The radio is playing at the front of the room. How does the sound get from the speakers to your ears? A: A breeze blows from the
1 answer:
You might be interested in
1) -0.5 m/s
We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.
Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:
After the shot, the momentum is:
where
M = 2000 kg is the mass of the cannon
m = 10 kg is the mass of the shell
v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)
V = ? is the velocity of the cannon
Since momentum is conserved, we can write
And solving for V, we find the velocity of the cannon:
where the negative sign indicates that the cannon moves in the direction opposite to the shell.
2) 0.5 m
The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:
where
v is the final velocity of the cannon
u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)
is the deceleration of the cannon
s is the distance travelled by the cannon
The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:
Answer:
The work done will be
Explanation:
The work equation is given by:
Where:
F is the force due to gravity (weight = mg)
x is the length of the ramp (3 m)
Now, the force acting here is the component of weight in the ramp direction, so it will be:
Therefore, the work done will be:
I hope it helps you!