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Anton [14]
2 years ago
14

According to the universal law of gravitation, the force due to gravity is b. directly proportional to the square of the distanc

e between objects. d. not dependent on the distance between objects. c. inversely proportional to the distance between objects. a. inversely proportional to the square of the distance between objects. e. directly proportional to the distance between objects.
Physics
1 answer:
Nataly [62]2 years ago
7 0

According to the universal law of gravitation, the force due to gravity is a. inversely proportional to the square of the distance between objects.

<h3>What is the universal law of gravitation?</h3>

Newton's law of universal gravitation states that objects are attracted to each other in the universe with a force directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

The equation that represents this law is;

F = G × M1M2 ÷ r∧2

Where,

F is the gravitational force acting between two objects

M1 and M2 are the masses of the objects

r is the distance between the centers of their masses

G is the gravitational constant.

Thus. according to the universal law of gravitation, the force of gravity is inversely proportional to the square of the distance between objects.

Learn more on universal law of gravitation here:

brainly.com/question/9373839

#SPJ1

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Answer:

A., B., and C.

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In contrast, a non-Ohmic material is one that does not obey Ohm's law.

Ohm's law states that the voltage across an electrical object is proportional to the current flowing through it, with the constant of proportionality being Resistance, R (in Ohm's).

The only Non-Ohmic material is the semiconductor, as semiconductors do not obey Ohm's law.

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If a woman weighs 125 lb, her mass expressed in kilograms is x kg, where x is
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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat
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a) T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds V are given by:

V=\omega L=\frac{2\pi}{T} L (1)

Where:

\omega is the angular frequency

L is the radius of the orbit of each satellite

T is the period of the orbit of each satellite

Isolating T:

T=\frac{2 \pi L}{V} (2)

Applying this equation to each satellite:

T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s (3)

T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s (4)

T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s (5)

T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s (6)

T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s (7)

T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s (8)

Ordering this periods from largest to smallest:

T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) Acceleration a is defined as the variation of velocity in time:

a=\frac{V}{T} (9)

Applying this equation to each satellite:

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a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2} (12)

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a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2} (14)

a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2} (15)

Ordering this acceerations from largest to smallest:

a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

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