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Ivan
2 years ago
8

A car with an initial velocity of 16.0 meters per second east slows uniformly to 6.0 meters per second east in 4.0 seconds. What

is the acceleration of the car during this 4.0-second interval?
Physics
1 answer:
pychu [463]2 years ago
5 0
(6-16)/4.0=-2.5 m/s²
Acceleration of the car is -2.5 m/s²
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A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
1 year ago
The image shows a pendulum that is released from rest at point A. Shari tells her friend that no energy transformation occurs as
Masja [62]
Is  D    the  right  answer
6 0
1 year ago
Read 2 more answers
The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.40 V is across it at a freq
White raven [17]

Answer:92

Explanation:

3 0
2 years ago
A 1000 kg roller coaster begins on a 10 m tall hill with an initial velocity of 6m/s and travels down before traveling up a seco
balu736 [363]

Answer:

10.6 meters.

Explanation:

We use the law of conservation of energy, which says that the total energy of the system must remain constant, namely:

\frac{1}{2}mv_i^2+mgh_i-1700j=\frac{1}{2}mv_f^2+mgh_f

In words this means that the initial kinetic energy of the roller coaster plus its gravitational potential energy minus the energy lost due to friction (1700j) must equal to the final kinetic energy at top of the second hill.

Now let us put in the numerical values in the above equation.

m=100kg

h_i=10m

v_i= 6m/s

v_f=4,6m/s

and solve for h_f

h_f= \frac{\frac{1}{2}mv_i^2+mgh_i-1700j-\frac{1}{2}mv_f^2}{mg} =\boxed{ 10.6\:meters}

Notice that this height is greater than the initial height the roller coaster started with because the initial kinetic energy it had.

6 0
2 years ago
An astronaut takes what he measures to be a 10-min nap in a space station orbiting Earth at 8000 m/s. A signal is sent from the
svet-max [94.6K]

Answer:

longer than

Explanation:

given,

time of nap = 10 min

speed of orbiting earth = 8000 m/s

c is the speed of light

using the equation of time dilation

t' = \dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}}}

now inserting all the values

t' = \dfrac{10}{\sqrt{1-\dfrac{8000^2}{3\times 10^8)^2}}}

t' = \dfrac{10}{0.9999}

t' = 10.001 s

on solving the above equation we will get a value greater than 10minutes.

hence, On earth time of nap measured will be longer than 10 min

3 0
2 years ago
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