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Butoxors [25]
1 year ago
10

If it were possible to move a star towards the earth then its apparent magnitude number would ______ while its absolute magnitud

e number would _______.
Physics
1 answer:
ArbitrLikvidat [17]1 year ago
5 0

If it were possible to move a star towards the earth then its apparent magnitude number would decrease while its absolute magnitude number would stay the same.

Definition of apparent magnitude:

The luminosity of a celestial body (such as a star) as observed from the earth compare absolute magnitude.

So for example, the apparent magnitude of the Sun is -26.7 and is the brightest celestial object we can see from Earth. However, if the Sun were 10 parsecs away, its apparent magnitude would be +4.7, only about as bright as Ganymede appears to us on Earth.

Definition of absolute magnitude:

Absolute magnitude is a measure of the luminosity of a celestial object on an inverse logarithmic astronomical magnitude scale.

To learn more about apparent magnitude here

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How will sunlight most likely affect a black shirt on a hot summer day? the temperature of the shirt will depend on how much sun
vladimir2022 [97]

The correct answer is option (C) the temperature of the shirt will increase because all wavelengths of light are absorbed by the shirt.

The relationship of heat and light

  • Heat is a measure of the movement of particles in the body, the more particles move, the warmer the body becomes.
  • When the body absorbs light radiation, its particles vibrate in accordance with the electromagnetic radiation's wavelengths, which causes an increase in the temperature with the increase in particle movement.
  • The more wavelengths of radiation absorbed by an object, produces more heat.

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5 0
11 months ago
Consider an isolated system, by which we mean a system free of external forces. If the sum of the particle energies in the syste
Airida [17]

Answer:

The system's potential energy is -147 J.

Explanation:

Given that,

Energy = 147 J

We know that,

System is isolated and it is free from external forces.

So, the work done by the external forces on the system should be equal to zero.

W=0

We need to calculate the system's potential energy

Using thermodynamics first equation

\Delta U=W-\Delta E

Put the value into the formula

\Delta U=0-147

\Delta U=-147\ J

Hence, The system's potential energy is -147 J.

5 0
2 years ago
a man of mass 50 kg on the top floor of skyscraper step into an elevator what is the max weight as the elevator moves downward​
shtirl [24]
514.5 N is the answer
6 0
2 years ago
a block weighing (Fg) 50 N is resting on a steel table (us = 0.74). The minimum force to start this block moving is what N
Montano1993 [528]

Answer:37

Explanation:

5 0
3 years ago
A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/
hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is 9.1\cdot 10^{-4} J

C) The work done by the electric field is -2.4\cdot 10^{-3} J

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the particle

\theta is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

  • The force is directed vertically upward (because the field is directed vertically upward)
  • The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

\theta=90^{\circ}

and cos 90^{\circ}=0: therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

\theta=0^{\circ}

and

cos 0^{\circ}=1

Therefore, the work done by the electric force is

W=Fd

and we have:

F=qE is the magnitude of the electric force. Since

E=4.30\cdot 10^4 V/m is the magnitude of the electric field

q=32.0 nC = 32.0\cdot 10^{-9}C is the charge

The electric force is

F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

\theta=90^{\circ}+45^{\circ}=135^{\circ}

Moreover, we have:

F=1.38\cdot 10^{-3} N (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J

And the work is negative because the electric force is opposite direction to the displacement of the charge.

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5 0
3 years ago
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