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AleksandrR [38]

3 years ago

7

A stationary hockey puck has a mass of 0.27 kg. A hockey player uses her stick to apply a 18 N force over a distance of 0.42 m.

How much work does this force do on the puck?
Group of answer choices

7.6 J

-43 J

-7.6 J

28 J

43 J

1 answer:

jeka57 [31]3 years ago

3 0

W=Fs
W=18(0.42)
W=7.56

Answer: 7.6 J (first option)

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<u>Answer:</u>

<h3>As electric current is carried in a cable, around it, a magnetic field is created. The lines of the magnetic fields form concentric circles around the wire. The direction of the magnetic field hinges on the direction of the current. It can be calculated by pointing the thumb of your right hand in the direction of the moment, using the "right hand law." The position of your curled fingers is in the magnetic field lines. The magnetic field magnitude depends on the sum of current, and the distance from the wire carrying the charge.</h3>

<u></u>

<u>Explanation:</u>

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Note: Refer the image attached below

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Склянка с воздухом закрыта «пробкой» из мыльной пены.
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Выбери правильный вариант ответа.
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Выбери правильный вариант ответа.
Давление не изменится.

и дальнейшее объяснение произошедшего

Может показаться, что воздух в вашей бутылке - просто пустое пространство, но на самом деле это смесь газов. Сила, с которой отдельные молекулы отскакивают от внутренней и внешней стороны бутылки, называется давлением.

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3 years ago

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is<u> positive.</u>

<u></u>

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

<h2><u>Part 1:</u></h2>

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is <u>positive.</u>

<u></u>

<h2><u>Part 2:</u></h2>

<u></u>

<u>Given:</u>

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

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The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

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3 years ago