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Ber [7]
3 years ago
6

The Andromeda Galaxy is faintly visible to the naked eye in the constellation Andromeda. Suppose instead it were located in the

same direction in space as the center of the Milky Way Galaxy (but still at its current distance). How would it appear to the eye in that case?
a) We could not see it at all.
b) It would look about the same, but would be in the constellation Sagittarius instead of Andromeda.
c) It would be much brighter, because it would be illuminated by the many stars in the center of our galaxy.
d) It would look about the same, but it would be harder to pick out because its cloud-like appearance would make it blend in with the cloud-like appearance of the Milky Way in our sky.
Physics
1 answer:
Kaylis [27]3 years ago
8 0

Answer:

a) We could not see it at all.

Explanation:

The most distant object that can be seen is the andromeda galaxy, which we may have a slight view of. The andromeda galaxy is a large galaxy that along with the previous two is also part of the local group. Spiral-type galaxy that is approximately 250,000 light years in diameter (more than twice the diameter of the Milky Way!) And is about 2.9 million light years away from our galaxy. Because of its distance, we have difficulty visualizing this galaxy, we would have this difficulty even if the andromeda galaxy was in the center of the Milky Way, but maintaining its current distance. That is, even if the andromeda galaxy were located in the same direction in space as the center of the Milky Way (but still at its current distance), we could not see it at all.

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As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases
frozen [14]
Bohr's equation for the change in energy is
\Delta E= \frac{hc}{\lambda}
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
     TRUE

b. As λ increases, f increases and ΔE increases.
    FALSE

c. As λ increases, f increases and ΔE decreases.
    FALSE

Answer: 
As the wavelength increases, the frequency decreases and energy decreases.

3 0
3 years ago
Read 2 more answers
A wall clock has a minute hand with a length of 0.55 m and an hour hand with a length of 0.26 m. Take the center of the clock as
Lemur [1.5K]

Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

Explanation:

Given that,

Length of minute hand = 0.55 m

Length of hour hand = 0.26 m

The time taken by the minute hand to complete one revelation is

T= 3600\ sec

We need to calculate the angular frequency

Using formula of angular frequency

\omega=\dfrac{2\pi}{T}

Put the value into the formula

\omega=\dfrac{2\pi}{3600}

\omega=0.001745\ rad/s

We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock

Using formula of acceleration

a=r\omega^2

Put the value into the formula

a=0.55\times(0.001745)^2

a=1.675\times10^{-6}\ m/s^2

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

3 0
3 years ago
A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of
Drupady [299]

Answer:

4.5\times 10^{-5} T

Explanation:

We are given that

Current in wire=40 A

Magnetic field=B_1=3.5\times 10^{-5} T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

B_{wire}=B_2=\frac{\mu_0I}{2\pi R}

We have R=29 cm=\frac{29}{100}=0.29 m

1 m=100 cm

Substitute the values in the given formula

B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T

The resultant magnetic field is given by

B=\sqrt{B^2_1+B^2_2}

Substitute the values then we get

B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}

B=4.5\times 10^{-5} T

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire=4.5\times 10^{-5} T

Hence, the resultant magnitude of the magnetic field 29 cm above  the wire=4.5\times 10^{-5} T

7 0
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What single property was the most important in jesseca's material
jek_recluse [69]

Answer:

Jesseca wanted to create a material that reflected most of the light that fell on it.

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6 0
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Select all of the statements that are true.
Strike441 [17]

I think its all four of them could be wrong but try all four !!!!!!

7 0
3 years ago
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