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:Your life is your responsibility.
Explanation:
The answer is A.
p=m/v
p= 240/60
p= 4 g/cm^3
Answer:
Gravitational Potential Energy
Explanation:
a ball is held rest at the top of hill
gravitational potential energy will store due to its height
it. and body will start move downward and its potential energy will convert into kinetic energy due to motion of body
at the ground level it will stop and potential energy will became zero and kinetic energy get convert into internal energy due to collisions
A spring that obeys Hooke's law has a spring force constant of 272 N/m. This spring is then stretched by 28.6 cm