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irina [24]
2 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]2 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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QveST [7]

Answer:

Please see attachment

Explanation:

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2 years ago
A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106 hz and 1hz=
AURORKA [14]
The frequency of the radio station is
f=88.7 fm= 88.7 MHz = 88.7 \cdot 10^6 Hz

For radio waves (which are electromagnetic waves), the relationship between frequency f and wavelength \lambda is
\lambda= \frac{c}{f}
where c is the speed of light. Substituting the frequency of the radio station, we find the wavelength:
\lambda= \frac{3 \cdot 10^8 m/s}{88.7 \cdot 10^6 Hz}=3.38 m
6 0
2 years ago
Read 2 more answers
A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a consta
-Dominant- [34]

The car would go from  zero to 58.0 mph in 2.6 sec.

Since the force on the car is constant, therefore the acceleration of the car would also be constant.

Now for constant acceleration we can use the equation of motion

Using first equation of motion to calculate the acceleration of the car

v=u+at

29=0+a×1.30       ...... Eq. (1)

Again using the first equation of motion

58=0+a*t             ....... Eq. (2)

Dividing eq. (2) with equation 1

t=2×1.3

t=2.6 sec

7 0
2 years ago
1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra
UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.  

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

F_r = mgsin\theta

F_r = \mu N

F_r = \mu mg cos\theta

Matching the two equation we have,

\mu N = \mu mg cos\theta

\mu = tan\theta

Applying energy conservation,

\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1

h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})

Frictionless surface

\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2

\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2

h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between h_1 and h_2 is given by

\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}

\frac{h_1}{h_2} = \frac{1}{2}

8 0
2 years ago
Tori experiments with pulleys in physics class. She applies 70 newtons of force to a single pulley to lift a bowling ball. By ad
e-lub [12.9K]

The minimum input force she'll need to lift the ball is 35 N.

Explanation:

Mechanical advantage of a single pulley is 1. As, she applies 70 N of force to lift the bowling ball, so the output force(weight of the ball) is also 70 N.

Now, adding another pulley gives a mechanical advantage of 2. We have,

M.A = (Output Force)/(Input Force)

Substituting the values we get,

2=\frac{70}{F_{i}}

F_{i}=\frac{70}{2} = 35 N

Input force equals to 35 N needs to be applied.

6 0
2 years ago
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