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natulia [17]
2 years ago
14

How many slices of bread did each climber have to eat to compensate for the increase of the gravitational potential energy of th

e system climbers-Earth? (One piece of bread releases about 1.0×106J of energy in the body.) Assume that the height of Mt. Everest is 8850 m, the weight of a person is about 79 kg , and efficiency in converting the body's chemical energy into mechanical work is 10.5 %.
Physics
1 answer:
N76 [4]2 years ago
3 0

Answer:

So No of slices to be consumed by each person = n = 65

Explanation:

Energy released by one slice = E1

E1=10^6\ J

h = 8850 m ; m = 79 kg ,η= 10.5%

We know that potential energy given as

u = m g h

u = 79 x 9.81 x 8850

u=6.8\times 10^6\ J

we know from the defination of efficiency that,  η= E(out)/E(in)

Now amount of PE has to be compensated, In our case, E(out) =u

0.105=\dfrac{E(out)}{E(in)}

0.105=\dfrac{6.8\times 10^6}{E(in)}

E(in)=64.76\times 10^6\ J

Let n be the number of bread slices to be consumed.

n = E(in)/E1

n=\dfrac{64.76\times 10^6}{10^6}

n=64.76

So No of slices to be consumed by each person = n = 65

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Answer:

mg=200.4 N.

Explanation:

This problem can be solved using Newton's law of universal gravitation: F=G\frac{m_{1}m_{2}}{r^{2}},

where F is the gravitational force between two masses m_{1} and m_{2}, r is the distance between the masses (their center of mass), and G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2}) is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting w_{s} be the weight on the surface:

w_{s}=mg,

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since we now that g=9.8m/s^{2} we get that the mass is

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Now we can use Newton's law of universal gravitation

F=G\frac{Mm}{r^{2}},  

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F=ma,

in this case the acceleration is the gravity so

F=mg, (<u>becarefull, gravity at this point is no longer</u> 9.8m/s^{2} <u>because we are not in the surface anymore</u>)

and this get us to

mg=G\frac{Mm}{r^{2}}, where mg is his new weight.

We need to remember that the mass of the earth is M=5.972*10^{24}kg and its radius is 6.37*10^{6}m.

The total distance between the astronaut and the earth is

r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6} meters.

Now we can compute his weigh:

mg=G\frac{Mm}{r^{2}},

mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}},

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Energy is 33750 Juoles,  v = 30m/s

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2 years ago
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Ancient Greek philosophers spent lots of time thinking about science and imaging explanations for the natural world. What part o
Illusion [34]

Answer:

Testing

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2 years ago
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Answer:

Explanation:

Given that:

For a first order instrument with a sensitivity of .4 mV/K

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y(t) = 109.2  + (189.2 - 109.2)( 1 - \mathbf{e^{-t/c}})mV

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The plot of y(t) as a function of time can be seen in the diagram  attached below.

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The unit for y(t) is mV.

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0.7635  mV = ( 1 - \mathbf{e^{t/25\times 10^{-3}}})) mV

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t = 0.036 s

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The error fraction = 0.1

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2 years ago
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