Answer:
T² ∝ R³
Explanation:
Given data,
The period of revolution of the planet around the sun, T
The mean distance of the planet from the sun, R
According to the III law of Kepler, " Law of Periods' states that the square of the orbital period to go around the sun once is directly proportional to the cube of the mean distance between the sun and the planet.
T² ∝ R³
From the above equation it is clear that T² varies directly as the R³.
Perpendicular acceleration:
F = ma
a = 4 / 2 = 2 m/s²
Perpendicular distance:
s = ut + 1/2 at²
s = 0 x 4 + 1/2 x 2 x 4²
s = 16 m
Horizontal distance:
s = ut
= 3 x 4
= 12 m
Total distance = √(12² + 16²)
= 20 m.
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Answer
given,
mass of ball, m = 57.5 g = 0.0575 kg
velocity of ball northward,v = 26.7 m/s
mass of racket, M = 331 g = 0.331 Kg
velocity of the ball after collision,v' = 29.5 m/s
a) momentum of ball before collision
P₁ = m v
P₁ = 0.0575 x 26.7
P₁ = 1.535 kg.m/s
b) momentum of ball after collision
P₂ = m v'
P₂ = 0.0575 x (-29.5)
P₂ = -1.696 kg.m/s
c) change in momentum
Δ P = P₂ - P₁
Δ P = -1.696 -1.535
Δ P = -3.231 kg.m/s
d) using conservation of momentum
initial speed of racket = 0 m/s
M u + m v = Mu' + m v
M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)
0.331 u' = 3.232
u' = 9.76 m/s
change in velocity of the racket is equal to 9.76 m/s
Answer:
the angular velocity of the car is 12.568 rad/s.
Explanation:
Given;
radius of the circular track, r = 0.3 m
number of revolutions per second made by the car, ω = 2 rev/s
The angular velocity of the car in radian per second is calculated as;
From the given data, we convert the angular velocity in revolution per second to radian per second.
Therefore, the angular velocity of the car is 12.568 rad/s.
