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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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Which of the following mathematical operations are and are not not allowed on two quantities with different unit dimensions?

Katyanochek1 [597]

The following mathematical operations are and are not allowed on two quantities with different unit dimensions are C) Allowed : Multiplication, Division , Not Allowed : Addition, Subtraction, Equality

To answer the question, we have to know what mathematical operations are.

<h3>What are Mathematical operations?</h3>

Mathematical operations are operations which are performed to change the value of a variable. The arithmetic mathematical operations we have are

addition,
subtraction,
multiplication,
subtraction and
equality which shows the relationship between two quantities.

Now, for two quantities with different unit dimensions, addition, subtraction and equality are not allowed because these mathematical operations require the quantities to be in the same unit dimensions

Also, for two different quantities with differnt unit dimensions, multiplication and division are allowed because these mathematical operations do not require the quantities to be in the same unit dimensions.

So, the mathematical operations that are allowed are multiplication and division while the mathematical operations that are not allowed are addition, subtraction and equality.

So, the following mathematical operations are and are not allowed on two quantities with different unit dimensions are C) Allowed : Multiplication, Division , Not Allowed : Addition, Subtraction, Equality

Learn more about mathematical operations here:

brainly.com/question/26112472

4 0

2 years ago

Lila is a track and field athlete. She must complete four laps around a circular track. The track itself is a 400 meter track an

Aleonysh [2.5K]

Answer:

Her speed is 1.1 m/s, and her velocity is 0 m/s

Explanation:

Speed = Distance covered/Time

Given

Distance = 400m

Time = 6minutes = 6*60 = 360 secs

Substitute the given parameter into the formula;

Speed = 400/360

Speed = 1.1m/s

Since the track is a circular track, the displacement will be zero. She is only moving in a circular path (no direction)

Velocity = Displacement/Time

Velocity = 0/3600

Velocity = 0m/s

Hence her speed is 1.1 m/s, and her velocity is 0 m/s

6 0

3 years ago

A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and

Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

$q_{out}=q_{2}-q_{1}$

$q_{out}=-6.43-(-4.00)$

$q_{out}=-2.43\ \mu C$

The charge on the inner surface is q.

$q+(-2.43)=-6.43$

$q=-6.43+2.43= 4.00\ \mu C$

We need to calculate the electric field outside the shell

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}$

$E=33927618.73\ N/C$

$E=3.39\times10^{7}\ N/C$

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

5 0

2 years ago

A closely wound circular coil has a radius of 6.00 cmand carries a current of 2.65 A. How many turns must it have if the magneti

ad-work [718]

Answer:

Given:

radius of the coil, R = 6 cm = 0.06 m

current in the coil, I = 2.65 A

Magnetic field at the center, B = $6.31\times 10^{4} T$

Solution:

To find the number of turns, N, we use the given formula:

$B = \frac{\mu_{o}NI}{2R}$

Therefore,

$N = \frac{2BR}{\mu_{o}I}$

$N = \frac{2\times 6.31\times 10^{4}\times 0.06}{4\pi \times 10^{- 7}\times 2.65}$

N = 22.74 = 23 turns (approx)

8 0

2 years ago

What happens if you move a bar magnet back and forth along the axis of the

nadezda [96]

C) A current is induced in the coiled wire, which lights the light bulb

The moving magnetic field creates electricity which lights the light bulb

Hope it helps!

8 0

1 year ago

Read 2 more answers