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3 years ago

Explain why the material surrounding a young star forms a disk

1 answer:

3 0

Everything in presence is moving. That is the most essential naturally visible thing to think about the universe. In this way, consider a youthful star and any sensible measure of material about it. Regardless of the possibility that it was by one means or another equally, arbitrarily separated in a circle about the youthful star and moving in any case, the star and its gravitational impacts will rule. So its immense mass turns in a given plane and its gravity field twists fundamentally spacetime in that plane and less so in whatever another plane. In any case, the development of the star drags the only things that are important around it step by step into its rotational plane.

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Which statement describes a question that can be answered by a scientific

goblinko [34]

It’s d! hope I can help

4 0

4 years ago

Read 2 more answers

Two small nonconducting spheres have a total charge of 93.0 μC . Part A

Travka [436]

Answer:

charge on each

Q1 = 2.06 × $10^{-5}$ C

Q2 = 7.23 × $10^{-5}$ C

when force were attractive

Q1 = 1.07 × $10^{-4}$ C

Q2 = -1.39 × $10^{-5}$ C

Explanation:

given data

total charge = 93.0 μC

apart distance r = 1.14 m

force exerted F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC = 93 × $10^{-6}$ C

and

According to Coulomb's law force between two sphere is

Force F = $\frac{K Q1 Q2 }{r^2}$ .........1

Q1Q2 = $\frac{F*r^2}{k}$

here F is force and r is apart distance and k is 9 × $10^{9}$ N-m²/C² put all value we get

Q1Q2 = $\frac{ 10.3*1.14^2}{9*10^9}$

Q1Q2 = 1.49 × $10^{-9}$ C²

and

we have Q2 = 93 × $10^{-6}$ C - Q1

put here value

Q1² - 93 × $10^{-6}$ Q1 + 1.49 × $10^{-9}$ = 0

solve we get

Q1 = 2.06 × $10^{-5}$ C

and

Q1Q2 = 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = 1.49 × $10^{-9}$

Q2 = 7.23 × $10^{-5}$ C

and

if force is attractive we get here

Q1Q2 = - 1.49 × $10^{-9}$ C²

then

Q1² - 93 × $10^{-6}$ Q1 - 1.49 × $10^{-9}$ = 0

we get here

Q1 = 1.07 × $10^{-4}$ C

and

Q1Q2 = - 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = - 1.49 × $10^{-9}$

Q2 = -1.39 × $10^{-5}$ C

5 0

3 years ago

Two resistors, one with 7.00 Ω of resistance and the other with 11.00 Ω of resistance, are connected in series to a 9.00 V batte

IRINA_888 [86]

Answer:

4.5 W

Explanation:

Applying,

P = V²/(R₁+R₂).................. Equation 1

Where P = Power, V = Voltage, R₁ and R₂ = values of the two resistor.

From the question,

Given: V = 9.00 V, R₁ = 7.00 Ω, R₂ = 11.00 Ω

Substitute these values into equation 1

P = 9²/(7+11)

P = 81/(18)

P = 4.5 Watt.

Hence the power dessipated by the two resistors is 4.5 watt

5 0

3 years ago

I need help with this someone help me?

Brilliant_brown [7]

Graph A so answer B also why isn’t answer a with graph A and B with graph B etc like that’s just confusing lol

3 0

3 years ago

Which of the following statements are true? Select all that apply.

beks73 [17]

Answer:

The following options are true based on the properties of electric field;

a) Electric field lines near positive point charges radiate outward.

b) The electric force acting on a point charge is proportional to the magnitude of the point charge.

d) In a uniform electric field, the field lines are straight, parallel, and uniformly spaced.

Explanation:

From option b) From coulomb's law F = Kq1q2r/r2

5 0

4 years ago