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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1. Roll a ball across a table into an object
2. Drop something
Answer: IM 95%sure that the answer is B jus took the test got the answer right
Explanation:
Explanation:
change 0.5 g to kg so 0.005kg then change 100 ml to m so 0.001m so density=mass over volume so from there you can continue
Answer:
4.44s
Explanation:
A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables
since the mass of the child and that of the swing is negligible, the masses wont be involved in the calculation
T=2π√L/g
g=acceleration due to gravity which is 9.81m/s2
the length of the supporting cable is 4.9m
T the period
period is the time required to make a complete oscillation
T=2*π√4.9/9.81
T=2*π*0.706
T=4.44s
4.44s
