Answer:
the kinetic energy of clown A is 0.444 times the kinetic energy of clown B.
Explanation:
Let the spring constant of the spring is k.
For clown A:
m = 40 kg
let the extension in the spring is y.
So, the spring force, F = k y
m g = k y
40 x g = k x y
y = 40 x g / k ..... (1)
For clown B:
m' = 60 kg
Let the extension in the spring is y'.
So, the spring force, F' = k y'
m' g = k y'
y' = 60 x g / k .....(2)
Kinetic energy for A, K = 1/2 ky^2
Kinetic energy for B, K' = 1/2 ky'^2
So, K/K' = y^2/y'^2 K / K' = (40 x 40) / (60 x 60) (from equation (1) and (2))
K / K' = 0.444
K = 0.444 K'
So the kinetic energy of clown A is 0.444 times the kinetic energy of clown B.
Answer:
42.11 years old
Explanation:
Given that:
In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040
To find her age we use:

Δtm is time interval for the observer stationary relative to the sequence of
events = 2040 - 2000 = 40 years
Δts is is the time interval for an observer moving with a speed v relative to the sequence of event
v = velocity = 2.5 x 10^8 m/s
c = speed of light = 3 x 10^8 m/s

Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old
Answer:
Explanation:
velocity of projection, vo = 381 m/s
angle of projection, θ = 73.5°
The formula for the range is


R = 8067.4 m
Range in shorten by 34.1 %
So, the new range is
R' = 8067.4 - 34.1 x 8067.4/100
R' = 5316.4 m