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yuradex [85]
3 years ago
5

As you can see from the Group 1 stars, the cooler or hotter a star is, the brighter it will be. Group 2 and Group 3 stars do not

follow this pattern. Hence, there must be something besides temperature that can affect how bright a star is. Describe your own hypothesis about these stars (Group 2 and Group 3). Why would their brightness not be strictly related to their temperature
Physics
1 answer:
forsale [732]3 years ago
8 0

Answer:

brightness that we observe from a star is related to the energy produced and the distance to the Earth

Explanation:

In stars, the color that we observe is directly related to the temperature of the star by the y of the Wien displacement.

             λ_{max} T = 2,898 10³

the brightness that we observe from a star is related to the energy produced and the distance to the Earth

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Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
3 years ago
Suppose we could shrink the Earth without changing its mass. At what fraction of its current radius would the free-fall accelera
spin [16.1K]

Answer:

R' = \frac{1}{\sqrt{3}}R

Explanation:

The acceleration due to gravity on the surface of the Earth is given by:

g=\frac{GM}{R^2}

where

G is the gravitational constant

M is the mass of the Earth

R is the radius of the Earth

Here we want to find the new Earth radius R' for which the gravitational acceleration at the surface, g', would be 3 times the current value of g:

g' = 3g

So we would have

\frac{GM}{R'^2}=3(\frac{GM}{R^2})

Solving the equation for R', we find

R'^2 = \frac{1}{3}R^2\\R' = \frac{1}{\sqrt{3}}R

7 0
3 years ago
Read 2 more answers
Calculate the initial (from rest) acceleration of a proton in a 5.00 x 10^6 N/C electric field (such as created by a research Va
julsineya [31]

Answer:

Acceleration of the proton will be equal to 4.79\times 10^{14}m/sec^2

Explanation:

We have given electric field E=5\times 10^6N/C

Mass of proton is equal to m=1.67\times 10^{-27}kg

And charge on proton is equal to e=1.6\times 10^{-19}C

Electrostatic force will be responsible for the motion of proton

Electrostatic force will be equal to F=qE=1.6\times 10^{-19}\times 5\times 10^6=8\times 10^{-13}N

According to newton law force on the proton will be equal to F = ma, here m is mass of proton and a is acceleration

This newton force will be equal to electrostatic force

So 1.67\times 10^{-27}\times a=8\times 10^{-13}

a=4.79\times 10^{14}m/sec^2

So acceleration of the proton will be equal to 4.79\times 10^{14}m/sec^2

5 0
3 years ago
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton
pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx  

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:


F = K * \Delta{x}

2\:N = 4\:N/cm*\Delta{x}

4\:N/cm*\Delta{x} = 2\:N

\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}

simplify by 2

\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}

\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark

Answer:

B.) 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

7 0
3 years ago
WHAT IS THE S.I UNIT OF MASS
ycow [4]

The SI unit of mass is kilogram (kg)....

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3 years ago
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