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Dima020 [189]
3 years ago
5

What name is used for a high-power variable resistor?

Physics
2 answers:
alexandr1967 [171]3 years ago
6 0
The Answer B:Potentiometer
const2013 [10]3 years ago
3 0
The answer would be C: Rheostat.  :)
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Sound waves travel at the rate of 343 m/s at 20°C. If a man standing 450 meters away from the wall of a canyon yells, “Hello,” h
nekit [7.7K]

Answer:

2.62seconds

Explanation:

Speed is defined as the ratio of the distance covered by a body with respect to time.

Speed v = Distance (s)/Time (t)

For a traveling sound wave, the distance between the source of a sound and the reflector is '2S'.

Speed v = 2 × distance (S)/Time (T)

V = 2S/t

2S = vt

Given speed of the wave = 342m/s

Distance covered = 450m

t = 2S/v

t = (2×450)/343

t = 900/343

t = 2.62seconds

It will take him 2.62seconds for him to hear his own voice echo off of the wall.

5 0
3 years ago
Read 2 more answers
A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds , how far does it move during this
krok68 [10]

Answer:

add 44m/s and 22m/s then multiply it by 11

Explanation:

8 0
2 years ago
Which of the following represents an upright image? <br> A. -do<br> B. +m<br> C. -m<br> D. +do
Tom [10]

Answer:

B. +m

Explanation:

The magnification of an image is defined as the ratio between the size of the image and of the object:

m = \frac{y'}{y}

where we have

y' = size of the image

y = size of the object

There are two possible situations:

- When m is positive, y' has same sign as y: this means that the image image is upright

- When m is negative, y' has opposite sign to y: this means that the image is upside down

Therefore, the correct option representing an upright image is

B. +m

5 0
3 years ago
Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C
miskamm [114]
<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

\boxed{q_{B}=+2q}

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential V, which is <em>the energy per unit charge, </em>so the potential V at any point in an electric field with a test charge q_{0} at that point is:

V=\frac{U}{q_{0}}

The potential V due to a single point charge q is:

V=k\frac{q}{r}

Where k is an electric constant, q is value of point charge and r is  the distance from point charge to  where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius r. Before the sphere A and B touches we have:

V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r

When they touches each other the potential is the same, so:

V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}

Therefore:

(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}

So after A and B touch and are separated the charge on sphere B is:

\boxed{q_{B}=+2q}

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

\boxed{q_{C}=+1.5q}

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

q_{A}=q_{B}=+2q

Second:  C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here q_{A}=+2q and C carries no net charge or q_{C}=0. Also, r_{A}=r_{C}=r

V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}

Applying the same concept as the previous problem when sphere touches we have:

k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}

For the principle of conservation of charge:

q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q

Finally, from the third step:

Here q_{B}=+2q \ and \ q_{C}=+q. Also, r_{B}=r_{C}=r

V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}

When sphere touches we have:

k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}

For the principle of conservation of charge:

q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q

So the charge that ends up on sphere C is:

q_{C}=+1.5q

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

+4q

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

+4q

8 0
3 years ago
Astronauts wear liquid cooled space suits to keep their body temperature moderate. One
Anton [14]

Answer:

D. Because they are using space technology on a shirt so people can wear it on earth as well

6 0
2 years ago
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