Answer:
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Explanation:
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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:
Option D is the correct answer.
Explanation:
Since value of angular acceleration is constant, the body has only centripetal acceleration.
Centripetal acceleration

We have radius = 7.112 cm = 0.07112 m
Frequency, f = 1975 rpm = 32.92 rps
Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s
Substituting in centripetal acceleration equation,

Option D is the correct answer.
Answer:
It is a instrument used to measure the distance traveled by a vehicle.
Explanation:
This is what it looks like ↓
Answer:
"1155 N" is the appropriate solution.
Explanation:
Given:
Acceleration,

Forces resisting motion,

Mass,

By using Newton's second law, we get
⇒ 
Or,
⇒ 
By putting the values, we get
⇒ 
⇒ 
⇒ 