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coldgirl [10]
3 years ago
11

In a circuit, a variable resistor is to be connected across a potential difference of 60 volts to allow the current to be adjust

ed in the range from 40 milliamperes to 100 milliamperes.
A variable resistor is a device in which the resistance can be varied.
mA = milliamperes, or 10-3 amperes
Over what range must the resistance vary?

plz help :')
Physics
1 answer:
sweet [91]3 years ago
4 0

Answer:

600 and 1500 [ohm

Explanation:

To solve this problem we must use ohm's law, which tells us that the voltage is the product of the current by the resistance, so we have:

V = I*R

where:

V = voltage [V]

I = current [amp]

R = resistance [ohm]

<u>Therefore:</u>

R = V/I

R1 = 60/(40*10^-3) = 1500 [ohm]

R2 = 60/(100*10^-3) = 600 [ohm]

So the resistance should be among 600 and 1500 [ohm]

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When talking about the speeds of atoms in a gas you must refer to the average speed because
cluponka [151]

Answer:

An Atom's individual speed will change as it collides with other atoms, so we have to use an average.

Explanation:

In a gas a single atoms does an assortment of things during its time in the gas—sometimes it collides with an other atom gaining a lot of speed, sometimes losing a lot of speed in the collision, and sometimes just moving freely. Therefore: the motion of one individual atom is unpredictable, and it cannot be  representative of all the the atoms in a gas, which is why we must average over all speeds of all atoms to find an average speed that allows us to calculate other quantities like temperature and pressure of the gas.

Hence, the second option <em>"an Atom's individual speed will change as it collides with other atoms, so we have to use an average" </em>stands correct.

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3 years ago
Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th
agasfer [191]

Answer:

T_1=T_3=\dfrac{2\pi}{21}

T_2=T_4=\dfrac{2\pi}{42}

Explanation:

Wave 1, y_1=0.12\ cos(3x-21t)

Wave 2, y_2=0.15\ sin(6x+42t)

Wave 3, y_3=0.13\ cos(6x+21t)

Wave 4, y_4=-0.27\ sin(3x-42t)

The general equation of travelling wave is given by :

y=A\ cos(kx\pm \omega t)

The value of \omega will remain the same if we take phase difference into account.

For first wave,

\omega_1=21

\dfrac{2\pi }{T_1}=21

T_1=\dfrac{2\pi}{21}

For second wave,

\omega_2=42

\dfrac{2\pi }{T_2}=42

T_2=\dfrac{2\pi}{42}

For the third wave,

\omega_3=21

\dfrac{2\pi }{T_3}=21

T_3=\dfrac{2\pi}{21}

For the fourth wave,

\omega_4=42

\dfrac{2\pi }{T_4}=42

T_4=\dfrac{2\pi}{42}

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

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3 years ago
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