All categories

3 years ago

How does an objects size affect the gravity exerted by that object

Physics

1 answer:

egoroff_w [7]3 years ago

4 0

Answer:

The size of an object is directly proportional to the gravity

Explanation:

The size of an object has significant impact on the gravity exerted by such a body.

The more massive a body is, the larger the gravity it exerts.

The reason for this is because of the newton's law of universal gravitation.

It states that "the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".
As such, gravity is directly proportional to mass

You might be interested in

Which of the following is NOT a type of model used in scientific inquiry? interactive, mathematical, conceptual, convergent

katrin2010 [14]

A conceptual model. Hope this helps.

5 0

3 years ago

Read 2 more answers

Bumper car A (281 kg) moving

emmainna [20.7K]

Answer:

1.88 m/s

Explanation:

from elastic collision

KE before collision = KE after collision

8 0

3 years ago

Read 2 more answers

An object has rotational inertia I. The object, initially at rest, begins to rotate with a constant angular acceleration of magn

butalik [34]

Answer:

Explanation:

initial angular velocity, ωo = 0

angular acceleration = α

time = t

let the angular velocity after time t is ω.

Use first equation of motion for rotational motion

ω = o + α t

ω = αt

The angular momentum is given by

Angular momentum = moment of inertia x angular velocity

L = I x ω

L = I x αt

L = I α t

3 0

3 years ago

A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o

Margarita [4]

Answer:

V_{average} = $\frac{1}{2} V_o$ , V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

V (t) = $\left \{ {{V=V_o \ \ \ t< \tau /2} \atop {V=0 \ \ \ \ t> \tau /2 } } \right.$

to substitute in the equation let us separate the into two pairs

V_average = $\int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt$

V_average = $V_o \ \int\limits^{1/2}_0 {} \, dt$

V_{average} = $\frac{1}{2} V_o$

we evaluate V₀ = 4 V

V_{average} = 4 / 2)

V_{average} = 2 V

6 0

3 years ago

Why are clouds considered to be a part of the hydrosphere?

blondinia [14]

It is considered parts of the Hydrosphere because water condenses in clouds and falls as precipitation so it is a vital part of the hydrosphere.

7 0

4 years ago

How does an objects size affect the gravity exerted by that object​

How does an objects size affect the gravity exerted by that object