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3 years ago

What is a possible diameter for a typical neutron star?

1 answer:

8 0

The <span>possible diameter for a typical neutron star is 20 kilometer. The answer is letter B. The rest of the choices do not answer the question above.</span>

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5. George walks to a friend's house. He walks 750 meters North, then realizes he walked too far.

dedylja [7]

Answer:

Average speed: approximately $76.9\; {\rm m\cdot s^{-1}}$ .

Average velocity: approximately $38.5\; {\rm m \cdot s^{-1}}$ (to the north.)

Explanation:

Consider an object that travelled along a certain path. Distance travelled would be equal to the length of the entire path.

In contrast, the magnitude of displacement is equal to distance between where the object started and where it stopped.

In this question, the path George took required him to travel $750\; {\rm m} + 250\; {\rm m} = 1000\; {\rm m}$ in total. Hence, the distance George travelled would be $1000\; {\rm m}$ . However, since George stopped at a point $(750\; {\rm m} - 250\; {\rm m}) = 500\; {\rm m}$ to the north of where he started, his displacement would be only $500\; {\rm m}$ to the north.

Divide total distance by total time to find the average speed.

Divide total displacement by total time to find average velocity.

The total time of travel in this question is $13\; {\rm s}$ .. Therefore:

$\begin{aligned}\text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &= \frac{1000\; {\rm m}}{13\; {\rm s}} \\ &\approx 76.9\; {\rm m\cdot s^{-1}}\end{aligned}$ .

$\begin{aligned}\text{average velocity} &= \frac{\text{total displacement}}{\text{total time}} \\ &= \frac{500\; {\rm m}}{13\; {\rm s}} && \genfrac{}{}{0px}{}{(\text{to the north})}{}\\ &\approx 38.5\; {\rm m\cdot s^{-1}} && (\text{to the north})\end{aligned}$ .

3 0

2 years ago

Pls help ASAP !! Phyics

mina [271]

Answer:

Explanation:

Given

mass (m)= 20 kg

acceleration (a)= 10 m/s^2

Force (f)= m a

= 20 * 10

= 200 N

7 0

3 years ago

Might it be possible to explain the interaction of the rod and pieces of paper as a gravitational interaction? please explain wh

mina [271]

It is not possible to explain the interaction of the rod and pieces of paper as a gravitational interaction.

<h3>What is Gravitational interaction?</h3>

This is defined as the interaction between a particle or body resulting from their mass. This type of interaction is usually weak and occurs in all distances possible.

It is not gravitational interaction, because the rod attracts paper only against the gravitational force of the earth and there is no attraction between both bodies under a different condition.

This is therefore the reason why it is not possible to explain the interaction of the rod and pieces of paper as a gravitational interaction.

Read more about Gravitational interaction here brainly.com/question/25624188

#SPJ1

5 0

2 years ago

You can also enter units that are combinations of other units. keep in mind that you have to indicate the multiplication of unit

omeli [17]

Watch this ! You've never seen anything like it !
No wait. I'm pretty sure you have.

Weight = (mass) · (gravity)

Weight = (10 kg) · (9.8 m/s²)

Weight = (10 · 9.8) kg·m/s²

Weight = 98 Newtons

7 0

4 years ago

Pin p is constrained to move along the curve defined by the lemniscate r=(4sin2θ)ft.if the slotted arm oa rotates counterclockwi

ruslelena [56]

position of the peg is given by the equation

$r = 4 sin2\theta$