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Elanso [62]
2 years ago
15

You throw a ball so that it hits the floor a few feet in front of you. Will it bounce back to you? Explain how the bounce is sim

ilar to the way waves behave, and what laws it should obey.
50 PTS PLS HELP FAST
Physics
1 answer:
Morgarella [4.7K]2 years ago
4 0

Answer:

Yes force will make it bounce depends on the amount of force given or sometimes the type of ball

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in which item is energy stored in the form of gravitational potential energy A)a slice of bread B)a compressed string C)an apple
olchik [2.2K]

For this case, in the next item we have gravitational potential energy:

An apple in a tree.

Suppose we define our reference system at the floor level.

Suppose the apple is at a height h from the floor and has mass m.

The gravitational potential energy of the apple is given by:

U = mgh

Where,

m: apple mass

h: height of the apple with respect to the floor

g: acceleration due to gravity

Answer:

C) an apple on a tree

7 0
4 years ago
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Which part of the eye is used to see things in high detail?
irina1246 [14]

Answer:

Retina is the part of eye which is used to see things in high details

7 0
3 years ago
The moon is 3x10^5 km away from Nepal and the mass of the moon is 7x10^22 kg. Calculate the force with which the Moon pulls ever
hammer [34]

Answer:

Approximately 5.19 \times 10^{-5}\; \rm N.

Explanation:

Let G denote the gravitational constant. (G \approx 6.67 \times 10^{-11} \; \rm N \cdot kg^{-2} \cdot m^{2}.)

Let M and m denote the mass of two objects separated by r.

By Newton's Law of Universal Gravitation, the gravitational attraction between these two objects would measure:

\displaystyle F = \frac{G \cdot M \cdot m}{r^{2}}.

In this question: M = 7 \times 10^{22}\; \rm kg is the mass of the moon, while m = 1\; \rm kg is the mass of the water. The two are r = 3\times 10^{5}\; \rm km apart from one another.

Important: convert the unit of r to standard units (meters, not kilometers) to reflect the unit of the gravitational constant G.

\displaystyle r = 3 \times 10^{5}\; \rm km \times \frac{10^{3}\; \rm m}{1\; \rm km} = 3 \times 10^{8}\; \rm m.

\begin{aligned} F &= \frac{G \cdot M \cdot m}{r^{2}} \\ &= \frac{6.67 \times 10^{-11}\; \rm N \cdot kg^{-2}\cdot m^{2} \times 7 \times 10^{22}\; \rm kg \times 1\; \rm kg}{(3 \times 10^{8}\; \rm m)^{2}} \\ &\approx 5.19 \times 10^{-5} \; \rm N\end{aligned}.

5 0
3 years ago
Scientific theories have strod the test of time and will not change
Nataly [62]
That is not a question but not all scientific theories have stood the test of time
8 0
3 years ago
You observe a spiral galaxy with a large central bulge and tightly wrapped arms. It would be classified a
miv72 [106K]

Answer:

Sa

Explanation:

Spiral Galaxies  -

It is a disk shaped galaxies which have spiral structure , is refereed to as spiral galaxies .

According to Hubble , these galaxies are classified as Sa , Sb , Sc .

Where ,

Sa - have the structure , which is bulged from the central portion , along with a tightly wrapped spiral structure .

Sb - have a lesser bulge and the spiral is looser .

Sc - It has very weak bulge with the open spiral structure .

Hence , from the question ,

The given information is about the Sa .

6 0
3 years ago
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