If gravity between the Sun and the Earth suddenly vanished, Earth would continue moving

How do I convert 14.8 cm into MEters?

stellarik [79]

Multiply it by a fraction equal to ' 1 ', like this:

(14.8 cm) x (1 meter/100 cm) = 14.8/100 = 0.148 meter

6 0

3 years ago

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Can somone pls help me??!! i’m very stuck

LenKa [72]

Answer:

R

Explanation:

Her distance from home and Time increase

the stops when she gets to the library

then her distance from home decreases while her time increases

7 0

3 years ago

If the bus had a mass of 1,021kg how much does the bus weigh?

goblinko [34]

Weight of the bus= mass x acceleration due to gravity = 1021x9.8 = 10005.8 N. Hope it helps.

4 0

3 years ago

ONLY ANSWER IF YOU KNOW 100% BRAINLIEST TO THE FIRST CORRECT ANSWER, AND ANSWER ASAP!

saul85 [17]

Answer: Meter and second

5 0

1 year ago

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During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

2 years ago