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3 years ago

What does the power source in a circuit provide?

Physics

1 answer:

Serjik [45]3 years ago

7 0

Explanation:

The primary function of a power supply is to convert electric current from a source to the correct voltage, current, and frequency to power the load. As a result, power supplies are sometimes referred to as electric power converters. Some power supplies are separate standalone pieces of equipment, while others are built into the load appliances that they power. Examples of the latter include power supplies found in desktop computers and consumer electronics devices. Other functions that power supplies may perform include limiting the current drawn by the load to safe levels, shutting off the current in the event of an electrical fault, power conditioning to prevent electronic noise or voltage surges on the input from reaching the load, power-factor correction, and storing energy so it can continue to power the load in the event of a temporary interruption in the source power (uninterruptible power supply).

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How are domains arranged in materials that are magnetic and in ones that are not

Valentin [98]

Answer:

In most materials, atoms are arranged in such a way that the magnetic orientation of one electron cancels out the orientation of another

8 0

3 years ago

A satellite with a mass of 110 kg and a kinetic energy of 3.0 ´ 109 J must be moving at a speed of

djverab [1.8K]

Answer: A satellite with a mass of 110 kg and a kinetic energy of 3.08×10^9 J must be moving at a speed of 7483 m/s.

Explanation: To find the answer we need to know about the kinetic energy of a body.

<h3>How to solve the problem the equation of kinetic energy?</h3>

We have the expression for kinetic energy of a body as,

$KE=\frac{1}{2}mv^2$

Given that,

$m=110kg\\KE=3.08*10^9J\\$

We have to find the speed of the satellite,

$v=\sqrt{\frac{2KE}{m} } =\sqrt{\frac{2*3.08*10^9}{110} } =7.483*10^3 m/s$

Thus, we can conclude that, the velocity of the satellite will be 7438m/s.

Learn more about Kinetic energy here:

brainly.com/question/28105739

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8 0

2 years ago

Read 2 more answers

"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and

spayn [35]

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

8 0

3 years ago

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?

IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as $m_{1} \ and\ m_{2}\ respectively$

$Let\ m_{1} =1 kg\ and\ m_{2} =2 kg$

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

$m_{1} g-T=m_{1} a$ [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction $[\mu]=0.13$

$Hence\ the\ frictional\ force\ F=\mu N$

$F=\mu m_{2} g$

Hence the net force acting on the body having mass 2 kg-

$T-\mu m_{2} g=m_{2} a$ [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

$m_{1} g-T=m_{1}a$ [1]

$T-\mu m_{2}g= m_{2} a$ [2]

---------------------------------------------------

$[m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a$

$a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g$

$a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2$

$a=\frac{0.74}{3} *9.8\ m/s^2$

$a=2.417 m/s$ [ans]

6 0

3 years ago

Read 2 more answers

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi

almond37 [142]

Answer:

A. Using

Pgauge= Pmanometer

And we know that

Pgauge= deta(hpg)

So deta h = Pgauge/density x g

= 7.5(6894.76/1psig)/ 13.6*10^3*9.8m/s²)

= 387.9mm

So to find height of pipe connected to the pipe we say

= h -deta h

= 900-387.97mm

=512.02mm

B. We use manometry principle

Pgas+density xg(25*10^3)-PX density{h-(H-0.25)=0

Finally Pgas= 6.54psig

3 0

4 years ago