The acceleration of the object is 11.11 m/[tex]s^2[/tex] (approximately).
What is acceleration?The rate of change of velocity of a body with respect to time is known as acceleration. Due to having both magnitude and direction, it is a vector quantity having SI unit m/s^2.
Given parameter,
The mass of the object, m = 45 Kg.
Net force acting on the object, F = 500 N.
Acceleration of the object, a= ?
From Newton's 2nd law of motion,
Force = mass × acceleration
⇒ F = ma
⇒ a = F/m
= 500 N /45 Kg
= 11.11 m/s^2. (approximate)
Hence, the approximate value of acceleration of the object, a = 11.11 m/s^2.
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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.
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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.
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
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 ️️
How does the eccentricity of a typical cometary orbit compare to that of a typical planet
Answer:
Explanation:
Comets are balls of ice and dust in orbit around the Sun. The orbits of comets are different from those of planets - they are elliptical. A comet's orbit takes it very close to the Sun and then far away again.
What is the speed of an object that travels 12 meters in 3
seconds?
Answer:
4 meters per second
Explanation:
12 divided by 3 = 4
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.
A man driving a car traveling at 20 m/sec slams on the brakes and decelerates at 3.25
m/s^2. How far does the car travel before it stops?
Recall that
[tex]{v_f}^2-{v_i}^2=2a\Delta x[/tex]
where [tex]v_i[/tex] and [tex]v_f[/tex] are initial and final velocities, respectively; [tex]a[/tex] is acceleration; and [tex]\Delta x[/tex] is the net displacement, or distance if the object is moving in a single direction.
The car has initial speed 20 m/s and acceleration -3.25 m/s². It comes to a stop, so it has 0 final speed. Then
0² - (20 m/s)² = 2 (-3.25 m/s²) ∆x
∆x = (20 m/s)² / (7.5 m/s²) ≈ 53.3 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
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What is the meaning of agility
Answer:
Agility or nimbleness is an ability to change the body's position efficiently, and requires the integration of isolated movement skills using a combination of balance, coordination, speed, reflexes, strength and endurance.
Explanation:
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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 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
what is the right way to sleep
Answer:
by lying down on a nice and soft quilted matress
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
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
Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of the moon’s surface, making what some consider the longest golf drive in history. Assume one of the golf balls was struck with initial velocity v0 = 31.75 m/s at an angle θ = 26° above the horizontal. The gravitational acceleration on the moon’s surface is approximately 1/6 that on the earth’s surface. Use a Cartesian coordinate system with the origin at the ball's initial position.
Randomized Variables
vo 32.75 m/s
theta 32 degrees
What horizontal distance, R in meters, did this golf ball travel before returning to the lunar surface?
Complete Question
Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of the moon’s surface, making what some consider the longest golf drive in history. Assume one of the golf balls was struck with initial velocity v0 = 32.75 m/s at an angle θ = 32° above the horizontal. The gravitational acceleration on the moon’s surface is approximately 1/6 that on the earth’s surface. Use a Cartesian coordinate system with the origin at the ball's initial position.
Randomized Variables
vo 32.75 m/s
theta 32 degrees
What horizontal distance, R in meters, did this golf ball travel before returning to the lunar surface?
Answer:
The horizontal distance is [tex]R = 590.2 \ m [/tex]
Explanation:
From the question we are told that
The initial velocity is [tex]v_o = 32.75 \ m/s[/tex]
The angle is [tex]\theta = 26^o[/tex]
The gravitational acceleration of the moon is [tex]g_m = \frac{1}{6} * 9.8 = 1.633 m/s^2[/tex]
Generally the distance traveled is mathematically represented as
[tex]R = \frac{v_o^2 sin 2(\theta)}{g_m}[/tex]
=> [tex]R = \frac{32.75^2 sin 2(32)}{1.633}[/tex]
=> [tex]R = 590.2 \ m [/tex]
Which statements are true for two oppositely charged, isolated parallel plates: C=capacitance, U=stored energy (Q and -Q = charge on the plates). Note: Isolated plates can not lose their charge.
1. Inserting a dielectric decreases U.
2. When the distance is doubled, C increases.
3. Increasing the distance increases the Electric field.
4. Inserting a dielectric increases Q.
5. Inserting a dielectric increases C.
6. When the distance is halved, Q stays the same.
7. When the distance is doubled, U increases.
Answer:
Explanation:
1) True. The stored energy (U) is proportional to the electric field strength (E). The electric field strength decreases when a dielectric is introduced hence inserting a dielectric decreases U.
