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dimulka [17.4K]
2 years ago
7

A 1 kg object sits on the earth’s surface. What is the force of gravity between the object and the earth? (mass of the earth = 5

.97 x 10 24kg, radius of the earth = 6.37 x 10 6m)
Physics
1 answer:
snow_lady [41]2 years ago
4 0

Answer:

9.81N

Explanation:

the force of attraction is given by

F=<u>GmM</u><u>/</u><u>R²</u><u> </u>

where m is mass of the body

M is mass of the earth

R is radius of the earth

G is the universal gravitational constant(6.67x10-¹¹)

hence we substitute the values in the formula.

<em> </em><em>you</em><em> </em><em>can</em><em> </em><em>ask</em><em> </em><em>questions</em>

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Determine if the data are qualitative or quantitative.
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Qualitative, Quantitative, Qualitative, Quantitative, and Qualitative.

Explanation:

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hodyreva [135]

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Zero

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W = F × s

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3 years ago
A solid weighs 16.5N on the surface of the moon. The force of gravity on the moon is 1.7N/Kg.
Fiesta28 [93]

Answer:

mass = 9.7 kg

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Physics help please!
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Answer:

student attach a save block to a horizontal spring so that the block spring system will oscillator with the block spring system released from rest horizontal position that is not the systems equilibrium position well this question regards about the energy used the answer may be 0.73 Joel ok you just try it ok verified

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6 0
3 years ago
The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta y = 434.791\,m

\Delta y = 0.435\,km

Lastly, the range of the bullet is 0.435 kilometers.

3 0
3 years ago
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