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Lina20 [59]
3 years ago
6

True or False: The more mass an object has, the faster it will fall.

Physics
2 answers:
Elza [17]3 years ago
8 0

Answer:

True

Explanation:

solong [7]3 years ago
6 0

Answer:yes yes yes yes YES!

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The weight of a body of certain mass becomes zero in space.why?write with reasons​
Anvisha [2.4K]

Answer:

Weight is what you get when a certain amount of gravity is acting on that mass, and something, like the surface of a planet, is resisting that action. In space, when falling freely, there's nothing resisting the pull of gravity so weight disappears. Mass however stays.

hope this helps u

Explanation:

7 0
3 years ago
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
Which best describes the way a sound wave is sent trough the radio?
horsena [70]
Sound wave > electric signal > radio wave > sound wave
4 0
3 years ago
Read 2 more answers
How many 1140 nm long molecules would you have to line up end to end to stretch a distance of 158 miles?
dezoksy [38]

Answer:

221754385964.9123

Explanation:

Convert miles to nanometer

1 mile = 1.6 km

1 km = 1×10³×10³×10³×10³ nm

1 mile = 1.6×10¹² nm

So,

158 miles = 158×1.6×10¹² = 252.8×10¹² nm

Length of each molecule = 1140 nm

Number of molecules = Total length / Length of each molecule

\text{Number of molecules}=\frac{252.8\times 10^{12}}{1140}\\\Rightarrow \text{Number of molecules}=221754385964.9123

There are 221754385964.9123 number of molecules in a stretch of 158 miles

3 0
3 years ago
According to Newton’s law of universal gravitation, which statement is true?
SIZIF [17.4K]

Option (a) is correct.

Falling objects accelerate as they approach the ground.This is because of the force of gravity acting on the falling objects. so the velocity of these objects increases continuously as they approach the ground. the acceleration acting on the falling objects is a constant ( close to the surface of earth) and is called as acceleration due to gravity denoted by g. value of g=9.8 m/s².

4 0
3 years ago
Read 2 more answers
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