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Arisa [49]
1 year ago
14

If a runner exerts 457 j of work to make 321 w of power then how long did it take the runner to do the work

Physics
1 answer:
MA_775_DIABLO [31]1 year ago
5 0

The time taken by the runner to do the work is 1.42 seconds.

Given the data in the question;

  • Work done; W = 457J
  • Power; P = 321W
  • Time elapsed; t = \ ?

<h3>Power</h3>

Power can be simply referred to as the quantity of energy transferred per unit time.

It is expressed as;

P = \frac{W}{t}

Where W is work done and t is time elapsed.

To determine the time it took the runner to do the work, we substitute our given values into the expression above.

P = \frac{W}{t} \\\\t = \frac{W}{P} \\\\t = \frac{457J}{321W} \\\\t = \frac{457 kgm^2/s^2}{321 kgm^2/s^3}\\\\t = 1.42s

Therefore, the time taken by the runner to do the work is 1.42 seconds.

Learn more about Power and Work: brainly.com/question/2962104

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What is the potential difference when the current in a circuit is 5mA and resistance is 30 Ohms
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<h2>\bf{ \underline{Given:- }}</h2>

\sf• \: The \:  current \:  in \:  a \:  circuit \:  is  \: 5 \: amps.  \: and  \: resistance \:  is \:  30 \:  Ohms.

\\

<h2>\bf{ \underline{To \:  Find :- }}</h2>

\sf{• \:  The  \: Potential  \: Difference. }

\\

\huge\bf{ \underline{ Solution:- }}

\sf According  \: to  \: the  \: question,

\sf•  \: Current \:  (I) = 5  \: Amps.

\sf• \:  Resistance  \: (R) = 30 \:  Ω

\sf{Potential \:  difference  \: means  \: Voltage \: ( V).}

\sf{We \:  know \: that, }

\bf \red{ \bigstar{ \: V = IR }}

\rightarrow \sf V =5 \times 30

\rightarrow \sf V =150

\\

\sf \purple{Therefore, \:  the \:  potential  \: difference  \: is  \: 150  \: v \: .}

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3 years ago
Which of the following paraphrases Hubble Law?Select one:A. The greater the distance to a galaxy, the greater its redshift. B. T
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The correct answer is:

~A. The greater the distance to a galaxy, the greater its redshift.

Hope this helps!!!

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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
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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.

Weight of the balloon, \rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.

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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