A 21 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar
1 answer:
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Refer to the diagram shown.
Because the surface is frictionless, the resistive for, R, is zero.
Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²
The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg
For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²
Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)
Because the surface is frictionless, the resistive for, R, is zero.
Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²
The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg
For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²
Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)
Answer:c
Explanation:
Given
object is falling Freely with an odometer
Suppose it falls with zero initial velocity
so distance fallen in time t is given by
here u=0 and t=time taken
for
for
distance traveled in 2 nd sec
for
distance traveled in 3 rd sec
so we can see that distance traveled in each successive second is increasing