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3 years ago

Which factor determines the strength of gravity between Earth and the sun?

Geography

1 answer:

cluponka [151]3 years ago

5 0

Answer:

(2)the distance between Earth and the sun

Explanation:

if the earth is closer to the sun we will have more gravity but if we are far away from the sun we will experience less gravity

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He measures the amount of radioactive uranium in the sample and compares it to the amount of radioactive lead. He then compares

soldi70 [24.7K]

Answer:

C. its age

Explanation:

Rex is most likely trying to figure out the age of the rock formation. It is usually done by comparing the isotopic ratios of the daughter to the parent and their decay rates.

This is known as radiometric dating.
Here an exact numerical age of the rock is usually obtained by measuring these isotopic ratio.
The rate of decay serves as a clock to measure how far the radioactive material has decayed in time

8 0

4 years ago

Which continent do the westerly winds<br> travel across before reaching Russia?

dem82 [27]

Answer:

asia my friend

Explanation:

or latin America

4 0

4 years ago

Select the correct answer.

Trava [24]

I think the answer is B.

8 0

2 years ago

. If the fastest passenger aircraft can fly 0.45 km/s (1000 mph), how long would it take to reach the sun? The galactic center?

nadezda [96]

Answer:

The passenger aircraft would take 10.542 years to reach the Sun from the Earth.

The passenger aircraft would take $1.733\times 10^{10}$ years to reach the gallactic center.

Explanation:

The distance to the sun from the Earth is approximately equal to $1.496\times 10^{8}\,km$ , if the passenger travels at constant speed, then the time needed to reach the sun is calculated by the following kinematic formula:

$\Delta t = \frac{s}{v}$ (1)

Where:

$s$ - Travelled distance, measured in kilometers.

$v$ - Speed of the passenger aircraft, measured in kilometers per second.

$\Delta t$ - Travelling time, measured in seconds.

If we know that $s = 1.496\times 10^{8}\,km$ and $v = 0.45\,\frac{km}{s}$ , then the travelling time is:

$\Delta t = \frac{1.496\times 10^{8}\,km}{0.45\,\frac{km}{h} }$

$\Delta t = 3.324\times 10^{8}\,s$

$\Delta t = 3847.736\,days$

$\Delta t = 10.542\,years$

The passenger aircraft would take 10.542 years to reach the Sun from the Earth.

The distance between the Earth and the galactic center is approximately equal to $2.460\times 10^{17}\,km$ . If the passenger travels at constant speed and if we know that $s = 2.460\times 10^{17}\,km$ and $v = 0.45\,\frac{km}{s}$ , then the travelling time is: