All categories

3 years ago

The image shows a model of the gravitational field around Earth. Which best describes the model?

Physics

2 answers:

Alekssandra [29.7K]3 years ago

6 0

Answer: the answer is D "The strength of the field increases as you get closer to Earth, as shown by the smaller spaces between the circles"

Explanation:

GuDViN [60]3 years ago

3 0

Answer: The model is described by :

The strength of the field increases as you get closer to Earth, as shown by the smaller spaces between the circles.

Explanation :

According to universal gravitational law, the force acting in the universe between two objects is given by :

$F=G\dfrac{m_1m_2}{r^2}$

It is clear that the gravitational force depends directly on the product of masses and inversely on the square of the distance between them.

The given image shows a model of the gravitational field around Earth. The arrows in the image show the direction of the force. Gravitational field depends on the position of objects.

As we move closer to the earth, the force increases and hence the field increase.

So, the correct option is (D) " The strength of the field increases as you get closer to Earth, as shown by the smaller spaces between the circles ".

You might be interested in

Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in

postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is $\rho = \frac{m}{V}$ , and the volume of a pyramid is (confusingly written on the proposal) $V=\frac{1}{3} Ah$ , so we can write:

$m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h$

Where s is the side of the base, being $s^2$ the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that $\frac{1m}{100cm}=1$ , and we can use this factor to convert anything written in cm to anything written in m. Example:

$500cm=500cm\frac{1m}{100cm}=5m$

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so it's not doing anything but changing the units.

We can use this tool like this:

$2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3$

Where we have used the fact that $1^3=1$ (we can elevate any conversion factor to any number and they still will be 1) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

$m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg$

And now we will convert to short tons using two conversion factors at the same time:

$m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons$

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0

2 years ago

When the distance between two charges is halved, the electrical force between them?

Llana [10]

If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
$F= k \frac{q_1 q_2}{r^2}$
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
$r'= \frac{r}{2}$
the magnitude of the force changes as follows:
$F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k \frac{q_1 q_2}{r^2}=4 F$
so, the force increases by a factor 4.

3 0

3 years ago

What is the relationship between force in velocity selector in a bain bridge.

kifflom [539]

Answer:

In a velocity selector, there are two forces namely;

» Electric field Intensity

» Magnetic field density

Relationship:

$E _{e}= B$

E is the electric field intensity

B is the magnetic flux density

4 0

3 years ago

During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 *

marta [7]

Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s

To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s

Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m

This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.

But anyways, good luck!

4 0

3 years ago

An automobile traveling on a straight, level road has an initial speed v when the brakes are applied. In coming to rest with a c

Molodets [167]

Answer:

Explanation:

Use $v^{2} = u^{2} +2as$ to do the question.