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Likurg_2 [28]
3 years ago
9

Find the magnitude of the gravitational force a 63.5 kg person would experience while standing on the surface of Earth with a ma

ss of 5.98 × 1024 kg and a radius of 6.37 × 106 m. The universal gravitational constant is 6.673 × 10−11 N · m2 /kg2 . Answer in units of N
Physics
1 answer:
Anastaziya [24]3 years ago
3 0

Answer: 3976N

Explanation:

Using the formula for calculating gravitational force between two masses, we have

F = GMm/r^2

Where G is the gravitational constant

M and m are the masses

r is the distance between the masses

F= 6.673 × 10-¹¹ × 5.98 × 10²⁴ × 63.5/ (6.37 × 10^6)^2

F= 2.533×10^16/6.37×10^12

F= 0.3976×10⁴N

F= 3976N

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The of the sound wave is given by;

I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}

Where;

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P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535

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