Question

an object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. the initial and the final points of the path of the object lie on the same line what is the work done by the force of gravity on the object​

Answer 1

Hola !

Let the work done by the force of gravity of Object be 'W'

It is equal to the negative of PE change on that Object, which is given as:

$W = -\triangle{P.E}$. ---------[1]

We know,

PE = mgh

Here, h differs for both of the height when the object thrown is at Initial and Final Points

Let it be denoted by $h_{1}$ and $h_{2}$ respectively.

•°• $\triangle{PE } = mgh_{2} - mgh_{1}$

Substituting in [1], we get

$W = - (mgh_{2} - mgh_{1})$

$W = - mg(h_{2} - h_{1})$

It is clearly mentioned that, the initial and final points of the object lie on same line

i.e., $h_{1} = h_{2}$

•°• $\mathbf{W = - mg \times 0 = 0}$

So, your answer is $\mathbf{\underline{Zero}}$

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Answer 2

Answer:

within the electric salt of the

battery the chemical energy change into electrical energy the electric energy on flowing through the filoment of the bulb first

change into heat energy and them into the light energy.

H☺️pe it helps Y☺️u..

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