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4 years ago

5

A transformer consists of 175 primary windings and 704 secondary windings. The emf of the primary coil is 28.0 V. What is the em

f across the secondary coil?
A. 56.0 V
B. 98.7 V
C. 189 V
D. 113 V

1 answer:

qwelly [4]4 years ago

5 0

Answer:

D. 113 v

if you need more help prior to this test let me know

Explanation:

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LekaFEV [45]

The purpose of the machine is to leverage its mechanical advantage such that the force it outputs to move the heavy object is greater than the force required for you to input.

But there's no such thing as a free lunch! When you apply the conservation of energy, the work the machine does on the object will always be equal to (in an ideal machine) or less than the work you input to the machine.

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The answer is: The work input required will equal the work output.

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3 years ago

The primary purpose of a mirror is to _________ light rays.

Effectus [21]

a. reflect (I found this using prior knowledge and process of illumination it can't be absorb cuz you wouldn't see anything it can't be refract because it doesn't reverse the image and it isn't transmit )

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3 years ago

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an

Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

$F=\dfrac{kQ^2}{r^2}$ ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

$F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}$ .....(2)

Dividing equation (1) and (2), we get :

$\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}$

Hence, the correct option is (d) i.e. " 4F/9"

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3 years ago

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0

otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

$F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$

W = 83 N

m = $\frac{83}{9.81}$

= 8.46

Now, we will balance the forces along the y-component as follows.

N = W + $F_{2}$

= 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

$F_{1} - F_{r}$ = ma

$20 - F_{r} = 8.46 \times (-0.9)$

$F_{r}$ = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

$F_{r} = \mu \times N$

$\mu = \frac{F_{r}}{N}$

= $\frac{7.614}{108}$

= 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

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4 years ago

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Question 7 of 25

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Explanation:

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3 years ago

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