where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.
Also, recall that
v² - u² = 2 a ∆x
where u is initial speed, v is final speed, a is acceleration, and ∆x is net displacement.
At the rocket's maximum height ∆x, the velocity is 0. So, the maximum height is
0² - (15 m/s)² = 2 (-g) ∆x
∆x = (15 m/s)² / (2 * (9.80 m/s²)) ≈ 11.48 m
But this assumes the rocket is launched from the ground. We're given that the rocket is launced from 3 m above the ground, so we need to add this to the height above. So the maximum height is closer to 14.48 m.
As mentioned before, this happens when vertical velocity is 0:
0 = 15 m/s - g t
t = (15 m/s) / (9.80 m/s²) ≈ 1.53 s
An object of mass 10kg is accelerated upward at 2 m/s2. What force is required
Answer:
The answer is 20 NExplanation:
The force acting on an object given the mass and acceleration we use the formula
force = mass × accelerationFrom the question
mass = 10kg
acceleration = 2 m/s²
We have
force = 10 × 2
We have the final answer as
20 NHope this helps you
A 4500 kg car accelerates from rest to 45.0
km/h over a distance of 90 m. Find the net force
acting on the car.
please help
The car undergoes an acceleration a such that
(45.0 km/h)² - 0² = 2 a (90 m)
90 m = 0.09 km, so
(45.0 km/h)² - 0² = 2 a (0.09 km)
Solve for a :
a = (45.0 km/h)² / (2 (0.09 km)) = 11,250 km/h²
Ignoring friction, the net force acting on the car points in the direction of its movement (it's also pulled down by gravity, but the ground pushes back up). Newton's second law then says that the net force F is equal to the mass m times the acceleration a, so that
F = (4500 kg) (11,250 km/h²)
Recall that Newtons (N) are measured as
1 N = 1 kg • m/s²
so we should convert everything accordingly:
11,250 km/h² = (11,250 km/h²) (1000 m/km) (1/3600 h/s)² ≈ 0.868 m/s²
Then the force is
F = (4500 kg) (0.868 m/s²) = 3906.25 N ≈ 3900 N
what is not a property that can be used to identify a mineral?
Answer:
Color.
Explanation:
Hello.
In this case, among the properties minerals have, we list: color , streak which is the color as a powder , hardness , cleavage or fracture , crystalline structure , diaphaneity or degree of transparency, tenacity which is the capacity of its molecules to be held together, magnetism which is the capacity to attract or repel other magnetic materials , luster which explains if its surface reflects light , odor , taste and specific gravity which is a relationship between its density and the density of water.
In such a way, since the most of them are known as measurements, we can distinguish among minerals by contrasting those data, nevertheless, color is not a suitable property to identify a material since there could exist minerals with similar colors.
Best regards.
who is the first person to reach in space
Answer:
cosmonaut Yuri Gagarin
Explanation:
On that day in 1961, Russian cosmonaut Yuri Gagarin (left, on the way to the launch pad) became the first human in space, making a 108-minute orbital flight in his Vostok 1 spacecraft.
i hoped this helped plz list me as a brainliest
A straight line is drawn on the surface of a 18-cm-radius turntable from the center to the perimeter. A bug crawls along this line from the center outward as the turntable spins counterclockwise at a constant 45 rpm. Its walking speed relative to the turntable is a steady 3.4 cm/s. Let its initial heading be in the positive x-direction, where the x- and y-directions are fixed relative to the laboratory. As the bug reaches the edge of the turntable (still traveling at 3.4 cm/s radially, relative to the turntable) what are the x and y components of the velocity of the bug?
Answer:
v_r = 3.4 cm / s , v_t = 84.82 cm / s
Explanation:
The speed of the insect has two components, one radial and the other tangential,
The radial component is the speed with which the insect moves with respect to the plate
v_r = 3.4 cm / s
The tangential component is given by
v = w r
Let's reduce the angular velocity to the SI system
w = 45 rpm (2π rad / 1 rev) (1 min / 60 s) = 4.712 rav / s
let's calculate
v = 4.712 18
v_t = 84.82 cm / s
A space vehicle is coasting at a constant velocity of 16.9 m/s in the y- direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.395 m/s^2 in the x direction. After 38.8 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off.
Find:
a. The magnitude.
b. The direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the y- direction
Answer:
See attached file for the answer
What are two differences between the dwarf planets and our traditional 8 planets?
Enter Answer Here
The water from a fire hose is directed at a building . The water leaves the pipe at a speed of 25 m/s at an angle of 60 deg to the horizontal. If the building is 45 m away, at what height does it hit the building ?
Answer:
Approximately [tex]14\; \rm m[/tex] above the height of the fire hose, assuming that air resistance is negligible and that [tex]g = -9.81\; \rm m \cdot s^{-2}[/tex].
