A 0.20 kg stone is dropped (from rest) from a height of 40.0 m above the ground. How far above the ground is the stone when it's
1 answer:
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Answer:
s = 20 m
Explanation:
given,
mass of the roller blader = 60 Kg
length = 10 m
inclines at = 30°
coefficient of friction = 0.25
using conservation of energy
u = 9.89 m/s
Using second law of motion
ma =μ mg
a = μ g
a = 0.25 x 9.8
a = 2.45 m/s²
Using third equation of motion ,
v² - u² = 2 a s
0² - 9.89² = 2 x 2.45 x s
s = 20 m
the distance moved before stopping is 20 m
Answer:
<em>Height = 5.65 km</em>
Explanation:
is the circumference or we can say measures the boundary of hemisphere of friction-less ice that he is sitting on.
So, the height will be = 2 x 3.14 x (30)^2 = 5654.7 m = 5.65 km
Answer:
mass = 9.7 kg
Explanation:
As we know that when object is at rest on the ground of flat base then we will have
so from here
now we have
N = 95 Newton
now from above equation we will have