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tresset_1 [31]
1 year ago
13

A 55 kg track and field athlete has an average power output of 5.4 kW during the 200 meter dash. How quickly did she finish the

race?
Group of answer choices

a. 20 s

b. 22 s

c. 23 s

d. 18 s
Physics
1 answer:
olga_2 [115]1 year ago
5 0

The time taken for the athlete to finish the race is 20 s (Option A)

<h3>What is power? </h3>

Power is simply defined as the rate at which work is done. It can be expressed mathematically as

Power (P) = work (W) / time (t)

But

Work = weight × distance

Therefore,

Power = (weight × distance ) / time

<h3>How to determine the time </h3>
  • Mass (m) = 55 Kg
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Weight = mg = 55 × 9.8 = 539 N
  • Power (P) = 5.4 KW = 5.4 × 1000 = 5400 W
  • Distance (d) = 200 m
  • Time (t) =?

Power = (weight × distance ) / time

5400 = (539 × 200) / t

5400 = 107800 / t

Cross multiply

5400 × t = 107800

Divide both side by 5400

t = 107800 / 5400

t = 20 s

Learn more about power:

brainly.com/question/5684937

#SPJ1

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a blackbody is radiating with a characteristic wavelength of 9 microns what is the blackbody temperature answer in kelvin
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This question involves the concepts of Wein's displacement law and characteristic wavelength.

The blackbody temperature will be "3.22 x 10⁵ k".

<h3>WEIN'S DISPLACEMENT LAW</h3>

According to Wein's displacement law,

\lambda_{max} T = c\\\\T=\frac{c}{\lambda_{max}}

where,

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Answer:

  • Current = 0.33 A

Explanation:

  • For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

\dashrightarrow \: \:  \sf V = 2 \times 5 = 10 V

Three resistor of 5\Omega, 10\Omega, 15\Omega are connected in Series, so the net resistance:

\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}

\dashrightarrow \: \:  \sf R = 5 + 10 + 15

{ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \:  \Omega}}}}

According to ohm's law:

\dashrightarrow  \sf\: \: V = IR

\dashrightarrow  \sf \: \: I = \dfrac{V}{R}

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 {\pmb{\sf{\Omega}}} we get:

\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}

{ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}

\thereforeThe electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>

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