Explanation:
Consider the motion of water particles in two directions: vertical and horizontal.
Assuming that air resistance on the water particles is negligible.
Vertically, water particles would accelerate at [tex](-9.81\; \rm m \cdot s^{-2})[/tex] (towards the ground.)Horizontally, water particles would travel at a constant velocity.Initial velocity: [tex]v_0 = 25\; \rm m \cdot s^{-1}[/tex]. Angle of elevation: [tex]\epsilon = 60^\circ[/tex]. Calculate the initial velocity of these water particles in these two components:
Initial horizontal velocity: [tex]v_0 (\text{horizontal}) = v_0 \cdot \cos\left(\epsilon\right) \approx 12.5\; \rm m \cdot s^{-1}[/tex].Initial vertical velocity: [tex]v_0 (\text{vertical}) = v_0 \cdot \sin\left(\epsilon\right) \approx 21.7\; \rm m \cdot s^{-1}[/tex].How much time would it take for a water particle reaches the building from the hose? That particle needs to travel [tex]x(\text{horizontal}) = 45\; \rm m[/tex] at a constant horizontal speed of [tex]v(\text{horizontal}) \approx 12.5\; \rm m \cdot s^{-1}[/tex]. Therefore:
[tex]\displaystyle t = \frac{x(\text{horizontal})}{v(\text{horizontal})} \approx 3.6\; \rm s[/tex].
In other words, that water particle would be approximately [tex]3.6\; \rm s[/tex] into its flight when it hits the building. What would be its height? Assuming that the air resistance on that particle is negligible. The height of that water particle at time [tex]t[/tex] may be modeled using the SUVAT equation:
[tex]\displaystyle x(\text{vertical}) = \frac{1}{2}\, a \, t^{2} + \left[v_0(\text{vertical})\right]\, t[/tex],
where:
[tex]x(\text{vertical})[/tex] gives the height of the water particle (relative to where it was launched.)[tex]a = g = -9.81\; \rm m \cdot s^{-2}[/tex] (negative because gravitational acceleration points towards the ground.)[tex]v_0(\text{vertical}) \approx 21.7\; \rm m \cdot s^{-1}[/tex] (see above.)[tex]t \approx 3.6\; \rm s[/tex] (that's the time when the water particle hits the building.)Calculate the height of the water particle when it hits the building:
[tex]\begin{aligned}x(\text{vertical}) &= \frac{1}{2}\, a \, t^{2} + \left[v_0(\text{vertical})\right]\, t \\ &\approx \frac{1}{2} \times \left(-9.81\; \rm m \cdot s^{-2}\right) \times (3.6\; \rm s)^2 + 21.7\; \rm m \cdot s^{-1} \times 3.6\; \rm s \\ &\approx 14\; \rm m\end{aligned}[/tex].
A net force of 40 N south acts on an object with a mass of 20 KG. What is the objects acceleration
Answer:
The answer is 2 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
f is the force
m is the mass
We have
[tex]a = \frac{40}{20} = \frac{4}{2} \\ [/tex]
We have the final answer as
2 m/s²Hope this helps you
If you throw an object straight up into the air with an initial velocity of 42m/s. What is it’s velocity at the peak of its flight?
A. 42m/s
B. -42m/s
C. 9.8m/s
D. -9.8m/s
E. 0m/s
F. 0m/s2
G. None of these
How to drawing the resultant vector of two vectors being added:
Answer:
Explanation:
you connect vector b with vector a making it head to tail
I just had to shove an eos in my mouth because I keep shouting random things.....
Answer:
oooop tehehehehehee
ahahhhahaaahahjabasa
Answer:
uhm- heh, wut lol ️️
Starting from rest, a car travels 18 meters as it accelerates uniformly for 3.0 seconds. What is the magnitude of
the car's acceleration?
Answer:
[tex]a=4\frac{m}{s^2}[/tex]
Explanation:
Hello.
In this case, for this uniformly accelerated motion in which the car starts from rest at 0 m/s and travels 18 m in 3.0 s, we can compute the acceleration by using the following equation:
[tex]x_f=x_0+v_0t+\frac{1}{2}at^2[/tex]
Whereas the final distance is 18 m, the initial distance is 0 m, the initial velocity is 0 m/s and the time is 3.0 s, that is why the acceleration turns out:
[tex]a=\frac{2(x_f-v_ot)}{t^2} =\frac{2(18m-0m/s*3.0s)}{(3.0s)^2}\\ \\a=4\frac{m}{s^2}[/tex]
Best regards.
Given the distance travelled and the time taken, the magnitude of the car's acceleration is 4m/s²
Given the data in the question;
Since the car starts from rest,
Initial velocity; [tex]u = 0m/s[/tex]Distance travelled; [tex]s = 18m[/tex]time taken; [tex]t = 3.0s[/tex]Acceleration; [tex]a = \ ?[/tex]
To determine the magnitude of the car's acceleration
We use the Second Equation of Motion:
[tex]s = ut + \frac{1}{2}at^2[/tex]
Where s is the speed, u is the initial velocity, a is the acceleration and t is the time.