2) False. From the formula [tex]C=\frac{Q}{V}=\frac{Q}{Vd}[/tex], capacitance is inversely proportional to distance hence if the distance is doubled, capacitance decreases.
3) False. As the distance between the electric field and the object increases, its electric field decreases.
4) False. If a dielectric is inserted, the plates are further separated. Q stays the same.
5) True. The electric field strength decreases when a dielectric is introduced and capacitance is inversely proportional to electric field hence Inserting a dielectric increases C
6) True. If a dielectric is inserted, the plates are further separated. Q stays the same.
7) True. When the distance is doubled, U increases
What would the kinetic energy and potential energy be with the snowboarder, before during and after a jump?
Answer: Before the jump, the snowboarder would carry potential energy.
During the jump he will carry kinetic energy.
And after the jump, assuming hes at a full stop, he will carry potential energy once again.
Answer:
Before the jump, the snowboarder would carry potential energy.During the jump he will carry kinetic energy.And after the jump, assuming hes at a full stop, he will carry potential energy once again. 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.
How do warm ocean currents affect the weather?
A. They make the equator warmer.
B. They make the tropical regions warmer.
C. They reduce precipitation in coastal areas.
D. They result in warmer air temperatures near the current.
If you get it right you get 100 points and a brainliest
Answer:
the answer is D I believe
Answer:
Answer: D
They result in warmer air temperatures near the current.
Explanation:
i took a test with the same question and got it right
Iu Metallic bonding is similar to iconic bonding because
Answer:
In an ionic bond the valence electrons are transferred from the metal
Explanation:
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
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
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 are two differences between the dwarf planets and our traditional 8 planets?
Enter Answer Here
A car is pushed and travels 12 m. The same car is then pushed and it travels 15m. What could of
happened? Select 2 answers.
The force was increased
The mass was decreased
The force was decreased
The mass was increased
Explanation:
The force was increased
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.
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11)A 1100 kg car travels on a straight highway with a speed of 30 m/s. The driver sees a red light ahead and applies her
brakes, which exert a constant braking force of 4000N. In how many seconds will the car stop?
Answer:
Time taken = 8.25 second
Explanation:
Given:
Force = 4000 N
Force = ma
4,000 = (1100)(a)
Acceleration = 3.6363 m/s²
v = u + at
0 = 30 + (3.6363)t
Time taken = 8.25 second
The time taken by the car to stop is 8.26 s.
Given data:
The mass of car is, m = 1100 kg.
The speed of car is, u = 30 m/s.
The magnitude of braking force is, F = 4000 N.
We need to first obtain the acceleration of car to get that, apply the Newton's second law as,
F = - ma (Negative sign shows that the force will resist the motion)
4000 = -(1100) a
[tex]a =-\dfrac{4000}{1100}\\\\a =-3.63 \;\rm m/s^{2}[/tex]
Now, apply the first kinematic equation of motion to obtain the time taken by the car to stop as,
v = u + at
Here, v is the final speed and v = 0, since car will stop finally.
So,
[tex]0=30+(-3.63)t\\\\t = \dfrac{30}{3.63}\\\\t=8.26 \;\rm s[/tex]
Thus , we can conclude that the time taken by the car to stop is 8.26 s.
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astronaut A can cover 10 meters per minute walking with the heavy shovel. What does
this sentence describe?
a) None of these things
b) Both speed and velocity
C) The velocity of the astronaut
d) The speed of the astronaut
Answer:
d) The speed of the astronaut
Explanation:
The sentence describes the speed of the astronaut. This speed value is 10meters per minute.
Now let us understand why;
Speed is the distance divided by time. It is a scalar quantity without regard for direction but it has magnitude. The value 10meters per minute clearly shows this instance. We do not know the direction the astronaut is moving towards. Velocity, like speed is the displacement of a body with time. It is a vector quantity and it shows the direction of motion. For example, 10m/s due west is a velocity value because we know the direction.Therefore, since there is no directional sense, the value indicates speed.
The given sentence describes the speed of the astronaut. Hence, option (d) is correct.
Given data:
The distance covered by the astronaut A is, d = 10 m.
the time interval is, t = 1 min = 60 s.
Clearly, the given sentence describes the speed of the astronaut. This speed value is 10 meters per minute. This is because of following reason:
Speed is the distance divided by time. It is a scalar quantity without regard for direction but it has magnitude. The value 10meters per minute clearly shows this instance. We do not know the direction the astronaut is moving towards. Velocity, like speed is the displacement of a body with time. It is a vector quantity and it shows the direction of motion. For example, 10m/s due west is a velocity value because we know the direction.Thus, we can conclude that the given sentence describes the speed of the astronaut. Hence, option (d) is correct.
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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
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