Hi there!
We can use the work-energy theorem to solve.
Recall that:

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

Now, we can define work:

Now, plug in the values:

Solve for theta:

Answer:
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Explanation:
Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:
(1)
Where:
- Explosive force, measured in newtons.
- Barrel length, measured in meters.
- Mass of the shell, measured in kilograms.
,
- Initial and final speeds of the shell, measured in meters per second.
If we know that
,
,
and
, then the explosive force experienced by the shell inside the barrel is:

![F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%281250%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cleft%28750%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D-%5Cleft%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D%5Cright%5D%7D%7B2%5Ccdot%20%2815%5C%2Cm%29%7D)

The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Answer:
2.12/R mW
Explanation:
The electrical power, P generated by the rod is
P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?
So, the induced emf on the conductor is
E = BLv
= 0.575 T × 0.2 m × 0.4 m/s
= 0.046 V
= 46 mV
The electrical power, P generated by the rod is
P = B²L²v²/R
= B²L²v²/R
So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²
= 0.002116/R W
= 2.12/R mW
Answer:
The work done by the drag force is given by 29.96 J
Explanation:
Given :
Thrust force
N
Displacement
m
Mass of rocket
Kg
From work energy theorem,


Where
thrust work
gravitational work

After cutoff kinetic energy is converted into potential energy,

Put value of KE

Work done by drag force is given by,

J
Therefore, the work done by the drag force is given by 29.96 J
Answer:50 miles per hour 50/1hr
Explanation:100 divided by 2 is 50, divide 2 by 2 thats 1