The height of a typical playground slide is about 6 ft and it rises at an angle of 30 ∘ above the horizontal.
1 answer:
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The ball was dropped from a height 20 meters
Explanation:
The given is
1. A ball is dropped from the top of a cliff
2. By the time it reaches the ground, all the energy in its gravitational
potential energy store has been transferred into its kinetic energy
store, that mean K.E = P.E
3. The ball is travelling at 20 m/s when it hits the ground
4. The gravitational field strength is 10 N/kg
We need to find the height that the ball dropped from it
The ball dropped from the top of a cliff means the initial speed is 0
→ K.E =
where v is the final speed, in the initial speed and m
is the mass
→ v = 20 m/s and = 0 m/s
→ K.E =
→ K.E =
→ K.E = 200 m joules ⇒ when the ball hits the ground
→ P.E = m g h
where g is the gravitational field strength, m is the mass and h is
the height
→ g = 10 N/kg
→ P.E = m(10)(h)
→ P.E = 10 m h joules
→ P.E = K.E
→ 10 m h = 200 m
Divide both sides by 10 m
→ h = 20 meters
The ball was dropped from a height 20 meters
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Refer to the diagram shown below.
When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.
At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.
Answer:
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.
When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.
At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.
Answer:
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.
Answer:
vB' = 0.075[m/s]
Explanation:
We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.
Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.
where:
mA = 0.355 [kg]
vA = 0.095 [m/s] before the collision
mB = 0.710 [kg]
vB = 0.045 [m/s] before the collision
vA' = 0.035 [m/s] after the collision
vB' [m/s] after the collison.
The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.