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2 years ago

The sun’s absolute magnitude is described as _______ in comparison to other stars.

Physics

2 answers:

neonofarm [45]2 years ago

7 0

Answer: average

Explanation:

Absolute magnitude can be defines as the apparent magnitude at a standard distance of 10 Parsecs.

The absolute magnitude of Sun is 4.83 where as its apparent magnitude is -26.74.

The absolute magnitude of Sun is neither very large nor very small. It lies in the limiting magnitude range.

Hence, the correct option is The sun’s absolute magnitude is described as average in comparison to other stars.

cupoosta [38]2 years ago

3 0

very small

Being close to the earth can be milseading. I think that if the sun were at the same distance as anohter star, it might be hard to see because it is small, relatiavely speaking

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2 years ago

What is the passenger's apparent weight at t=1.0s?

Fittoniya [83]

Answer:

For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.

Explanation:

The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.

From the slope of graph it is clear that acceleration at t = 1 sec is given as:

Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2

Now, there are two cases:

1- Elevator moving up

2- Elevator moving down

For upward motion:

Apparent Weight = m(g + a)

Apparent Weight = (75 kg)(9.8 + 4)m/s^2

Apparent Weight = 1035 N

For downward motion:

Apparent Weight = m(g - a)

Apparent Weight = (75 kg)(9.8 - 4)m/s^2

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A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

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Answer:

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