Answer: 11 m/s
vinitial=2 m/s
time=3 s
acceleration = 3 m/s^2
vfinal = ?
The key here is that it is a constant acceleration, so we can use the constant acceleration equations. The easiest one to use would be:
vfinal=vinitial + a*t
We need vfinal, so algebraically we are ready to put in numbers into the equation:
vfinal=vinitial + a*t = 2 m/s + (3 m/s^2)*(3 s ) = 11 m/s is the final velocity
Answer:
Ft
Explanation:
We are given that
Initial velocity=u=0
We have to find the magnitude of p of the momentum of the particle at time t.
Let mass of particle=m
Applied force=F
Acceleration, 
Final velocity , 
Substitute the values

We know that
Momentum, p=mv
Using the formula

Answer:
Explanation:
40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.
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.