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1 year ago

A process generates 250 watts of useful energy and 600 watts of waste energy. WhT is the efficiency of the process?

Physics

1 answer:

Cloud [144]1 year ago

3 0

The efficiency of the process is 42%

From the question given above, the following data were obtained:

Useful energy = 250 watt

Wasteful energy = 600 watt

<h3>Efficiency =? </h3>

The efficiency of the process can be obtained as illustrated below:

$efficiency \: = \frac{useful \: energy}{wasteful \: energy} \times 100 \\ \\ efficiency \: = \frac{250}{600} \times 100 \\$

<h3>Efficiency = 42%</h3>

Therefore, the efficiency of the process is 42%

Learn more: brainly.com/question/2300041

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A child uses a rubber band to launch a bottle cap at an angle of 37.0° above the horizontal. The cap travels a horizontal distan

zavuch27 [327]

Answer:

Initial velocity will be 1.356 m/sec

Explanation:

Let the initial speed = u

Angle at which rubber band is launched = 37°

Horizontal component of initial velocity $u_x=ucos\Theta =ucos37^{\circ}=0.7986u$

Time is given as t = 1.20 sec

Distance in horizontal direction = 1.30 m

We know that distance = speed × time

So time $t=\frac{distance}{speed}$

$1.20=\frac{1.3}{0.7986u}$

$u=1.356m/sec$

So initial velocity will be 1.356 m/sec

3 0

2 years ago

50POINTS! Find the orbital speed of a satellite in a circular orbit 1700km above the surface of the Earth. M_earth 5.97e24kg, r_

ivolga24 [154]

<h2>Answer: 7020.117 m/s</h2>

Explanation:

The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:

$V=\sqrt{G\frac{M}{R}}$ (1)

Where:

$G=6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}}$ is the gravity constant

$M_{Earth}=5.97{10}^{24}kg$ the mass of the massive body around which the satellite is orbiting, in this case, the Earth .

$R=r_{Earth}+h=8080000m$ the radius of the orbit (measured from the center of the planet to the satellite).

This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth $r_{Earth}$ and the altitude of the satellite above the Earth's surface $h$ .

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

$V=\sqrt{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})\frac{5.97{10}^{24}kg}{8080000m}}$