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4 years ago

The measure of energy transferred between two points in an electric circuit is?

1 answer:

6 0

An Ohm or a electrical resistance is a measure of the amount of electrical energy transferred by an electric charge as it moves from one point to another circuit. The voltage between two points in a circuit can be measured with a voltmeter.

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What is the net Force needed to get a 16 kg box moving 4 m/s^2?

igomit [66]

Answer:

The net force should be of a magnitude of 64 N

Explanation:

We use Newton's second Law for this:

$F_{net} = m\,*\,a$

which for our case gives:

$F_{net} = m\,*\,a=16\,(4)\,N= 64\,\,N$

3 0

3 years ago

Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then

Andreas93 [3]

Answer:

The displacement is $\Delta H = - 1 \ m$

The distance is $D = 4 \ m$

Explanation:

From the question we are told that

The height from which the ball is dropped is $h = 1 \ m$

The height attained at the first bounce is $h_1 = 0.8 \ m$

The height attained at the second bounce is $h_2 = 0.5 \ m$

The height attained at the third bounce is $h_3 = 0.2 \ m$

Note : When calculating displacement we consider the direction of motion

Generally given that upward is positive the total displacement of the ball is mathematically represented as

$\Delta H = (0 - h ) + ( h_1 - h_1 ) + (h_2 - h_2 )+ (h_3 - h_3)$

Here the 0 show that there was no bounce back to the point where Billy released the ball

$\Delta H = (0 - 1 ) + ( 0.8- 0.8 ) + (0.5 - 0.5 )+ (0.2 - 0.2)$

=> $\Delta H = - 1 \ m$

Generally the distance covered by the ball is mathematically represented as

$D = h + 2h_2 + 2h_3 + 2h_3$

The 2 shows that the ball traveled the height two times

$D = 1 + 2* 0.8 + 2* 0.5 + 2* 0.2$

=> $D = 4 \ m$

5 0

3 years ago

Work done by a force forum X to X

Ksivusya [100]

Answer:

W = 68 J

Explanation:

On a force vs displacement chart, work is the area under the curve.

The area under the curve can be divided into a rectangle and a triangle

W = Fd = (2 N)(12 - 0 m) + ½(13 - 2 N)(12 - 4 m) = 68 N•m = J

5 0

3 years ago

What would happen to the two balls if one of them were kept positively charged and the charge on the other ball were slowly incr

Whitepunk [10]

Answer:

The force of repulsion between the two balls will increase

Explanation:

The electrostatic force between two charged objects is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the distance between the centres of the two objects

We see that the magnitude of the force is directly proportional to the charges on the two objects. in this problem, we have two positively charged balls (so, there is a force of repulsion between them, since like charges repel each other and unlike charges attract each other), and the positive charge in one of them is slowly increased: this means that either $q_1$ or $q_2$ in the formula is increasing, and so the magnitude of the force is increasing.