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andrezito [222]
3 years ago
7

The acceleration due to gravity on the earths surface is?​

Physics
2 answers:
Gennadij [26K]3 years ago
8 0

Acceleration due to gravity depends directly on the mass of Earth and inversely to the square of radius of Earth.

marta [7]3 years ago
8 0

Answer:

9.8 m/s2

Explanation:

Acceleration due to gravity is the acceleration gained by an object due to the gravitational force. Its SI unit is m/s 2. It has both magnitude and direction, hence, it’s a vector quantity. Acceleration due to gravity is represented by g. The standard value of g on the surface of the earth at sea level is 9.8 m/s2.

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Newton's first law of motion says (in part) that an object in motion will remain in motion at a constant speed and in a straight
kolbaska11 [484]

Your answer is C. Unbalanced

unbalanced forces change the direction and speed of objects, therefore it would change the constant speed of an object in motion.

Hope this helps u!

6 0
3 years ago
Read 2 more answers
Ian walks 2 km to his best friend's house, then walks 0.5 km to the library. He then makes a 2.5 km walk home. The entire walk t
earnstyle [38]

Average speed = (total distance covered) / (time to cover the distance)

Ian's total distance covered = (2km + 0.5km + 2.5km) = 5 km.

His time to cover the distance = 3 hours.

Average speed = (5 km) / (3 hrs)

Average speed = (5/3) (km/hr)

<em>Average speed = 1.67 km/hr</em>

5 0
3 years ago
(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o
IgorLugansk [536]

Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J

So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.

3 0
3 years ago
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The distance of Saturn from the sun is:<br><br> &lt; 1 A.U.<br> &gt; 1 A.U.<br> = 1 A.U.
sertanlavr [38]

Answer:

1 astronomical unit is the average distance from the Earth to the Sun; approximately 150 million km. At its closest point, Saturn is 9 AU, and then at its most distant point, it's 10.1 AU. Saturn's average distance from the Sun is 9.6 AU. We have written many articles about Saturn for Universe Today.

Explanation:

6 0
3 years ago
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2. The force between charges is 7 N. The distance between the charges is 4 x 10-6 m. If one of the charges is 2 x 10-8 C, what i
g100num [7]

Answer:

2) 6.22 × 10^-13

3) 4.56 × 10^-4

Explanation:

4 0
3 years ago
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