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viktelen [127]
3 years ago
5

A galaxy that is a featureless spherical ball of stars would be called a type

Physics
1 answer:
satela [25.4K]3 years ago
7 0

Answer:

E0

Explanation:

Yes. The "0" indicates that it is spherical

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What are the conditions of the thermonuclear fusion reaction in the sun?​
Verizon [17]

High temperature gives the hydrogen atoms enough energy to overcome the electrical repulsion between the protons. Fusion requires temperatures of about 100 million Kelvin (approximately six times hotter than the sun's core).

study \: well

8 0
2 years ago
What is the period of a sound wave having a frequency of 340 Hz
sertanlavr [38]

Answer:

lambda = 343 m/s divided by 340 Hz = 1.009 seconds

Hope it helps and have a wonderful day!

4 0
3 years ago
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Newton’s first law relates motion to balanced and unbalanced forces.
BlackZzzverrR [31]
True, an object at rest stays and rest and an object in motion stays in motion
8 0
3 years ago
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A playground merry-go-round has a radius of 4.6 m and a moment of inertia of 200 kg-m2 and turns with negligible friction about
tankabanditka [31]

Answer:

8050 J

Explanation:

Given:

r = 4.6 m

I = 200 kg m²

F = 26.0 N

t = 15.0 s

First, find the angular acceleration.

∑τ = Iα

Fr = Iα

α = Fr / I

α = (26.0 N) (4.6 m) / (200 kg m²)

α = 0.598 rad/s²

Now you can find the final angular velocity, then use that to find the rotational energy:

ω = αt

ω = (0.598 rad/s²) (15.0 s)

ω = 8.97 rad/s

W = ½ I ω²

W = ½ (200 kg m²) (8.97 rad/s)²

W = 8050 J

Or you can find the angular displacement and find the work done that way:

θ = θ₀ + ω₀ t + ½ αt²

θ = ½ (0.598 rad/s²) (15.0 s)²

θ = 67.3 rad

W = τθ

W = Frθ

W = (26.0 N) (4.6 m) (67.3 rad)

W = 8050 J

6 0
3 years ago
A fluid, with a density of rho = 1165 kg/m3, flows in a horizontal pipe. In one segment of the pipe the flow speed is v1 = 4.53
Elza [17]

Answer:

The pressure difference between two pipe is 1.01 \times 10^{4} Pa

Explanation:

Density \rho = 1165 \frac{kg}{m^{3} }

Speed in one pipe v_{1} = 4.53 \frac{m}{s}

Speed in second pipe v_{2} = 1.77 \frac{m}{s}

According to the bernoulli equation,

The pressure difference is given by,

     P = \frac{1}{2} \rho v^{2}

P_{2} - P_{1} = \frac{1}{2}  \rho (v_{1}^{2}  - v_{2}^{2}  )

P_{2} - P_{1} = \frac{1}{2} \times 1165 \times[ (4.53)^{2}- (1.77)^{2}]

P_{2} - P_{1} = 10128.51

P_{2} - P_{1} = 1.01 \times 10^{4} Pa

Therefore, the pressure difference between two pipe is 1.01 \times 10^{4} Pa

6 0
3 years ago
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