Answer:
44.1 m
Explanation:
Given:
v₀ = 0 m/s
a = -9.8 m/s²
t = 3 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (3 s) + ½ (-9.8 m/s²) (3 s)²
Δy = -44.1 m
The ball should be dropped from 44.1 meters.
The answer is d
F=ma. so by substituting we find that a=4
and a=vi-vf/t
by substiting we find vf=8
so the difference is vf-vi = 8m/s
The required spring constant:
The spring constant of the spring is
.
Calculation:
The mass of the car is m=1200 kg, the speed of the car is v=25 m/s, and after colliding the spring is brought to rest at a distance of x=2.5m. Let the spring constant of the spring is, k.
From the conservation of energy,
Total initial kinetic energy= Total final potential energy of the spring
Therefore,

Now, substituting the values of the mass of the car, speed of the car, and displacement, we get:

To know more about spring constant, refer to:
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Answer: distance = 24 meters
Explanation:
speed = distance / time
distance = speed x time
speed = 160 cm/s = 1.6 m /s [ 100 cm = 1 m, 160 cm = 1.6 m ]
time = 15 s
distance = 1.6 m/s x 15 s = 24 m
Answer: The gravitational acceleration on planet X is 5 N/kg
On Earth (with the gravitational accelartion g_E) the mass of 2kg will correspond to

On planet X we are told the same measure is only 10N. Since there is a proportional relationship between g and F, we can calculate g_X:
