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4 years ago

A young child hold a string attached to a balloon. What is the reaction force to the balloon pulling up on the earth?

Physics

1 answer:

Alina [70]4 years ago

3 0

Answer:

As the Ballon pulls up, the distance between the Ballon and the center of the earth increases, the gravitational pull reduces and the gravity potential energy increases.

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Juan was wearing a bright red shirt in a very dark room. What color did his shirt appear to the people with him in the room? A)

ikadub [295]

It would appear black.

Hope I helped.

8 0

3 years ago

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How fast would you have to jump off the ground in order to jump at least 10 meters in the air?

ivanzaharov [21]

Answer:

30.5

Explanation:

because you will basically have to be like flash

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4 years ago

Which of the following are vector quantities?

pshichka [43]

Answer:

Vector Quantity: A physical quantity is said to be a vector quantity when it has both magnitude and direction. The scalar quantities are distance, mass, time, volume, density, speed, temperature, and energy, The vector quantities are weight, velocity, acceleration, and force.

Explanation:

Mark me brainleist PLZZZZ

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3 years ago

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A mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 30. cm3. If the combustion of this mixtu

Y_Kistochka [10]

Answer:

Explanation:

Given

Original volume V1=30cm^3 converting to L

=30/1000=0.03L

Constant pressure P= 648 tors

Converting to atm; 648 tors*1atm/760 torr=0.853 atm

Work=984J= 984**1L/101.33=9.7L.atm

Note before

W= -P(Vfinal-Vinitial)

-9.7/0.853+0.03L=11.68L

5 0

4 years ago

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A high voltage transmission line with a resistance of 0.51 Ω/km carries a current of 1099 A. The line is at a potential of 1300

Illusion [34]

<h2>Answer:</h2>

$\boxed{P=96.09MW}$

<h2>Explanation:</h2>

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

$R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega$

So we can calculate the power loss as:

$P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}$

Finally, the power loss due to resistance in the line is 96.09MW

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4 years ago