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mr_godi [17]
3 years ago
10

A heater gives off heat at a rate of 330 kj/min. what is the rate of heat output in kilocalories per hour? (1 cal 4.184 j)

Physics
1 answer:
Cerrena [4.2K]3 years ago
4 0
Answer: 4732.3135 kcal/hour

To answer this question you need to convert the kj into kcalorie then convert the min into hour. Becareful because 1 kilocalories = 1000 calories = 4.184 kjoule.
The calculation would be: 
330kj/min x (1kilocalorie/4.184kj) x 60 min/hour= 4732.3135 kcal/hour
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Answer:

Explanation:

change in flux = no of turns x area of loop x change in magnetic field

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= 3.9 x 10⁻³ / .10

= 39 x 10⁻³ V

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The Earth orbits the Sun at a speed of 30 km/s. At that speed it completes one path around the Sun every year. Of course, as tha
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Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = \frac{t}{\sqrt{1-\frac{v^2}{c^2} } }

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 20

t₁= 33.4   years

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