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Law Incorporation [45]
3 years ago
15

Briefly describe the formation of the planets from the solar nebula

Physics
1 answer:
saul85 [17]3 years ago
6 0
<span>The formation of the Solar System began 4.6 billion years ago with the gravitational collapse of a small part of a giant molecular cloud. Most of the collapsing mass collected in the center, forming the Sun, while the rest flattened into a proto-planetary disk out of which the planets, moons, asteroids, and other small Solar System bodies formed.


Hope it helped
</span>
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What causes a solar eclipse? Select all that apply.
brilliants [131]

I think it is either D) or E)

But i am going to go with E)

4 0
3 years ago
Read 2 more answers
What effect or effects would be most significant if the Moon's orbital plane were exactly the same as the ecliptic plane?
Lilit [14]

Answer:

c. Solar eclipses would be much more frequent.

Explanation:

The <u>ecliptic plane</u> is the apparent orbit that the sun describes around the earth (although it is the earth that orbits the sun), is the path the sun follows in earth's sky.

A <u>solar eclipse</u> occurs when the moon gets between the earth and the sun, so a shadow is cast on the earth because the light from the sun is blocked.

The reason why solar eclipses are not very frequent is because the moon's orbital plane is not in the same plane as the orbit of the earth around the sun, but rather that it is somewhat inclined with respect to it.

So <u>if both orbits were aligned, the moon would interpose between the sun and the earth more frequently, producing more solar eclipses.</u>

So, if the moon's orbital plane were exacly the same as the ecliptic plane solar eclipses would be more frequent.

the answer is: c.

8 0
4 years ago
What is the potential energy of an object 20 m in the air with a<br> mass of 600 kg?
Lana71 [14]

Answer:

Ep = 117600 J

Explanation:

Data:

  • Mass (m) = 600 kg
  • Height (h) = 20 m
  • Gravity (g) = 9.8 m/s²
  • Potential Energy (Ep) = ?

Use formula:

  • Ep = m * g * h

Replace:

  • Ep = 600 kg * 9.8 m/s² * 20 m

Multiply operations, and units:

  • Ep = 117600 J

What is the potential energy?

The potential energy is <u>117600 Joules.</u>

7 0
3 years ago
A person wearing a shoulder harness can survive a car crash if the acceleration is smaller than -300 m/s . assuming constant acc
mars1129 [50]

To solve this problem, we use the equation:

<span>d = (v^2  - v0^2) / 2a</span>

 

where,

d = distance of collapse

v0 = initial velocity = 101 km / h = 28.06 m / s

v = final velocity = 0

a = acceleration = - 300 m / s^2

 

d = (-28.06 m / s)^2 / (2 * - 300 m / s^2)

<span>d = 1.31 m</span>

3 0
3 years ago
A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)
wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

7 0
3 years ago
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