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1 year ago

Question 9 of 27

Physics

1 answer:

Ksju [112]1 year ago

6 0

The power of man performing 500 J of work in 8 seconds is 62.5 J/s.

Power can be defined as the pace at which work is completed in a given amount of time.

Horsepower is sometimes used to describe the power of motor vehicles and other machinery.

The pace at which work is done on an item is defined as its power. Power is a temporal quantity.

Which is connected to how quickly a project is completed.

The power formula is shown below.

Power = Energy / Time

Power = E / T

Because the standard metric unit for labour is the Joule and the standard metric unit for time is the second, the standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W.

Here we have given Energy as 500 J and Time as 8 second.

Power = Energy / Time

Power = 500 / 8 Joule / sec

Power = 250 / 4 Joule / sec

Power = 125 / 2 Joule / sec

Power = 62.5 Joule / sec or 62.5 watt

Power came out to be 62.5 J/s when the man performed 500 Joule of work in 8 seconds.

So we can conclude that the power in the Energy transmitted per unit of time, and can be find out by dividing Energy by time. In our case the Power came out to be 62.5 Joule / Second.

Learn more about Power here:

brainly.com/question/1634438

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A rock is thrown horizontally from a building at 15 m/s. It hits the ground 45 m from the base of the building. How high was the

evablogger [386]

Answer:

Explanation:

hope this helps!

7 0

2 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$