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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A will be the fastest and c the slowest because of the dip it has a is a straight line fastest way to get from a to b is a straight line b is the second fastest and d is last
What are the "following" devices ? ?
I think they're a list of choices that you have but aren't sharing.
A few devices associated with the reception of various types of
radio signals include the resonant tank, the local oscillator, the
mixer, the detector, the coherer, the discriminator, the parabolic
reflector, the lecher wires, the audio transducer, the demultiplexer,
and ... my personal guess ... the 'antenna' or 'aerial'.
Well, the surface of still water has surface tension. If there isn't enough mass or weight to break the surface tension, the object will float.
Answer:
The body has negative acceleration PR a deceleration.
Explanation:
HOPE THAT THIS IS HELPFUL.
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