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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Given speed and the distance that must be covered, the time it will take the ultraviolet light to reach the earth is 3.7 × 10⁴ hours.
<h3>
What is Speed?</h3>
Speed is simply referred to as distance traveled per unit time.
Mathematically, Speed = Distance ÷ time.
Given the data in the question;
- Speed of the Ultraviolet light c = 3.0 × 10⁸m/s = 1.08 × 10⁹km/h
- Distance it must cover d = 4.0 × 10¹³km
We substitute our given values into the expression above.
Speed = Distance ÷ time
1.08 × 10⁹km/h = 4.0 × 10¹³km ÷ t
t = 4.0 × 10¹³km ÷ 1.08 × 10⁹km/h
t = 3.7 × 10⁴ hrs
Therefore, given speed and the distance that must be covered, the time it will take the ultraviolet light to reach the earth is 3.7 × 10⁴ hours.
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The gentleman bug's angular speed is the same as the ladybug's (1 rev/s)
Answer:25000/23 N/M^2
Explanation:
P=f/A=30/(0.23×0.12)=25000/23
