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:
Explanation:
We have given initial length of the steel guitar l = 1 m
Cross sectional area 
Young's modulus 
Force F = 1500 N
So stress 
We know that young's modulus 
So 
Now strain 
Answer:
Explanation:
Current, I = 6 A
diameter of wire, d = 2.05 mm
number of electrons per unit volume, n = 8.5 x 10^28
If the diameter is doubled,
The resistance of the wire is inversely proportional to the square of the diameter of the wire, so the resistance is one forth an the current is directly proportional to the diameter of the wire so the current is four times the initial value.
Answer:
Vx = 10.9 m/s , Vy = 15.6 m/s
Explanation:
Given velocity V= 19 m/s
the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°
θ = 55°
to Find Vx = ? and Vy= ?
Vx = V cos θ
Vx = 19 m/s × cos 55°
Vx = 10.9 m/s
Vx = V sin θ
Vy = 19 m/s × sin 55°
Vy = 15.6 m/s
