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3 years ago

What two factors determine how bright a star appears to be in the sky?

Physics

1 answer:

Delicious77 [7]3 years ago

7 0

Answer:

1. Luminosity

2.Apparent brightness

Explanation:

There are two factors on which brightness of star appear to be in the sky

The two factors are

1. Luminosity

2.Apparent brightness

1.Luminosity :It is defined as the total energy emitted by the object in a given time.Luminosity vary with the distance of observer from the star.Luminosity is a intrinsic property which depends on the fundamental chemical composition and structure of the material.Luminosity is depends on the size of star.Lager the star luminosity will be more.

2.Apparent brightness: It is defined as how bright a star appears from an observer on the earth and the amount of starlight reaching the earth.if the distance is large then the brightness decreases.When the distance of star from us small then the brightness of star increases.Distance is inversely proportional to brightness of the star.

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Reactance Frequency Dependence: Sketch a graph of the frequency dependence of a resistor, capacitor, and inductor. RLC Circuit R

jolli1 [7]

Answer:

$f=\frac{1}{2\pi \sqrt{LC}}$

Explanation:

We know that impedance of a RLC circuit is given by $Z=R+J(X_L-X_C)$

So $Z=\sqrt{R^2+(X_L-X_C)^2}$ here R is resistance $X_L$ is inductive reactance and $X_C$ is capacitive reactance

To minimize the impedance $X_L-X_C$ should be zero we know that $X_L=\omega L\ and \ X_C=\frac{1}{\omega C}$

So $\omega L-\frac{1}{\omega C}=0$

$\omega ^2=\frac{1}{LC}$

$\omega =\sqrt{\frac{1}{LC}}$

We know that $\omega =2\pi f$

So $\omega =2\pi f=\frac{1}{\sqrt{LC}}$

$f=\frac{1}{2\pi \sqrt{LC}}$

Where f is resonance frequency

8 0

3 years ago

Why do photons take so much longer than neutrinos to emerge from the sun?

Alex777 [14]

Answer:

Because of the strong interaction of photons with matter,neutrinos rarely interact with matter

Explanation:

In the core of the sun,nuclear fusion which is a nuclear reaction,produces the photons and neutrinos,for photons to emerge from the sun's core it passes through denser particles colliding and losing energy as it moves at such it can take it up to a million years to emerge from the sun core.Unlike the neutrinos that has more easier path,with little or no collission.

8 0

3 years ago

Why do we have thunder storms

Over [174]

Thunderstorms happen when cold air sinks and hot air rises in rapid succession. There has to be an unstable body of hot air for a thunderstorm to happen. Cool air mingles with the hot air causing condensation (that means that raindrops form) This makes a cumulous cloud.
The moisture in the cloud is either positively charged or negatively charged. Negatively charged particles sink to the bottom of the cumulous cloud while positively charges particles rise. The positive and negative charges rub together and spark causing lightning.
Thunder is formed by the lightning, a bolt of lightning is hollow, and has no air inside it. When the lightning dissipates the air rushes to fill the space breaking the sound barrier in the process, this creates the boom.

3 0

3 years ago

Areas near baton rouge louisiana recently received 36.2 cm of rain in a single day. how many meters of rain was this?

sergey [27]

100 cm is 1 meter. So your answer would be 0.362 meters.

5 0

3 years ago

Read 2 more answers

In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact

OLga [1]

Answer:

v=3.73m/s, $a=3.35m/s^{2}$

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

$K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}$

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

$K_{A}+U_{Ag}=U_{Cs}+W_{Cf}$

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}$

so we can now solve this for k, which is the constant of the spring.

$k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}$

and now we can substitute the respective data:

$k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}$

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

$K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}$

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

$\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}$

and we can solve for the Velocity in B, so we get:

$V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}$

we can now input all the data directly, so we get:

$V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}$

which yields:

$V_{B}=3.73m/s$

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

$\sum F=ma$