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3 years ago

Which statement best summarizes Newton's law of action-reaction? *

Physics

1 answer:

Tom [10]3 years ago

4 0

Answer:

The correct answer is the third

Force equals mass times acceleration

Explanation:

Newton's second law says that force equals mass times the acceleration of the body

F = m a

This is the most used way to solve problems

The correct answer is the third

Newton's third law states that forces always work in pairs, by action and reaction

Newton's first law states that an object per macerate with constant speed unless some force modifies this state.

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A rocket being thrust upward as the force of the fuel being burned pushes downward is an example of which of Newton's laws?

RoseWind [281]

The correct answer was

3rd Law

Thanks Mimi55555

4 0

4 years ago

Read 2 more answers

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

3 years ago

Read 2 more answers

a bullet is fired horizontally with a speed of 524 m/s from a height of 22m above the ground. calculate where it will hit the gr

lara31 [8.8K]

Given data:

* The height of the bullet is 22 m.

* The speed of bullet in the horizontal direction is 524 m/s.

Solution:

By the kinematics equation, the time taken by the bullet to reach the ground is,

$h=u_yt+\frac{1}{2}gt^2$

where u_y is the vertical velocity component, t is the time taken to reach the ground, g is the acceleration due to gravity, and h is the height of the bullet,

Substituting the known values,

$\begin{gathered} 22=0+\frac{1}{2}\times9.8\times t^2 \\ t^2=\frac{22\times2}{9.8} \\ t^2=4.49 \\ t=2.12\text{ s} \end{gathered}$

Thus, the time taken by the bullet to reach the ground is 2.12 seconds.

By the kinematics equation for the horizontal motion, the horizontal range of the bullet is,

$R=u_xt+\frac{1}{2}at^2$

where u_x is the horizontal component of the velocity, a is the acceleration along the horizontal direction, t is the time taken to reach the ground and R is the horizontal range,

Substituting the known values,

$\begin{gathered} R=524\times2.12+0 \\ R=1110.9\text{ m} \end{gathered}$

Thus, the horizontal range of the bullet is 1110.9 meters.

Hence, the bullet hit the ground at 1110.9 meters.

8 0

1 year ago

Saki is ice-staking to the left with velocity of 5 m/s. A steady breeze starts blowing. Causing saki to accelerate leftward at a

Pavlova-9 [17]

Answer:

Final velocity will be equal to 14 m/sec

Explanation:

We have given initial velocity u = 5 m/sec

Constant acceleration is given $a=1.5m/sec^2$

Time t = 6 sec

We have to find the final velocity

From first equation of motion $v=u+at$ , here v is final velocity, u is initial velocity , a is acceleration and t is time

So $v=5+1.5\times 6=14m/sec$

So equal final velocity will be equal to 14 m/sec

4 0

3 years ago

A truck skids for a distance of 25m with the road pushing on its tires with force of 1500N as its brakes are applied. What is th

Daniel [21]

Answer:

37500 J

Explanation:

Law of conservation of energy: Its states that energy can neither be created nor destroyed but it can be transformed from one form to another.

Applying the law of conservation of energy.

The change in kinetic energy for the truck = Net work done by the truck skids

ΔK.E = F×d.................... Equation 1

Where ΔK.E = Change in kinetic energy, F = Force, d = distant.

Given: F = 1500 N, d = 25 m

Substitute these values into equation 1

ΔK.E = 1500×25

ΔK.E = 37500 J

3 0

3 years ago