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3 years ago

What fills the void between stars and galaxies.

1 answer:

6 0

Answer: The voids between stars in our galaxy can be filled with tenuous clouds of gas and other molecules. ... That material gets "ripped away" from the galaxies by the force of gravity, and often enough it collides with other material.

HOPE IT HELPED:) HAVE A NICE DAY

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A resistor with an unknown resistance is connected in parallel to a 13 Ω resistor. When both resistors are connected in parallel

larisa86 [58]

Answer:

R2 = 10.31Ω

Explanation:

For two resistors in parallel you have that the equivalent resistance is:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\\$ (1)

R1 = 13 Ω

R2 = ?

The equivalent resistance of the circuit can also be calculated by using the Ohm's law:

$I=\frac{V}{R_{eq}}\\\\R_{eq}=\frac{V}{I}$ (2)

V: emf source voltage = 23 V

I: current = 4 A

You calculate the Req by using the equation (2):

$R_{eq}=\frac{23V}{4A}=5.75\Omega$

Now, you can calculate the unknown resistor R2 by using the equation (1):

$\frac{1}{R_2}=\frac{1}{R_{eq}}-\frac{1}{R_1}\\\\R_2=\frac{R_{eq}R_1}{R_1-R_{eq}}\\\\R_2=\frac{(5.75\Omega)(13\Omega)}{13\Omega-5.75\Omega}=10.31\Omega$

hence, the resistance of the unknown resistor is 10.31Ω

8 0

3 years ago

Read 2 more answers

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago

Coal burns in a furnace, producing light and heat. This reaction is

olasank [31]

The reaction is Exothermic

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3 years ago

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What describes how fast an object is moving and in what direction

Mashcka [7]

Speed is a description of how fast an object moves; velocity is how fast and in what direction it moves. In physics, velocity is speed in a given direction. When we say a car travels at 60 km/h, we are specifying its speed.

6 0

3 years ago

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What effect would decreasing the distance between objects have on their gravitational attraction to each other?

Lady_Fox [76]

Decreasing the distance between two objects having a considerable mass would increase the attraction on gravitation. The reverse is true that if you separate or inrease the objects distance would substantially decrease their gravitational attraction. Most object in our planet is held by its gravitational force towards it's center.

5 0

3 years ago

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