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3 years ago

A hydrogen line in a star's spectrum

Physics

1 answer:

Serhud [2]3 years ago

5 0

Answer: V = 4.35 × 10^12 m/s

Explanation: To calculate Rigel velocity relative to us, we should use doppler effect formula.

Fo = (c ) / ( c - Vs) × f

Fo/f = (c ) / ( c - Vs)

4.26 × 10^10 = (3×10^8) 6.17×10^14/(3×10^8 - V)

4.26×10^10/6.17×10^14 = (3×10^8)/(3×10^8 - V)

6.9×10^-5 = (3×10^8)/(3×10^8 - V)

Cross multiply

20713.13 - 6.9×10^-5V = 3×10^8

Collect the like terms

6.9×10^-5V = 3×10^8 - 20713.13

6.9×10^-5V = 299979286.9

V = 299979286.9/6.9×10^-5

V = 4.3475 × 10^12 m/s

Therefore, Rigel's velocity relative to us is 4.3475 × 10^12 m/s

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Find the heat energy is required to change 2Kg of ice at 0 C to water at 20 C ( specific latent heat of fusion of water = 336000

katrin2010 [14]

We want to find the energy that we need to transform 2kg of ice at 0°C to water at 20°C.

We will find that we must give 840,000 Joules.

First, we must change of phase from ice to water.

We use the specific latent heat of fusion to do this, this quantity tells us the amount of energy that we need to transform 1 kg of ice into water.

So we need 336,000 J of energy to transform 1kg of ice into water, and there are 2kg of ice, then we need twice that amount of energy:

2*336,000 J = 672,000 J

Now we have 2kg of water at 0°C, and we need to increase its temperature to 20°C.

Here we use the specific heat, it tell us the amount of energy that we need to increase the temperature per mass of water by 1°C.

We know that:

specific heat of capacity of water = 4200 J/kg°C

This means that we need to give 4,200 Joules of energy to increase the temperature by 1°C of 1kg of water.

Then to increase 1°C of 2kg of water we need twice that amount:

2*4,200 J = 8,400 J

And that is for 1°C, we need to give that amount 20 times (to increase 20°C) this is:

20*8,400 J = 168,000 J

Then the total amount of energy that we must give is:

E = 672,000 J + 168,000 J = 840,000 J

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3 years ago

Think about the atmosphere of Venus. How is it different from that of other terrestrial planets?

Digiron [165]

Because one is thicker

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4 years ago

Any ideas for the science fair? I am a 7th grader but looking for a first-place type of project.Thanks!

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U can do what is the strongest paper towel

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4 years ago

Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and

kow [346]

A) Acceleration: $a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2$

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

Acceleration is defined as:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

$a_1 = \frac{11-8}{4}=0.75 m/s^2$ (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: $a_2 = 0$

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

$a=\frac{0-11}{7}=-1.57 m/s^2$

And the negative sign means the acceleration is south, opposite to the direction of motion.

In a uniformly accelerated motion, the displacement can be calculated as:

$s=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

$u = 0\\a = 0.75 m/s^2\\t=4 s$

So the displacement is

$s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m$

- For the second segment, we have

$u = 11 m/s\\a = 0\\t=15 s$

So the displacement is

$s_2 = (11)(15)+0=165 m$

- For the third segment, we have

$u = 11\\a = -1.57 m/s^2\\t=7 s$

So the displacement is

$s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m$

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

The average velocity is given by:

$v=\frac{d}{t}$

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

$v=\frac{209.5}{26}=8.06 m/s$ (north)

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4 years ago

A 8.03-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial dis

Lynna [10]

Answer:

Answer for a is 656.56N and for b is 157.4N

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3 years ago