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S_A_V [24]
3 years ago
15

Through what vertical height is a 50 N object moved if 250 J of work is done lifting it against the gravitational field of Earth

?
Physics
1 answer:
zimovet [89]3 years ago
6 0

5m

Explanation:

Given parameters:

Weight of object = 50N

Work done in lifting object = 250J

Unknown:

Vertical height = ?

Solution:

The work done on an object is the force applied to lift a body in a specific direction.

   Work done = force x distance

  Weight is a force in the presence of gravity;

  Work done = weight x height of lifting

Height of lifting = \frac{work done }{weight}

 Height of lifting = \frac{250}{50} = 5m

The vertical height through which the object was lifted is 5m

learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

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solong [7]

Due to difference in electric potential.

<h3>What is Potential difference?</h3>

The amount of work in an electrical field to move a unit charge from one place to another is known as the electrical potential difference.

Here , the potential difference is the difference in electric potential between two charged substances.

Current is the flow of charges from one end end to another end of a conducting wire connected across a two terminals of a battery.

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6 0
2 years ago
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Explanation:

3 0
3 years ago
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How do you solve for a net force
AVprozaik [17]

Answer:

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8 0
3 years ago
(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord ca
Aleks04 [339]

(a) Let v be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

a_{\rm rad} = \dfrac{v^2}R

where R is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

F = (1.500\,\mathrm{kg}) a_{\rm rad}

so that

(1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N

Solve for v :

v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}

(b) The net force equation in part (a) leads us to the relation

F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}

so that v is directly proportional to the square root of R. As the radius R increases, the maximum linear speed v will also increase, so the cord is less likely to break if we keep up the same speed.

6 0
2 years ago
How is voltmeter connected in the circuit to measure the potential difference between two points?
34kurt

Voltmeter is used to find the potential difference between two points.

We always connect it in parallel to the points where we need the potential difference.

Here in order to make the reading accurate we can increase the resistance of voltmeter so that it can not withdraw any current from the circuit.

7 0
4 years ago
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