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Soloha48 [4]
3 years ago
13

What is the function of a resistor in a circuit? How does it affect the amount of charge that flows? How does it affect the rate

at which charge flows? How does it affect the initial and final voltage across the capacitor?
Physics
1 answer:
ipn [44]3 years ago
7 0

Answer:

Explanation:

The function of resistance in an electronic circuit is that it helps to limit the current in the circuit, so as to prevent it from high currents.

The resistor helps to limit the charge flowing through the circuit and also limits the flow rate of the charge.

The resistor opposes the flow of current through the circuit.

The initial and final voltage across the capacitor decreases.

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Un the way to the moon, the Apollo astro-
kherson [118]

Answer:

Distance =  345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2}  } \\a=3.33*10^{19} [m/s^2]

6 0
3 years ago
Gasoline burns inside a car’s engine. how does this fuel enable a car to move?
Svetllana [295]
<span> The purpose of a gasoline car engine is to convert gasoline into motion so that your car can move. Currently the easiest way to create motion from gasoline is to burn the gasoline inside an engine.
Therefore, a car engine is an internal combustion engine -- combustion takes place internally.
There is such a thing as an external combustion engine. A steam engine in old-fashioned trains and steam boats is the best example of an external combustion engine. The fuel (coal, wood, oil, whatever) in a steam engine burns outside the engine to create steam, and the steam creates motion inside the engine. Internal combustion is a lot more efficient (takes less fuel per mile) than external combustion, plus an internal combustion engine is a lot smaller than an equivalent external combustion engine. This explains why we don't see any cars using steam engines.

To understand the basic idea behind how a reciprocating internal combustion engine works, it is helpful to have a good mental image of how "internal combustion" works.

One good example is an old Revolutionary War cannon. You have probably seen these in movies, where the soldiers load the cannon with gun powder and a cannon ball and light it. That is internal combustion, but it is hard to imagine that having anything to do with engines.

A potato cannon uses the basic principle behind any reciprocating internal combustion engine: If you put a tiny amount of high-energy fuel (like gasoline) in a small, enclosed space and ignite it, an incredible amount of energy is released in the form of expanding gas. You can use that energy to propel a potato 500 feet. In this case, the energy is translated into potato motion. You can also use it for more interesting purposes. For example, if you can create a cycle that allows you to set off explosions like this hundreds of times per minute, and if you can harness that energy in a useful way, what you have is the core of a car engine! </span>
8 0
3 years ago
What combination of variables will make an experiment the most reliable
Virty [35]

Answer:

Explanation:

A combination of two controlled variables will make an experiment the most reliable.

Variables are the values that we take under any circumstance while doing an experiment so that we can keep on changing  and get new results at the end.

It is important to have them in pair so that two values can be kept on changing in terms of any constant condition. This will help to get better results in over all experiment data.

4 0
3 years ago
Is it possible for a population with a high birth rate to decrease in size.
matrenka [14]
Sure, if the mortality (death) rate is even higher than the birth rate.
7 0
3 years ago
Calculate the strength and direction of the electric field 300 mm to the right of a -200 C electric
Norma-Jean [14]

Answer: -1.91*10^13 v/m

Explanation:

6 0
3 years ago
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