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brilliants [131]
4 years ago
5

What causes interstellar dust and clouds to condense in the planets and stars

Physics
1 answer:
lorasvet [3.4K]4 years ago
3 0
That would be a nebula, which is an interstellar cloud of hydrogen gas, dust, and plasma. It is the first stage of a star's cycle.
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You ride on an elevator that is moving with constant upward acceleration while standing on a bathroom scale. the reading on the
Blababa [14]

The reading on the scale is greater than your actual weight.

4 0
4 years ago
Which of these best describes the relationship between the incident ray, the reflected ray, and the normal for a curved mirror?(
lions [1.4K]

For a curved mirror, all points have the same normal and the angle of incidence is also equal to the angle of reflection.

According to the laws of reflection, the incident ray, reflected ray and normal all lie on the same plane. For a curved mirror, the normal remains the same at all points along the curved mirror.

Again, the angle made between the incident ray and the normal is the same as the angle made between the reflected ray and the normal. Therefore, the angle of reflection is equal to the angle of incidence.

Learn more: brainly.com/question/17638582

8 0
2 years ago
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7
Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

v = \frac{I}{nqA}

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}                  

Now, we can find the drift speed:

v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

4 0
3 years ago
the physics concept corresponding to oomph is momentum. building on your above answers, figure out a formula for momentum (oomph
topjm [15]

Momentum = mass * velocity

I need parts A and B to explain the intuitions.

7 0
3 years ago
Consider the system consisting of two blocks connected by a rope and a pulley. The coefficient of static friction between the ra
RSB [31]

Answer:

1.2 kg

Explanation:

Let UP ramp be the positive direction

                                                  F = ma

   T     -   Wt ||      -         Ff            = m(0)

  mg   -  Μgsinθ -      μΜgcosθ    = 0

m(9.8) - 13sin35 - 0.36(13)cos35 = 0        

                                                 m = 13(sin35 + 0.36cos35) / 9.8

                                                 m = 1.15205... ≈ 1.2 kg

5 0
3 years ago
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