The distance at which the man slips is 0.3 m
Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.
Given-
mass of man= 70 kg
frictional coefficient μ=0.02
mass of body thrown= m2 = 3kg
let s be the stopping distance
we know that frictional force = F= μN
=μMg= 0.02 x 70 x 10
=14 N
∴acceleration, a= 14/70 = 0.2 m/s²
now on applying conservation of linear momentum
pi=pf pi=0 (initially at rest)
0=m1v1-m2v2 (v1= velocity of man) (v2=velocity of body= 8m/s
v1= m2v2 /m1= 0.3 m/s
we know,
v²- u² = -2as
0- (0.3) ²= -2 x 0.2 x 5
s= 0.09/0.4 ≈ 0.3 m
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Answer:
11.3 m/s
Explanation:
KE₁ = KE₂
½m₁v₁² = ½m₂v₂²
½ (2 kg) v² = ½ (4 kg) (8 m/s)²
v ≈ 11.3 m/s
Answer:
14 hours 18 minutes.
Explanation:
ratio of number of orbits, so it completes 7 orbits in the time Janus does 6.
(16*60+41)*6/7=858 minutes or 14 hours 18 minutes
The answer is :78 I think
Answer:
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Explanation:
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