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11Alexandr11 [23.1K]
3 years ago
9

A rock excerts a pressure of 20 N/cm^2 on the ground. what does 20N/cm^2 mean?

Physics
1 answer:
Lelechka [254]3 years ago
6 0

Answer:

A force of 20N acts on each cm^2 of area

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Let's take two balls, each of radius 0.1 meters and charge them each to 3 microCoulombs. Then, let's put them on a flat surface
salantis [7]

Explanation:

First, we will calculate the electric potential energy of two charges at a distance R as follows.

                   R = 2r

                      = 2 \times 0.1 m

                      = 0.2 m

where,    R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.

                U = \frac{k \times Q \times Q}{R}

                    = \frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}

                    = 0.081 J

As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, \frac{U}{2}.

Hence,   K.E = \frac{U}{2}

                     = \frac{0.081}{2}

                     = 0.0405 J

Now, we will calculate the speed of balls as follows.

                 V = \sqrt{\sqrt{\frac{2 \times K.E}{m}}

                     = \sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}

                     = 0.142 m/s

Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.

6 0
3 years ago
Life cycle of a medium mass star
miskamm [114]

Our sun is a medium mass star, so it wouldn't be too different from the sun's life cycle. It is born, lives for about 10 billion years and then dies. ... As a medium mass star nears the end of its life, it runs out of hydrogen which it has been fusing onto helium in its core for its whole life.

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3 years ago
I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to twice its length an
Nikitich [7]

Answer:

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Explanation:

Fundamental frequency = wave velocity/2L

where;

L is the length of the stretched rubber

Wave velocity = \sqrt{\frac{T}{\frac{M}{L}}}

Frequency (F₁) = \frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ = \frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).

7 0
3 years ago
A 1569 kg car moves with a velocity of 15 m/s. What is its kinetic energy? Joules
Crank
Kinetic energy = 1/2 * mass * velocity^2

In this case,
KE = 1/2 * 1569 kg * (15 (m/s))^2 = 176,5 kN
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3 years ago
Based on Kepler's work, which best describes the orbit If a planet around the Sun?
krek1111 [17]

Answer:

an ellipse with the Sun at one focus  or D

Explanation:

Answer for edgenuity

7 0
3 years ago
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