Power is calculated as work per unit time, and work in turn is calculated as force multiplied by distance. In this case, the force required is equivalent to the weight of the barbell multiplied by acceleration due to gravity.
P = W/t = Fd/t = mgd/t = (200 kg)(9.81 m/s^2)(2 m)/2.2 s = 1783.64 Watts.
Answer:
Option 3 and 6
Explanation:
If the particle is under the action of only one external force, both its kinetic energy and the velocity must vary.
When the particle is exerted by an external force, then this external force tends to accelerate the particle.
When the particle accelerates, its velocity must vary for that acceleration to occur.
When the velocity of the particle changes then its Kinetic energy must vary with the velocity of the particle.
The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
To find the answer, we have to know about the pressure.
<h3>How to find the weight of a column of air?</h3>
- As we know that the expression of pressure as,

where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.
- It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,

- From this, the value of weight will be,

Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
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The tension in the string corresponds to the gravitational attraction between the Sun and any planet.