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rewona [7]
4 years ago
11

Two objects attract each other gravitationally with a force of 2.5 x 10^-10N when they are 0.25 m apart. Their total mass is 4.0

0 kg. Find their individual masses.
Physics
1 answer:
In-s [12.5K]4 years ago
6 0

Answer:

M = 3.9406 kg and m = 0.0594 kg

Explanation:

The gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically it is expressed as follows:

Fg = (G×M×m)/r²    Formula (1)

Where:

Fg is the gravitational force (N)

G is the universal gravitation constant, G = 6.67 × 10⁻¹¹ (N×m²)/kg²

M and m are the masses of the bodies that interact  (kg).

r is the distance that separates them (m).

Known Data

Fg = 2.5 × 10⁻¹⁰ N

r = 0.25 m

G = 6.67 × 10⁻¹¹ (N×m²)/kg²

Problem development

We propose 2 equations

M + m = 4kg

M = 4 - m equation (1)

We replace in formula (1)

2.5 × 10⁻¹⁰ = (6.67 × 10⁻¹¹ × M × m)/(0.25)²

2.5 × 10⁻¹⁰ × (0.25)² = (6.67 × 10⁻¹¹ × M × m)

(2.5 × 10⁻¹⁰ × (0.25)²)/(6.67 × 10⁻¹¹) = M × m

M × m = 0.234 equation (2)

We replace M = 4 - m in equation (2)

(4 - m) × m = 0.234

4m - m² = 0.234

m² - 4m + 0.234 = 0 (quadratic equation)

We apply the formula for the quadratic equation and obtain 2 values for m that meet the conditions:

m = 3.9406 kg or m =  0.0594 kg

We replace m in equation (1)

M = 4 - 3.9406 = 0.0594 kg or M = 4 - 0.0594 = 3.9406

To meet the condition that M + m must give 4 kg, one mass must be equal 3.9406 and the other must equal 0.0594, then:

M = 3.9406 kg and m = 0.0594 kg

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An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the ce
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h = \dfrac{\omega^2 r^2}{2g} + C

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replacing the value of h in equation (2); we have:

V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr

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Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then V_f  =  V_i

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Finally, the angular velocity of the tank at which the bottom of the tank is exposed  = 10.48 rad/s

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