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3 years ago

Power output of an engine is with work (w=60,000) during 60 s Calculate the power of this work?*

Physics

1 answer:

igomit [66]3 years ago

6 0

Answer:

The power output of the engine is 1,000 Watt

Explanation:

Mechanical Power

Power is the amount of energy transferred or converted per unit of time. In the SI, the unit of power is the watt, equal to one joule per second.

The power can be calculated as:

$\displaystyle P=\frac {W}{t}$

Where W is the work and t is the time.

The engine does a work of W=60,000 J during t=60 s. Calculate the power:

$\displaystyle P=\frac {60,000}{60}$

P = 1,000 Watt

The power output of the engine is 1,000 Watt

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a man is standing on the edge of a 20.0m high cliff. he throws a rock vertically with an initial velocity of 10.0m/s. how high d

marin [14]

Answer:5m

Explanation:a=-10m/s (since it moved upwards against gravity)

V=0

U=10m/s

v2=u2+2as

0= 100-20s

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4 years ago

A heat engine is 20% efficient. If it loses 800 J to the cooling system and exhaust during each cycle, the work done by the engi

Maurinko [17]

Answer: A. 200J

Therefore, the workdone by the heat engine is 200J

Explanation:

Given ;

The efficiency of the heat engine is E = 20% = 0.2

Heat loss L= 800J

For an heat engine the efficiency is measured by the amount of workdone by the heat engine when compared to the heat generated.

Efficiency E = workdone/heat generated × 100%

Heat generated G= workdone W + heat loss L

G = W + L

According to the question.

W = 20% of G

W = 0.2G ......1

L = 80% of G

L = 0.8G

G = L/0.8 ......2

Substituting equation 2 to 1

W = 0.2(L/0.8)

And L = 800J

W = 0.2(800/0.8)

W = 200J

Therefore, the workdone by the heat engine is 200J

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3 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

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0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

Explanation:

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

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