We substitute our values into the equation and solve for "a"
[tex]18m = (0m/s\ * 3.0s) + (\frac{1}{2}\ *\ a\ *\ (3.0s)^2) \\\\18m = 4.5s^2 \ *\ a\\\\a = \frac{18m}{4.5s^2} \\\\a = 4 m/s^2[/tex]
Therefore, the magnitude of the car's acceleration is 4m/s²
Learn more: https://brainly.com/question/10428597
A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s.
What is the final velocity after 8.0
Two planes are about to drop an empty fuel tank. At the moment of release each plane has the same speed of 195 m/s, and each tank is at the same height of around 1.10 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15° above the horizontal (Plane A) and the other is flying at an angle of 15° below the horizontal (Plane B).
1. Find the:
a. magnitude.
b. direction of the velocity with which the fuel tank hits the ground if it is from plane A.
2. Find the:
a. magnitude
b. direction of the velocity with which the fuel tank hits the ground if it is from plane B.
In each part, give the direction as a positive angle with respect to the horizontal.
Answer:
376.71
Explanation:
1.a
To find the angle between them we add 15 +15=30°
Since they aren't perpendicular we use cosine law
[tex] \alpha = 180 - 30[/tex]
[tex] \sqrt{ {a}^{2} + {b}^{2} - 2ab \cos( \alpha ) } [/tex]
[tex] \sqrt{ {195}^{2} + {195}^{2} - 2 \times 195 \times 195 \times \cos(150) } [/tex]
=376.711
A 6.16-g bullet is moving horizontally with a velocity of 342 m/s, where the sign indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1155 g, and its velocity is 0.728 m/s after the bullet passes through it. The mass of the second block is 1517 g.
(a) What is the velocity of the second block after the bullet imbeds itself?
(b) Find the ratio of the total kinetic energy after the collisions to that before the collisions.
Answer:
a
The value is [tex]v = 0.8351 \ m/s [/tex]
b
The ratio is [tex]\frac{K_a}{K_b} = 5.94 11 *10^{-6}[/tex]
Explanation:
From the question we are told that
The mass of the bullet is [tex]m_b = 6.16 \ g = 0.00616 \ kg[/tex]
The velocity is [tex]u_b = 342 \ m/s[/tex]
mass of the first block is [tex]m__{{B_1}}} = 1155 \ g = 1.155 \ kg[/tex]
The velocity of the first block after bullet passes is [tex]v__{{B_1}}} = 0.728 m/s[/tex]
The mass of the second block is [tex]m__{{ B_2}}} = 1517 g = 1.517 \ kg[/tex]
Gnerally according to the law of momentum conservation
[tex]m_b * u_b + m__{{B_1}}} * u__{{B_1}}} = m__{{B_1}}} * v__{{B_1}}} + v_b * m_b[/tex]
Here [tex]v_b[/tex] is the velocity of the bullet emerging from the first block
and [tex]u__{{B_1}}}[/tex] is zero because initial the first block was at rest
So
[tex]0.00616 * 342 + 1.155* 0 = v_b * 0.00616 + 1.155 * 0.728 [/tex]
[tex]v_b = 206.5 \ m/s [/tex]
Considering the second block
Gnerally according to the law of momentum conservation
[tex]m_b * v_b + m__{{B_2}}} * u__{{B_2}}} = [m__{{B_2}}} +m_b] v[/tex]
Here [tex]u__{{B_2}}}[/tex] is zero because initial the second block was at rest
=> [tex]0.00616 * 206.5 + 1.517* 0= [1.517 +m_b] v [/tex]
=> [tex]0.00616 * 206.5 = [1.517 +0.00616] v [/tex]
=> [tex]v = 0.8351 \ m/s [/tex]
The kinetic energy of the bullet before collision is
[tex]K_b = \frac{1}{2} * m * u_b^2[/tex]
=> [tex]K_b = 0.5 * 0.00616 * 342^2[/tex]
=> [tex]K_b = 360.2 \ J [/tex]
The kinetic energy of the bullet after collision is
[tex]K_a = \frac{1}{2} * m * v^2[/tex]
=> [tex]K_a = 0.5 * 0.00616 * 0.8351^2[/tex]
=> [tex]K_a = 0.00214 \ J [/tex]
Generally the ratio of the kinetic energy is mathematically represented as
[tex]\frac{K_a}{K_b} = \frac{0.00214}{360.2 }[/tex]
=> [tex]\frac{K_a}{K_b} = \frac{0.00214}{360.2 }[/tex]
=> [tex]\frac{K_a}{K_b} = 5.94 11 *10^{-6}[/tex]