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dolphi86 [110]
3 years ago
5

if an object starts at rest and moves 60 meters north along a straight line in 2 seconds, what is the average velocity?

Physics
2 answers:
VashaNatasha [74]3 years ago
8 0
30 m/s because you would divide 60 meters by 2 seconds. 60 divided by two equals 30
Thepotemich [5.8K]3 years ago
5 0

Velocity =
       (distance between start-point and end-point) / (time to run the course)
in the direction of
       (direction from the start-point to the end-point).

Distance from start-point to end-point = 60 meters
Time to run the course = 2 seconds
Direction = north

Velocity = 30 m/s north

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The magnitude of the gravitational field strength near Earth's surface is represented by
Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

F = G\cdot \frac{M\cdot m}{r^{2}}

Where:

M - Mass of the planet Earth, measured in kilograms.

m - Mass of the person, measured in kilograms.

r - Radius of the Earth, measured in meters.

G - Gravitational constant, measured in \frac{m^{3}}{kg\cdot s^{2}}.

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

F = m \cdot g

Where:

m - Mass of the person, measured in kilograms.

g - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

g = \frac{G\cdot M}{r^{2}}

Given that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972 \times 10^{24}\,kg and r = 6.371 \times 10^{6}\,m, the magnitude of the gravitational field near Earth's surface is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}

g \approx 9.82\,\frac{m}{s^{2}}

The magnitude of the gravitational field strength near Earth's surface is represented by approximately 9.82\,\frac{m}{s^{2}}.

4 0
3 years ago
Prove that the weight of an object on moon is 1/6th of that on earth​
Elena L [17]

Answer:

The mass of moon is 1/100 times and its radius 1/4 times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is 1/6th its weight on the earth.

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Drag the right word to the word that means the opposite
Ainat [17]
What do you mean? Is there a picture ?
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Suppose two objects are gravitationally attracted to each other with some force F. If the mass of object 1 is multiplied by a fa
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If F = Gm₁m₂/d², and we change m₁ to 5m₁ and m₂ to 2m₂, then the new magnitude of the gravitational force is

F' = G (5m₁) (2m₂) / d²

F' = 10  Gm₁m₂ / d²

but this is really just F' = 10F. So J is the correct choice.

8 0
2 years ago
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