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irina [24]
3 years ago
6

If the distance between us and a star is doubled, with everything else remaining the same, the luminosity Group of answer choice

s remains the same, but the apparent brightness is decreased by a factor of two. is decreased by a factor of four, and the apparent brightness is decreased by a factor of four. is decreased by a factor of four, but the apparent brightness remains the same. is decreased by a factor of two, and the apparent brightness is decreased by a factor of two. remains the same, but the apparent brightness is decreased by a factor of four.
Physics
1 answer:
Savatey [412]3 years ago
8 0

Answer:

remains the same, but the apparent brightness is decreased by a factor of four.

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).

The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).

The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.

The apparent brightness of a star is measured in watts per square meter.

If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).

Some of the examples of stars are;

- Canopus.

- Sun (closest to the Earth)

- Betelgeuse.

- Antares.

- Vega.

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A water line enters a house 2.0 m below ground. A smaller diameter pipe carries water to a faucet 5.0 m above ground, on the sec
nirvana33 [79]

Answer:

 P₁- P₂ = 91.1 10³ Pa

Explanation:

For this exercise we will use Bernoulli's equation, where point 1 is at the bottom of the house and point 2 on the second floor

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

          P1-P2 = ½ ρ (v₂² - v₁²) + ρ g (y₂-y₁)

In the exercise they give us the speeds and the height of the turbid, so we can calculate the pressure difference

For heights let's set a reference system on the ground floor of the house, so we have 5m for the second floor and an entrance at -2m

 

        P₁-P₂ = ½ 1.0 10³ (7² - 2²) + 1.0 10³ 9.8 (5 + 2)

       P₁-P₂ = 22.5 10³ + 68.6 10³

       P₁- P₂ = 91.1 10³ Pa

3 0
3 years ago
4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistanc
Anon25 [30]

a) 15.2 m/s^2

b) 1.96 N

c) 1.96 N

Explanation:

a)

To find the box's acceleration, we have to find first the net force acting on the box in the horizontal direction.

We have:

- The forward force of 3.55 N

- The backward, resistive force of 0.52 N

So, the net force forward is

\sum F=3.55-0.52=3.03 N

Now we can find the acceleration by using Newton's second law of motion, which states that:

\sum F=ma

where

m = 0.200 kg is the mass of the box

a is its acceleration

And solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2

b)

The gravitational force on an object is the force with which the object is pulled towards the ground by the Earth.

It is given by

W=mg

where

m is the mass of the object

g is the gravitational field strength

In this problem we have

m = 0.200 kg is the mass of the box

g=9.8 m/s^2 is the gravitational field strength

So, the gravitational force on the box is

W=(0.200)(9.8)=1.96 N

c)

The normal force is the reaction force exerted by the floor on the box, in the upward direction.

In order to find the magnitude of this force, we apply Newton's second law of motion along the vertical direction.

We have two forces in this direction:

- The gravitational force, W, downward

- The normal force, N, upward

So the net force is

\sum F=N-W

According to Newton's second law,

\sum F=ma

However, the box is at rest in the vertical direction, so the vertical acceleration is zero:

a=0

This means that the net force is zero:

\sum F=0

And so, we can find the normal force:

N-W=0\\N=W=1.96 N

4 0
3 years ago
Help again i need it quick
Romashka [77]

Answer:

it seems that the answer should be 7.5 mph as the average

Explanation:

sorry i did (s x t) when it was (s/t)

3 0
2 years ago
Read 2 more answers
If the mass is displaced 0.270 m from equilibrium and released, what is its speed when it is 0.130 m from equilibrium?
puteri [66]

Answer:

The speed is 1.52 m.

Explanation:

Given that,

Displacement =0.270 m

Distance = 0.130 m

Suppose a 0.321-kg mass is attached to a spring with a force constant of 13.3 N/m.

We need to calculate the angular velocity

Using formula of angular velocity

\omega=\sqrt{\dfrac{k}{m}}

Put the value into the formula

\omega=\sqrt{\dfrac{13.3}{0.321}}

\omega=6.43\ rad/s

We need to calculate the velocity

Using formula of velocity

v=\omega\sqrt{A^2-x^2}

Put the value into the formula

v=6.43\times\sqrt{0.270^2-0.130^2}

v=1.52\ m/s

Hence, The speed is 1.52 m.

6 0
3 years ago
What are the different skills and strategies were you aware of in this situation that indicated to you that it would be appropri
const2013 [10]

Answer:

dad

Explanation:

daddy

4 0
2 years ago
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