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myrzilka [38]
1 year ago
11

A ball is thrown vertically upward. As it rises, what happens to its total energy?.

Physics
1 answer:
diamong [38]1 year ago
3 0

The total energy of a ball stays constant as it is thrown upward because potential energy increases while kinetic energy decreases. When the ball reaches its maximum height, the velocity is zero. Therefore, only potential energy exists rather than kinetic energy.

The thrower's movement imparts kinetic energy to a ball thrown vertically. The maximum height that can be achieved after leaving the hand will depend on the actual velocity. Air resistance causes some of this energy to be lost to the air as frictional dissipation, which warms the air in the area as well as the ball's surface.

We can just talk about how the ball moves when it is in the gravitational field of the Earth if we ignore this for the purposes of this discussion. The ball's total energy as it is released is comprised of both its gravitational potential energy and its kinetic energy, which result from the ball's velocity (due to its position).

The gravitational potential energy begins to rise as the ball moves vertically upward at precisely the same pace as it loses kinetic energy. The ball experiences a steady downward acceleration of 9.81 m/s2, which causes it to initially decline until it briefly comes to a stop at its highest point.

Due to its current position in the Earth's gravitational field relative to its initial position, all of the energy at this point is gravitational potential energy. As the ball experiences constant downward acceleration, its motion immediately becomes apparent in that direction because the acceleration easily transforms gravitational potential energy back into kinetic energy.

As a result, at every point along the trajectory, the total of these interchangeable forms of energy remains constant.

To learn more about what happens when a ball is thrown vertically upward:

brainly.com/question/1121850

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Uncertainty of any measuring instrument is the inaccuracy or lost of precision that occur due to its less divisions in an instrument

Uncertainty= Least Count /2

Uncertainty= 1 mm /2

Uncertainty= 0.5 mm

it means that measurement can be 2.3cm ± 0.5mm

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Makala arrives late to class and misses the first few minutes of the lecture. These are the notes she takes:
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4 years ago
A diverging lens has a focal length of magnitude 22.8 cm. (a) Locate the images for each of the following object distances. 45.6
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(a) -15.2 cm

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f = -22.8 cm is the focal length of the lens (negative because it is a diverging lens)

p = 45.6 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving the equation for q, we find the position of the image:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{45.6 cm}=-0.066 cm^{-1}

q=\frac{1}{-0.066 cm^{-1}}=-15.2 cm

and the negative sign means that the image is virtual.

(b) -11.4 cm

In this case, the distance of the object from the lens is

p = 22.8 cm

Substituting into the lens equation, we can find the new image distance, q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{22.8 cm}=-\frac{2}{22.8 cm}

q=\frac{-22.8 cm}{2}=-11.4 cm

and the negative sign means that the image is virtual.

(c) -7.6 cm

In this case, the distance of the object from the lens is

p = 11.4 cm

Substituting into the lens equation, we can find the new image distance, q:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-22.8 cm}-\frac{1}{11.4 cm}=-0.132 cm^{-1}

q=\frac{1}{0.132 cm^{-1}}=-7.6 cm

and again, the negative sign means that the image is virtual.

3 0
3 years ago
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