The acceleration of body is given 16.3m/s2 and the force is given 4.6 N then
We know,
Force=mass*acceleration
Then,
Mass=force/acceleration
Mass=4.6/16.3
Mass=0.28kg
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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time=distance/speed
1.6/100 secs = 0.016secs=16millisecs
Answer:
11.4 m/s
Explanation:
The expression for the Centripetal acceleration is :
Where, a is the accleration
v is the velocity around circumference of circle
R is radius of circle
In the given question,
a = g = Acceleration due to gravity as the car is at top = 
v = ?
R = 13.2 m
So,
<u>v = 11.4 m/s</u>
