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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Av Speed = total distance / time time = 32+ 46 / 2.7 = 28 m/sec
Av velocity = total displacement / time total = S / t
S = sqrt( 32^2 +46^2) = 56 m
Av Velocity = 56/ 2,7 = 20.75 m/sec
with angle tan^-1 = 0.7 north west ( about 35 degrees north west)
the answer should be:
When the buoyant force is equal to the force of gravity
Based on the law of conservation of energy, we know that we can't create energy, machines can only convert one type of energy into another. So, if we want to improve a machines's ability then we need to reduce it's loss energy (part of energy which is useless). Out of all the options only Option C fits best with it.
In short, Your Answer would be Option C
Hope this helps!
Answer:
(a) A = m/s^3, B = m/s.
(b) dx/dt = m/s.
Explanation:
(a)

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.
(b) 
This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.