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 
, 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 = 
= 
= ![[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}](https://tex.z-dn.net/?f=%5B14000x%20%2B%205000x%5E%7B2%7D%20-%208666.7x%5E%7B3%7D%5D%5E%7B0.54%7D_%7B0%7D)
= 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.
Answer:
Final speed of the crate is 15 m/s
Explanation:
As we know that constant force F = 80 N is applied on the object for t = 12 s
Now we can use definition of force to find the speed after t = 12 s
so here we know that object is at rest initially so we have
Now for next 6 s the force decreases to ZERO linearly
so we can write the force equation as
now again by same equation we have
put t = 6 s
Answer: a) angular acceleration, a = 5.24rad/s^2
b) time taken for the wheel to stop, ∆t = 0.30s
Explanation:
All shown in the attachment.
<span>Two objects move toward each other because of gravitational attraction. As the objects get closer and closer, the force between them increases. </span>
We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2
