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ahrayia [7]
2 years ago
14

How is power defined? Question 8 options: the quantity of work accomplished the direction of the work the total distance an obje

ct is moved the rate at which work is accomplished.
Physics
1 answer:
kifflom [539]2 years ago
7 0

Power is defined by the rate of work done by the object. Statement D represents the correct definition of Power.

<h3>What is Power?</h3>

Power can be defined as the transfer of energy from one place to another place. The rate at which this energy is transferred is called power.

In other words, the rate at which the work is done by an object is called its power. The mathematical representation of power is given as,

P = \dfrac {\Delta W}{\Delta t}

Here, P is the power, \Delta W is the change in energy or work done and  \Delta tchange in time.

The Power can be written in SI units as Watt or J/s.

Hence we can conclude that the power is the rate at which work is accomplished. Statement D is the correct representation of Power.

To know more about power, follow the link given below.

brainly.com/question/1618040.

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What is the momentum of a 5 kg object that has a velocity of 1.2 m/s? 3.8 kg • m/s 4.2 kg • m/s 6.0 kg • m/s 6.2 kg • m/s
Gnesinka [82]

Answer:

Your answer will be 6.0kg•m/s

Explanation:

In the given question all the required details d given. Using these information's a person can easily find the momentum of the object. In the question it is already given that the mass of the object is 5 kg and the velocity at which it is traveling is 1.2 m/s.We know the equation of finding momentum asMomentum = mass * velocity                   = 5 * 1.2                    = 6So the momentum of the object is 6 Newton.

4 0
4 years ago
Read 2 more answers
The equation for the speed of a satellite in a circular orbit around the earth depends on mass. Which mass?
katovenus [111]
<h3><u>Question: </u></h3>

The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?

a. The mass of the sun

b. The mass of the satellite

c. The mass of the Earth

<h3><u>Answer:</u></h3>

The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.

Option c

<h3><u> Explanation: </u></h3>

Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence F_{G} = F_{C}.

Gravitational force between Earth and Satellite: F_{G} = \frac{G \times M_e \times M_s}{R^2}

Centripetal force of Satellite :F_C = \frac{M_s \times V^2}{R}

Where G = Gravitational Constant

M_e= Mass of Earth

M_s= Mass of satellite

R= Radius of satellite’s circular orbit

V = Speed of satellite

Equating  F_G = F_C, we get  

Speed of Satellite V =\frac{\sqrt{G \times M_e}}{R}

Thus the speed of satellite depends only on the mass of Earth.

6 0
4 years ago
Which single force acts on an object in free fall
ICE Princess25 [194]

Answer:

gravity

Explanation:

4 0
3 years ago
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Can someone help me ASAP!!!!
madam [21]
The US English System of measurement grew out of the manner in which people secured measurements using body parts and familiar objects. For example, shorter ground distances were measured with the human foot and longer distances were measured by paces, with one mile being 1,000 paces. Capacities were measured with household items such as cups, pails (formerly called gallons) and baskets.
4 0
3 years ago
According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about
NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0
3 years ago
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