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:
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Answer:
magnitude of A − B = 15.81 km
Explanation:
Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a magnitude of 15 km.
According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.
A(0,-5)
B(15,0)
A - B = (-15 i - 5 j )
Magnitude of the vector is given by
|A - B| = 
|A - B| = 
|A - B| = 15.81 km