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valina [46]
3 years ago
5

consider the hypothetical observation irregular galaxies outside the local group are moving toward us. from part a, this observa

tion would contradict the idea of an expanding universe. why?
Physics
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

spiral galaxies move away from us 10% faster than elliptical galaxies at the same distances;  irregular galaxies outside the Local Group are moving toward us;  galaxy speeds are faster in summer than in winter

Explanation:

This may help I'm not sure.

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Describe what happens to the energy of a wave if the frequency decreases and the frequency increased
Nata [24]

"As frequency increases, wavelength decreases. Frequency and wavelength are inversely proportional. This basically means that when the wavelength is increased, the frequency decreases and vice versa. Wavelength is described as the distance between a trough to a trough or a crest to a crest."

I'd recommend paraphrasing it tho.

4 0
3 years ago
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Where might someone with a pharmaceutical manufacturing career work?
klemol [59]

Answer:  People with careers in pharmaceutical manufacturing can expect to work in: research laboratories or manufacturing plants, and to maintain work hours during the day.

Explanation:

8 0
3 years ago
The power output of a tuba is 0.35 W. At what distance is the sound
STALIN [3.7K]

The distance of the sound from the tuba is 4.82 m.

<h3>Area of the tube</h3>

The area of the tuba is calculated as follows;

I = P/A

where;

  • I is intensity of sound
  • P is power
  • A is area

A = P/I

A = 0.35 / (1.2 x 10⁻³)

A = 291.67 m²

<h3>Distance of the sound</h3>

Area = 4πr²

r = \sqrt{\frac{A}{4\pi} } \\\\r = \sqrt{\frac{291.67}{4\pi} } \\\\r = 4.82 \ m

Thus, the distance of the sound from the tuba is 4.82 m.

Learn more about intensity of sound here: brainly.com/question/4431819

8 0
3 years ago
An object is dropped from rest from a height of 4.4 107 m above the surface of the Earth. If there is no air resistance, what is
kondaur [170]

Answer:

Explanation:

An object is drop from a height, then it is in direction of gravity

g is +ve

When an object drop from a height, the initial velocity is 0,

U=0

Given that h=4.4×10^7m

V=?

g=9.81m/s^2

Then using equation of motion

V^2=U^2+2gh

V^2=0+2×9.81×4.4×10^7

V^2=86.33×10^7m/s

Take square root of both side

V=29381.63m/s

Now to km/s, divide by 1000

Since 1km=1000m

V=29.38km/s

6 0
3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction tim
miss Akunina [59]

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

a)

  • In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
  • The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
  • We can find this distance simply applying the definition of average velocity, as follows:

       \Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)

  • The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
  • We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

       v_{1f} ^{2}  - v_{1o} ^{2} = 2* a* \Delta x  (2)

  • where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
  • Solving for Δx, we get:

       \Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)

  • So, the total distance traveled was the sum of (1) and (3):
  • Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
  • Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

b)

  • We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
  • The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
  • ⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
  • Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

       \Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)

  • Δx₂, is the distance traveled while decelerating, and can be obtained  using (2):

        v_{omax} ^{2} = 2* a* \Delta x_{2} (7)

  • Solving for Δx₂, we get:

       \Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a}  = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)

  • Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
  • Taking the positive root in the quadratic formula, we get the following value for vomax:
  • v₀max = 24.3 m/s.
6 0
3 years ago
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