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3 years ago

Strong solar winds blew dust and gas out of the solar system during which phase of the development of the Sun?

2 answers:

8 0

The choices can be found elsewhere and as follows:

a. Alpha Centauri 
c. T-tauri 
b. The Big Bang 
d. Nebular

I believe the correct answer from the choices listed above is option D. Strong solar winds blew dust and gas out of the solar system during Nebular phase. This seems to be the most logical option from the choices. Hope this helps. Have a nice day.

sergejj [24]3 years ago

6 0

its T-tauri on Edge nuity

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A star is a ___ that emits energy produced by nuclear reactions in its interior.

grigory [225]

A star is a large ball of gas that emits energy produced by nuclear reactions in the star's interior. Much of this energy is emitted as electromagnetic radiation, including visible light. Light emitted by stars enables other objects in the universe to be seen by reflection.

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3 years ago

Which of the following statements correctly describes the models put forth by Ptolemy and Copernicus to explain the universe? A.

FrozenT [24]

The correct option is D.
The model developed by Ptolemy has a lot of inconsistency and during the middle age additional explanation was offered for the claims made by the model. The model was very complicated because it was based on erroneous assumptions.
Copernicus model was simpler and some of his claims were correct.

8 0

3 years ago

Read 2 more answers

It takes you 5 min to walk with an average velocity of .75 m/s to the north from the parking lot to the entrance of the amusemen

skad [1K]

A displacement is a vector quantity that takes into account the shortest distance from the starting point to the endpoint.

The given above gave a time interval in minutes which needs to be converted to seconds. Given that each minute is 60 seconds, 5 minutes equal 300 seconds. To determine the distance, multiply time with speed. The product is 225 m.

Thus, the displacement is 225 m.

3 0

3 years ago

A 30-mm-diameter copper rod is 1 m long with a yield strength of 70 MPa. Determine the axial force necessary to cause the diamet

ivolga24 [154]

Explanation:

Given data:

d = 30 mm = 0.03 m

L = 1m

S $_{y}$ = 70 Mpa

Δd = -0.0001d

Axial force = ?

validity of elastic deformation assumption.

Solution:

O'₂ = Δd/d = (-0.0001d)/d = -0.0001

For copper,

v = 0.326 E = 119×10³ Mpa

O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶

∵δ = F.L/E.A and σ = F/A so,

σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa

F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN

S $_{y}$ = 70 MPa > σ = 36.5 MPa

∵ elastic deformation assumption is valid.

so the answer is

F = 25800 K N and S $_{y}$ > σ

3 0

3 years ago

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

wariber [46]

Answer:

For left = 0 N/C

For right = 0 N/C

At middle = $-7.6836 * 10^{-11} \vec{i}$ N/C

Explanation:

Given data :-

б = $6.8 * 10^{-22}$ C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

$E = \frac{\sigma }{2\varepsilon }$

1) To the left of the plate

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C

5 0

3 years ago