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2 years ago

Which is the last stage in the life cycle of an average star? (1 point)

Physics

1 answer:

Shkiper50 [21]2 years ago

4 0

Answer:

The answer is A. White Dwarf

Explanation:

average stars like sun end their life as white dwarf surrounded by disappearing planetary nebula.

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When a wire is made thicker its resistance what?

NeTakaya

Making a wire thicker has the same effect as making a road wider. It makes it easier for the electron traffic to flow. The resistance decreases, and the current (traffic) increases.

7 0

3 years ago

How can I convert this? Please answer with solution. Thank you.

fomenos

Answer:

1 hr 45 min

Explanation:

8 0

3 years ago

which of the folloing statements about ionization energy is true? A elements toward the bottom of a group periodic table general

beks73 [17]

accept the flow of electrons.resist the flow of electrons.accept the flow of protons.resist the flow of protons.It is one of these

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3 years ago

Two children who are bored while waiting for their flight at the airport decide to race from one end of the 20-m-long moving sid

QveST [7]

Answer: Phillipe wins the race

Explanation:

Given

Length of track=20 m

Velocity of Phillippe=2 m/s w.r.t sidewalk

Velocity of sidewalk=1.5 m/s

Rena velocity=2 m/s

Therefore absolute velocity of P is 2+1.5=3.5 m/s

Time taken by phillippe $\frac{20}{3.5}+\frac{20}{3.5-1.5}$

$t_P$ =5.714+10=15.714 s

Time taken by rena

$t_r=\frac{20}{2}+\frac{20}{2}=20 s$

as the time taken by phillippe is less than rena therefore Phillippe wins the race.

8 0

3 years ago

Read 2 more answers

A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di

siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

$E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\$

part a)

Electric Field strength at point P: a = 0.75 m

$E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C$

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

$E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o$

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

6 0

3 years ago