Answer:
Q=1670J
Explanation:
Mass of ice: m=5g=0.005kg
Latent heat: lambda=3.34×10⁵J/kg
Heat received by ice: Q=m×lambda
Q=0.005×3.34×10⁵=5×334=1670J
The particles always move parallel and perpendicular to the waves. The waves which are in the water moves a circle. Both up and down and back and forth.
Good luck :)
<h3>
Answer: 130 newtons</h3>
===============================================================
Explanation:
We'll need the acceleration first.
- The initial speed (let's call that Vi) is 8.0 m/s
- The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
- This happens over a duration of t = 4.0 seconds
The acceleration is equal to the change in speed over change in time
a = acceleration
a = (change in speed)/(change in time)
a = (Vf - Vi)/(4 seconds)
a = (0 - 8.0)/4
a = -8/4
a = -2
The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.
Here's a further break down of Sam's speeds at the four points of interest
- At 0 seconds, he's going 8 m/s
- At the 1 second mark, he's slowing down to 8-2 = 6 m/s
- At the 2 second mark, he's now at 6-2 = 4 m/s
- At the 3 second mark, he's at 4-2 = 2 m/s
- Finally, at the 4 second mark, he's at 2-2 = 0 m/s
Next, we'll apply Newton's Second Law of motion
F = m*a
where,
- F = force applied
- m = mass
- a = acceleration
We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg
So the force applied must be:
F = m*a
F = 65*(-2)
F = -130 newtons
This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.
The magnitude of this force is |F| = |-130| = 130 newtons
Answer:
715 N
Explanation:
Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.
Let g = 9.8 m/s2
Gravity and equalized normal force is:
N = P = mg = 107*9.8 = 1048.6 N
Kinetic friction force and equalized tension force on the rope is
