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r-ruslan [8.4K]
3 years ago
11

The Earth has a mass of 6.0 x 1024 kg, and the Moon has a mass of 7.3 x 1023

Physics
1 answer:
Anna35 [415]3 years ago
6 0

The amount of gravitational force the two bodies exert on one another is 2.0 \times 10^{27} \;Newton.

<u>Given the following data:</u>

  • Mass of Moon = 7.3 \times 10^{23}\; kg
  • Mass of Earth = 6.0 \times 10^{24}\; kg
  • Radius = 3.84 \times 10^5\; kilometers

<u>Scientific data:</u>

  • Gravitational constant = 6.67\times 10^{-11}

To determine the amount of gravitational force the two bodies exert on one another, we would apply Newton's Law of Universal Gravitation:

Mathematically, Newton's Law of Universal Gravitation is given by the formula:

F = G\frac{M_1M_2}{r^2}

<u>Where:</u>

  • F is the gravitational force.
  • G is the gravitational constant.
  • M is the mass of object.
  • r is the distance between centers of the masses.

Substituting the given parameters into the formula, we have;

F = 6.67\times 10^{-11} \times \frac{7.3 \times 10^{23} \; \times \;6.0 \times 10^{24}}{(3.84 \times 10^5)^2}\\\\F = 6.67\times 10^{-11} \times \frac{4.38 \times 10^{48} }{1.48 \times 10^{11} }\\\\F=\frac{2.92 \times 10^{38} }{1.48 \times 10^{11}} \\\\F= 2.0 \times 10^{27} \;Newton

Read more: brainly.com/question/11359658

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How much GPE is stored when an 80kg astronaut climbs to the top of a 5m high lunar lander? The gravity strength on the moon is 1
8_murik_8 [283]

Answer:

The GPE, stored is 640 Joules

Explanation:

The given parameters are;

The given mass of the astronaut, m = 80 kg

The height of the top of the lunar lander to which the astronaut climbs, h = 5 m

The gravity strength on the moon, g = 1.6 N/kg

The Gravitational Potential Energy, GPE, stored is given according to the following equation;

GPE stored = m·g·h

Therefore, by substituting the known values, we have;

GPE Stored = 80 kg × 1.6 N/kg × 5 m = 640 Joules

The GPE, stored = 640 Joules.

6 0
3 years ago
A mass of 0.30kg is attached to a spring and is set into motion with a period of 0.24s. what is the spring constant of the sprin
Alenkinab [10]

Answer:

The spring constant of the spring is 205.42 N/m.

Explanation:

Springs have their own natural "spring constants" that define how stiff they are. The letter k is used for the spring constant, and it has the units N/m.

                                                         k = -F/x

The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.

Given:

mass of object in SHM = m = 0.30 kg

Time period of the spring mass system = 0.24s

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We know that  

6 0
3 years ago
A student calculates the density of a copper cube to be 4.15 g/cm . If the accepted value is 8.64 g/cm the percentage error in h
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 The percentage error in his experimental value is -51.97%.

<h3>What is percentage error?</h3>

This is the ratio of the error to the actual measurement, expressed in percentage.

To calculate the percentage error of the student, we use the formula below.

Formula:

  • Error(%) = (calculated value-accepted value)100/(accepted............. Equation 1

From the question,

Given:

  • Calculated value = 4.15 g/cm
  • accepted value = 8.64 g/cm

Substitute these values into equation 1

  • Error(%) = (4.15-8.64)100/8.64
  • Error(%) = -4.49(100)/8.64
  • Error(%) = -449/8.64
  • Error(%) = -51.97 %

 

Hence, The percentage error in his experimental value is -51.97%.

Learn more percentage error here: brainly.com/question/5493941

8 0
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Answer:

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