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3 years ago

Astronauts on the Moon can jump higher than if they were on Earth. This is because...

2 answers:

4 0

This is because of Gravity. Your muscles or your entire body is at ease or whatever because the moon has more mass than earth.

AnnZ [28]3 years ago

3 0

The Moon has less mass than the Earth, which means it has less gravity than the Earth as well. (Also, there is WAY less atmosphere—as in, pretty much none—on the Moon than we're used to having press down on us constantly.)

Since the gravity isn't pulling on us as strongly on the moon as it is on Earth, we can jump higher and further with more ease. That's because our muscles (and all other body parts involved, like our bones and joints and heart and lungs) all were developed and conditioned to work in the stronger gravity on Earth's surface.

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Which of the following are peninsulas in Southeast Asia?

xxTIMURxx [149]

Answer:

Explanation:

5 0

3 years ago

The orbital period of planet Venus is 0.62 years. What is its distance from the sun?

Vikki [24]

Kepler's third law shows the relationship between the orbital period of an object and the distance between the object and the object it orbits.

The simplified version of this law is: P^2 = a^3

Where,

P = period of the orbit in years = 0.62 years

a = average distance from the object to the object it orbits in AU. The astronomical unit AU is a unit of length which is roughly equivalent to the distance from Earth to the Sun.

Therefore calculating for a:

0.62 ^ 2 = a ^ 3

a = 0.62 ^ (2/3)

a = 0.727 AU = 0.72 AU

Therefore we can interpret this as: The distance from Venus to the Sun is about 72% of the distance from Earth to Sun.

Answer: B. 0.72 AU

5 0

3 years ago

What is the mass of 4 moles of helium, He?

aniked [119]

I am pretty sure it is 4.002602

4 0

3 years ago

How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of

alina1380 [7]

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

$E = k \frac{q}{r^{2} }$

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

$E = k \frac{q}{r^{2} }$

q = E x r²

q = 1250 N/C x 0.205m²

8.99 x 10⁹ Nm²/C²

q = 5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

$n = \frac{q}{e}$

n = 5.84 x 10⁻⁹ C

1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

3 0

3 years ago

Question 6 How many molecules are in 4.48 liters of CO 2?

Dovator [93]

22.4 molecules are in 4.48 liters of CO 2

8 0

3 years ago