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Lina20 [59]
3 years ago
10

Consider a 5kg rock and a 4.5g piece of paper held at the same height above the ground. What is true about the gravitational fie

ld at their location and the gravitational force on both objects ?
Physics
1 answer:
Volgvan3 years ago
4 0

Answer:

The gravitational field at any point is independent of the objects placed at that point such as a stone or paper.

And, also the gravitational force acting on a body is determined by the intensity of the field at that point.

Explanation:

Given,

The gravitational force between two bodies separated by a distance r is given by the relation,

                              F =  GMm / r²

Where,

                    M - the mass of the Earth

                    m - the mass of the object

                     r - the distance between the center of masses

Let m₁ be the mass of rock. Then force acting on the rock is given by

                                    F = m₁a

The gravitational force between the stone and Earth is given by

                                    F = GMm₁ / r²

Equating these two forces

                                     m₁a = GMm₁ / r²

Canceling out the mass of the stone.

This shows that the force component acceleration is independent of the mass of the stone.

Hence, it implies that the gravitational field at any point is independent of the objects placed at that point such as a stone or paper.

And, also the gravitational force acting on a body is determined by the intensity of the field at that point.

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Is the distance traveled during a specific unit of
Anton [14]

Answer:

velocity

Explanation:

I think i dont really know

4 0
3 years ago
An automobile of mass 1.46 cross times to the power of blank 10 cubed kg rounds a curve of radius 25.0 m with a velocity of 15.0
Vsevolod [243]

The centripetal force exerted on the automobile while rounding the curve is 1.31\times10^4 N

<u>Explanation:</u>

given that

Mass\ of\ the\ automobile\ m  =1.46\times 10^3 kg\\radius\ of\ the\ curve\ r =25 m\\velocity\ of\ the\ automobile\ v=15m/s\\

Objects moving around a circular track will experience centripetal force towards the center of the circular track.

centripetal\ force=mv^2/r\\=1.46\times10^3\times15^2/25\\=1.46\times 10^3\times 225/15\\=1.31\times 10^4 N

4 0
3 years ago
Which lists the main components of Darwin’s theory of evolution?
Nataliya [291]

Hi!


The correct option is B.


<h3>Explanation</h3>

Evolution is a biological phenomena which descries how the heritable characteristics of a species change over successive generations. These characteristics change as a result of a mutation, natural selection or genetic recombination (to list a few).

Darwin's theory of evolution explained how this phenomena occurs with respect to natural selection. The theory goes on to explain how a species diverges from a common ancestor into two or more different species (meaning different species may have common ancestor). This is a slow process as it can take thousands of years.


Hope this helps!

4 0
3 years ago
Read 2 more answers
A 600 kg car is at rest, and then accelerates to 5 m/s.
Crank

Answer:

0

7500J

7500J

Explanation:

Given parameters:

Mass of car  = 600kg

Velocity  = 5m/s

Unknown:

Original kinetic energy  = ?

Final kinetic energy  = ?

Work used  = ?

Solution:

The kinetic energy of a body is the energy due to the motion of a body.

It can be solve mathematically using expression below;

                K.E  = \frac{1}{2}  m v²

  where m is mass

              v is velocity

original kinetic energy;

  The car started at rest and v = 0, therefore K.E  = 0

Final kinetic energy;

           K.E  = \frac{1}{2}  x 600 x 5²   = 7500J

Work done:

   Work done  = Final K.E  - Initial K.E  = 7500 - 0 = 7500J

   

3 0
3 years ago
Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The
Brut [27]

Answer:

The resultant vector is \vec R = \vec A + \vec B = 7.196\,i + 6.464\,j.

Explanation:

First, each vector is determined in terms of absolute coordinates:

6-meter vector with direction: 30º north of east.

\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)

\vec A = 5.196\,i + 3\,j

4-meter vector with direction: 30º east of north.

\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)

\vec B = 2\,i + 3.464\,j

The resultant vector is obtaining by sum of components:

\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j

The resultant vector is \vec R = \vec A + \vec B = 7.196\,i + 6.464\,j.

5 0
4 years ago
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