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3 years ago

A device that has the capacity to receive and store electrical energy is a(n) .

Physics

1 answer:

Dmitriy789 [7]3 years ago

7 0

A capacitor is used to receive and store electrical energy.

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At a playground, a child slides down a slide that makes a 42° angle with the horizontal direction. The coefficient of kinetic fr

kumpel [21]

Answer:

$a = 5.1\ m/s^2$

Explanation:

Given,

The angle of the slide= $42^\circ$

The mass of the child is= m

coefficient of friction = 0.20

when she slides down now apply Newton's law

$ma =mg \sin\theta - f$

$ma = mg\sin \theta -\mu mg \cos \theta$

therefore the acceleration

$a=g \sin\theta -\mu gcos\theta$

$a=g[\sin \theta -\mu \cos \theta]$

$a=9.8\times [\sin 42^\circ -0.2\times \cos 42^\circ]$

$a = 5.1\ m/s^2$

hence, the magnitude of acceleration during her sliding is equal to $a = 5.1\ m/s^2$

4 0

4 years ago

If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a result?

creativ13 [48]

Mechanical energy i think will form

4 0

3 years ago

Read 2 more answers

Air entering the potential space of the pleural cavity is called

PolarNik [594]

Air entering the potential space of the pleural cavity is called pneumothorax.

4 0

3 years ago

Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3

sweet [91]

Given that,

Rank in order from largest to smallest the magnitude of the electric field at block dot.

Electric field :

Electric field is proportional to the charge divided by square of distance.

In mathematically,

$E\propto\dfrac{q}{r^2}$

Where, q = charge

r = distance

If the charge is greater then electric field will be greater.

If the distance is greater then electric field will be smaller.

We need to find the electric field at black dot

According to figure,

(I). The electric field at black dot due to positive charge point q to left. the distance is r.

The electric field will be

$E=\dfrac{kq}{r^2}$

The electric field will be largest.

(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.

Then, the electric field will be

$E=\dfrac{k2q}{(2r)^2}$

$E=\dfrac{kq}{2r^2}$

The electric field will be smallest.

(III). The electric field at black dot due to positive charge point 2q to left. The distance is r.

Then, the electric field will be

$E=\dfrac{k2q}{(r)^2}$

The electric field will be very largest.

(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.

Then, the electric field will be

$E=\dfrac{kq}{(2r)^2}$

$E=\dfrac{kq}{4r^2}$

The electric field will be very smallest.

So, The electric field from largest to smallest will be

$E_{3}>E_{1}>E_{2}>E_{4}$

Hence, The ranking will be 3, 1, 2, 4.

(D) is correct option.

4 0

3 years ago

Help????????????????

Naddik [55]

Answer:

23.5

Explanation:

Dunno how 2 explain but this is correct 4 sureeeeee.

6 0

3 years ago