Answer:
The final velocity of the bullet is 9 m/s.
Explanation:
We have,
Mass of a bullet is, m = 0.05 kg
Mass of wooden block is, M = 5 kg
Initial speed of bullet, v = 909 m/s
The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,
So, the final velocity of the bullet is 9 m/s.
1) 26.2 m/s
The mechanical energy of the divers at any point of their vertical motion is sum of the kinetic energy and the gravitational potential energy:
where
m is the mass of the diver
v is the speed
g = 9.8 m/s^2 is the acceleration due to gravity
h is the height above the water
When the diver is on the cliff, v = 0 (he is at rest), so K=0 and the initial mechanical energy is just potential energy:
where h=35 m is the height of the cliff.
When the diver hits the water above, h = 0, so U=0 and the final mechanical energy is just kinetic energy:
since the total mechanical energy is conserved, we have
And solving the equation for v, we find the speed when they reach the surface of the water:
2) It is converted into thermal energy of the water
When the diver enters the water, he suddenly feels another force acting against the motion of the diver: the resistance of the water. The resistance of the water acts upward, slowing down the diver until he stops.
In this process, the speed of the diver (v) decreases, and therefore the kinetic energy of the diver decreases as well, until it becomes zero.
However, this does not mean that the conservation of energy has been violated. In fact, the kinetic energy of the diver has been converted into thermal energy of the molecules of water surrounding the diver.
Answer:
A)
B)
C)
Explanation:
This question is incomplete. The full question was:
<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s. </em>
<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement. </em>
<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder. </em>
<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>
For part A), we make a balance of energy to calculate the work done by the friction force:
For part B), we use our previous value for the work:
Solving for friction force:
For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:
Solving for a:
Now, by dynamics:
