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Alenkinab [10]
3 years ago
14

suppose you press a basketball against a wall. what happpens to the basketball when you are pressing it?

Physics
1 answer:
AURORKA [14]3 years ago
5 0
It hardens because you are pressing it against something.
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What happens when you drop a penny from a height of 20 meters
Jet001 [13]
I'm not entirely sure but I believe that it will hit the ground and bounce back up
3 0
3 years ago
Which describes the properties reflection, absorption, and transmission when the light is shining on the oval end of a silver sp
I am Lyosha [343]

B. It reflects a lot of light, transmits almost no light, and absorbs some light.


5 0
3 years ago
Plzz help! A stationary speed gun emits a
baherus [9]

Answer:

The speed of the baseball is approximately 19.855 m/s

Explanation:

From the question, we have;

The frequency of the microwave beam emitted by the speed gun, f = 2.41 × 10¹⁰ Hz

The change in the frequency of the returning wave, Δf = +3190 Hz higher

The Doppler shift for the microwave frequency emitted by the speed gun which is then reflected back to the gun by the moving baseball is given by 2 shifts as follows;

 \dfrac{\Delta f}{f} = \dfrac{2 \cdot v_{baseball}}{c}

\therefore{\Delta f}{} = \dfrac{2 \cdot v_{baseball}}{c} \times f

Where;

Δf = The change in frequency observed, known as the beat frequency = 3190 Hz

v_{baseball} = The speed of the baseball

c = The speed of light = 3.0 × 10⁸ m/s

f = The frequency of the microwave beam = 2.41 × 10¹⁰ Hz

By plugging in the values, we have;

\therefore{\Delta f} = 3190 \ Hz =  \dfrac{2 \cdot v_{baseball}}{3.0 \times 10^8 \ m/s} \times 2.41 \times 10^{10} \ Hz

v_{baseball} = \dfrac{3190 \ Hz \times 3.0 \times 10^8 \ m/s }{2.41 \times 10^{10} \ Hz \times 2} \approx 19.855 \ m/s

The speed of the baseball, v_{baseball} ≈ 19.855 m/s

3 0
3 years ago
Suppose earth's mass increased but earth's diameter didn't change. Describe how the gravitational force between Earth and the ob
zheka24 [161]

Answer:

It increases proportionally

Explanation:

The gravitational force between the Earth and an object on its surface is given by

F=G\frac{Mm}{R^2}

where

G is the gravitational constant

M is the Earth's mass

m is the mass of the object

R is the Earth's radius

In this problem, the Earth's mass is increased, while the diameter (and therefore, the radius) doesn't change. From the equation, we see that the gravitational force is directly proportional to the Earth's mass: therefore, if the mass is increased, the force will increase as well by the same proportion (for example, if the mass is doubled, the force will double as well)

7 0
3 years ago
a golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0. if the hang time of the golf ball is 33.4 s
Kobotan [32]
Solving this using the time, we know that range = horizontal velocity x time of flight 

since there are no horizontal forces acting on the ball, there are no horizontal accelerations and the initial horizontal velocity of 36 cos 28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have 

range = 36 cos 28 x 3.44 s = 109.3 m

6 0
3 years ago
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