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3 years ago

At what distance must a star be to have its apparent magnitude equal to its absolute magnitude?

1 answer:

6 0

10 parsecs<span>Another way to look at this is that the apparent magnitude (brightness from true distance) is brighter than the absolute magnitude (brightness from 10 parsecsaway). So Ross 128 must be closer than 10 parsecs, making star look brighter than it "really" is.</span>

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(show your work)

Artist 52 [7]

Answer:

I don't do physics , I'm sorry can't help you

4 0

3 years ago

A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

sleet_krkn [62]

The electric potential V(z) on the z-axis is : V = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq = б( 2πR ) dr

dV $\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }$ ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = $\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,$

V = $\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]$

= $\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]$

Therefore the electric potential V(z) = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

Also

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

7 0

3 years ago

19,792,000,000 in scientific notation will have how many significant figures

xz_007 [3.2K]

Answer:

= 1.9792 × 10^10

Significant Figures= 5

Explanation:

Look at the attachment below

Hope this helps (:

8 0

4 years ago

What is the electrical consumption in KVA of a motor powered by a 3-phase, 60 Hz, 460 VAC supply that continuously draws 17 A

MrRa [10]

Answer:

15.34 kVA

Explanation:

A motor is a device that converts electrical energy into mechanical energy. It takes in electrical energy at the input and produce torque (motion) at the output.

The power consumption for a three phase motor is the product of voltage and current and √3. The √3 is because it is a three phase supply.

Hence Power (P) =√3 × voltage (V) × current (I)

P = √3 × V × I

Given that voltage (V) = 460 V, current (I) = 17 A. Hence:

P = √3 × V × I = √3 × 460 × 17 = 13544.64 VA

But 1000 VA = 1 kVA. Hence:

$P=13544.64\ VA*\frac{1\ kVA}{1000\ VA}=13.54\ kVA$

8 0

3 years ago

Which is normally greater, the energy in ordinary sound or the energy in ordinary light? how does the speed of sound compare to

Thepotemich [5.8K]

The energy in ordinary light is greater than the energy in ordinary sound

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.

Energy is a physical system's ability to perform labor. The capital letter E is a typical sign for energy. The joule, denoted by the letter J, is the common unit. The energy produced by one newton's (1 N) worth of force acting over one meter's (1 m) worth of displacement is measured in joules (1 J). Because it is a fundamental human requirement, energy plays a significant role in our daily lives.

To learn more about Energy please visit-
brainly.com/question/1932868
#SPJ4

8 0

2 years ago