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Bingel [31]
3 years ago
13

How many grams of butter, which has a usable energy content

Physics
1 answer:
VARVARA [1.3K]3 years ago
4 0

Answer:

m_{butter} = 156.93 grams

Explanation:

Given data;

energy content  = 6.0 Cal/g = 6000 cal/g

mass of man 60 kg

climb height = 6.70 km

g = 9.80 m/s^2

we know that

potential energy = mgh

E_t = mU

wehre U - usable energy constant

equating both energy

60 \times 9.8 \times 6.70\times 10^3 = m_{butter} 6000\times \frac{4.184 j}{cal}

m_{butter} = 156.93 grams

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Answer:

yes

Explanation:

Nuclear fusion is a reaction in which two or more atomic nuclei are combined to form one or more different atomic nuclei and subatomic particles.

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The attraction between earth and the moon is an example of ________ force. the attraction between earth and the moon is an examp
Korolek [52]

The answer is gravitational force. The gravitational force between the earth and the moon is the similar as between any other two masses in space.

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2 years ago
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Write a rule for the sequence. 3, -3, -9, -15. A. Start with 3 and add -6 repeatedly B. Start with -6 and add 3 repeatedly C. St
faust18 [17]

Substract two consecutive terms of the sequence to see if there is a common difference:

\begin{gathered} (-3)-(3)=-3-3=-6 \\ (-9)-(-3)=-9+3=-6 \\ (-15)-(-9)=-15+9=-6 \end{gathered}

As we can see, there is a common difference of -6.

Then, if a number of the sequence is given, the next one can be found by adding -6 (which is the same as subtracting 6).

Notice that the first term of the sequence is 3.

Then, the rule for the sequence is to start with 3 and add -6 repeatedly.

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7 0
1 year ago
The CERN particle accelerator is circular with a circumference of 7.0 km.
Contact [7]

Answer:

a_c=2.0196\times 10^{13}\ m/s^2

F=3.37273\times 10^{-14}\ N

Explanation:

m = Mass of proton = 1.67\times 10^{-27}\ kg

v = Speed of proton = 0.5c = 0.5\times 3\times 10^8=1.5\times 10^8\ m/s

Circumference of the colider is 7 km

P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m

a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2

Centripetal acceleration is 2.0196\times 10^{13}\ m/s^2

F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N

Force on protons is 3.37273\times 10^{-14}\ N

8 0
3 years ago
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