Answer:
a.) the speed at the bottom is greater for the steeperhill
Explanation:
since the energy at the bottom of the steeper hilis greater
As we can see from above that v is higher when h ishigher.
A 513 g ball strikes a wall at 12.1 m/s and
1 answer:
The average force on the ball is 287.3 N.
Explanation:
The impulse exerted on an object, which is equal to the product between the force exerted and the duration of the collision, is equal to the change in momentum of the object.
If we apply this to the ball, we can write:
where
F is the force exerted on the ball
is the duration of the collision
m = 513 g = 0.513 kg is the mass of the ball
u = 12.1 m/s is the initial velocity of the ball
v = -13.1 m/s is the final velocity (negative since the ball rebounds in the opposite direction)
And solving for F, we find:
So, the magnitude of the average force is 287.3 N.
Learn more about impulse and change in momentum:
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Answer:
a) W = 46.8 J and b) v = 3.84 m/s
Explanation:
The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy
W = ΔK = -K₀
a) work is the scalar product of force by distance
W = F . d
Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.
W = F d cos θ
W = 39.0 1.20 cos 0
W = 46.8 J
b) zero initial kinetic language because the package is stopped
W - =
-K₀
W - fr d= ½ m v² - 0
W - μ N d = ½ m v
on the horizontal surface using Newton's second law
N-W = 0
N = W = mg
W - μ mg d = ½ m v
v² = (W -μ mg d) 2/m
v = √(W -μ mg d) 2/m
v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3 ]
v = √(31.63 2/4.3)
v = 3.84 m/s