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NNADVOKAT [17]
2 years ago
11

A box of mass 3.6 kg is lifted 5.4 m above the

Physics
1 answer:
horsena [70]2 years ago
7 0

So, the energy change that occurs is 190.512 J.

<h3>Introduction</h3>

Hello ! I am Deva from Brainly Indonesia will help you regarding energy and its transformation. In this case, it's the use of energy from the lifter to be equivalent to the change in the object's potential energy. Why potential energy? Because the box undergoes a change in height and the potential energy specializes at a certain height. Work (W) due to change in potential energy (\sf{\Delta PE}) can be realized in the equation :

\sf{W = \Delta PE}

\sf{W = m \cdot g \cdot h_2 - m \cdot g \cdot h_2}

\boxed{\sf{\bold{W = m \cdot g \cdot (h_2 - h_1)}}}

With the following condition :

  • W = work of subject (J)
  • \sf{\Delta PE} = change of potential energy (J)
  • m = mass (kg)
  • g = acceleration of the gravity (m/s²)
  • \sf{h_2} = final height (m)
  • \sf{h_1} = initial height (m)

<h3>Problem Solving</h3>

We know that :

  • m = mass = 3.6 kg
  • g = acceleration of the gravity = 9.8 m/s²
  • \sf{h_2} = final height = 5.4 m
  • \sf{h_1} = initial height = 0 m

What was asked :

  • W = work of subject = ... J

Step by step :

\sf{W = \Delta PE}

\sf{W = m \cdot g \cdot (h_2 - h_1)}

\sf{W = 3.6 \cdot 9.8 \cdot (5.4 - 0)}

\sf{W = 3.6 \cdot 9.8 \cdot 5,4}

\boxed{\sf{W = 190.512 \: J}}

So, the energy change that occurs is 190.512 J.

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3 years ago
A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.
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Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

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now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight=\sqrt{\frac{2H}{g}}

substituting value

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<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u*\sqrt{\frac{2H}{g}}

  • =8.25*2.26
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<h3><u>The ball strikes 18.63m far from building</u>. </h3>
7 0
2 years ago
In a certain chemical process, a lab technician supplies 292 J of heat to a system. At the same time, 68.0 J of work are done on
vekshin1

Answer:

The increase in the internal energy of the system is 360 Joules.

Explanation:

Given that,

Heat supplied to a system, Q = 292 J

Work done on the system by its surroundings, W = 68 J

We need to find the increase in the internal energy of the system. It can be given by first law of thermodynamics. It is given by :

dE=dQ+dW\\\\dE=292\ J+68\ J\\\\dE=360\ J

So, the increase in the internal energy of the system is 360 Joules. Hence, this is the required solution.

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