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beks73 [17]
3 years ago
8

If you wanted to increase the voltage, which would you use?

Physics
2 answers:
Liula [17]3 years ago
5 0
B) step-up transformer
bixtya [17]3 years ago
4 0

The correct answer of the question is :  B) Step-up transformer.

EXPLANATION :

Before coming into any conclusion, first we have to understand the step-up and step-down transformer.

The step-up transformer is the transformer whose transformer ratio is greater than one. It means it will increase the voltage of the source or decrease its current.

On the other hand, the transformer ratio of step-down transformer is less than one. Hence, it will decrease the higher voltage to lower voltage or increase the lower current into higher current of the given source.

As per the question, we want to increase the voltage of the source.

Hence, we have to use step-up transformer.

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3 years ago
Hey scooters dragging of 520 kg walk-through forest at a constant speed of 3.5 m/s. If the scooter is applying a force of 1850 N
Simora [160]

The coefficient of friction is 0.363

Explanation:

There are two forces acting on the scooter in the horizontal direction:

- The applied force, F = 1850 N, forward

- The frictional force, F_f, backward

Since the scooter is moving at constant speed, the acceleration is zero, so the net force acting on the scooter must be zero. Therefore we can write:

F-F_f = 0\\F_f = F = 1850 N

The frictional force can be written as

F_f = \mu R (1)

where

\mu is the coefficient of friction

R is the normal reaction of the ground on the scooter

For a flat horizontal surface, there is equilibrium along the vertical direction, so the normal reaction is equal to the weight:

R = W = mg

where

m = 520 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

Substituting into (1),

F_f = \mu mg = 1850 N

and solving for \mu,

\mu=\frac{F_f}{mg}=\frac{1850}{(520)(9.8)}=0.363

Learn more about friction:

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5 0
4 years ago
You are pushing an 80.0 N wheelbarrow as shown in (Figure 1). You lift upward on the handle of the wheelbarrow so that the only
AleksandrR [38]

Answer:

Wl = 1740 N

Explanation:

maximum lift weight unaided = force exerted (F) = 650 N

length of the wheelbarrow (L) = 1.4 m

weight of the wheelbarrow (w) = 80 N

distance of center of gravity of the wheel barrow from the wheel = 0.5 m

distance of center of gravity of the load from the wheel = 0.5 m

find the weight of the load (Wl)

from the diagram attached we can see that there is going to be a rotation about the axis A. let clockwise rotation be positive

ΣT = (F x 1.4) - ((Wl x 0.5) + (w x 0.5) = 0

(F x 1.4) = ((Wl x 0.5) + (w x 0.5)

Wl =

Wl =

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6 0
2 years ago
Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1
Artemon [7]

Answer:

\frac{F_1}{F_2} =\frac{1}{2}

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

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F= Kx

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x= spring displacement

Case 1:

F_1=2k\times d

case 2:

F_2=2k\times 2d

therefore, \frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}

\frac{F_1}{F_2} =\frac{1}{2}

4 0
4 years ago
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