Answer:
The centripetal acceleration changed by a factor of 0.5
Explanation:
Given;
first radius of the horizontal circle, r₁ = 500 m
speed of the airplane, v = 150 m/s
second radius of the airplane, r₂ = 1000 m
Centripetal acceleration is given as;

At constant speed, we will have;

a₂ = 0.5a₁
Therefore, the centripetal acceleration changed by a factor of 0.5
Answer:
the answer is false :)
Explanation:
Science is based on theories and tests, if someone has a theory they will test and if it is correct yippie! and if it isn't they will redo the test over until they get it correct or unless it is unsolvable.
The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
Explanation:
Consider the vertical motion of ball,
We have equation of motion v = u + at
Initial velocity, u = u sin θ
Final velocity, v = 0 m/s
Acceleration = -g
Substituting
v = u + at
0 = u sin θ - g t

This is the time of flight.
Consider the horizontal motion of ball,
Initial velocity, u = u cos θ
Acceleration, a =0 m/s²
Time,
Substituting
s = ut + 0.5 at²

This is the range.
In this problem
u = 30 m/s
g = 9.81 m/s²
θ = 45° - For maximum range
Substituting

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.
So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
(a) 
The frequency of a wave is given by:

where
v is the wave's speed
is the wavelength
For the red laser light in this problem, we have
(speed of light)

Substituting,

(b) 427.6 nm
The wavelength of the wave in the glass is given by

where
is the original wavelength of the wave in air
n = 1.48 is the refractive index of glass
Substituting into the formula,

(c) 
The speed of the wave in the glass is given by

where
is the original speed of the wave in air
n = 1.48 is the refractive index of glass
Substituting into the formula,
