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Alina [70]
4 years ago
15

Two 2.5 kg physical science textbooks on a bookshelf are 0.14 m apart. What is the magnitude of the gravitational attraction bet

ween the books?
Physics
2 answers:
finlep [7]4 years ago
5 0

Answer:

Explanation:

According to Newton's laws of gravitation force of attraction is given by

F = GMm/R²

G = 6.67 X 10⁻¹¹

,M = 2.5 Kg ,

m = 2.5 Kg ,

R = .14 m

=

F = 6.67 X 10⁻¹¹ X 2.5 X 2.5 / 0.14²

= 2126.9 X 10 ⁻¹¹ N.

steposvetlana [31]4 years ago
5 0

Answer:

2.13 x 10^-8 N

Explanation:

m1 = 2.5 kg

m2 = 2.5 kg

r = 0.14 m

According to the Newton's law of gravitation, the force of attraction between them is given by

F = G m1 x m2 / r^2

Where, G is the Universal Gravitational constant and the value of G

is 6.67 x 10^-11 Nm^2/kg^2

So, the force between the text books is

F = (6.67 x 10^-11 x 2.5 x 2.5) / (0.14 x 0.14)

F = 2.13 x 10^-8 N

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<u>Answer:</u> The number of photons are 3.7\times 10^8

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m = mass of water = 165 g

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\Delta T = change in temperature = T_2-T_1=100^oC-23^oC=77^oC

Putting values in above equation, we get:

q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J

This energy is the amount of energy for 'n' number of photons.

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Putting values in above equation, we get:

n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}

Hence, the number of photons are 3.7\times 10^8

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