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GalinKa [24]
3 years ago
6

Stars form from clouds of gas and dust. As a protostar gravitationally contracts within its parent cloud, "conservation of angul

ar momentum" says thatA) the protostar rotates more slowly.b) the protostar\'s rotation does not change.c) the protostar\'s axis of rotation will change direction.d) the protostar rotates more quickly.
Physics
1 answer:
vredina [299]3 years ago
5 0

Answer:

D) True. the protostar rotates more quickly.

Explanation:

If the system is isolated, the angular momentum must be retained.

Initial

        L₀ = I w₀

Final

       L_{f} = I_{f}  w_{f}

      L₀ = L_{f}

      I w₀ = I_{f}w_{f}

     w_{f}  = I /I_{f} w₀

In general, the radius of the cloud decreases significantly to form the star, the moment of inertia must decrease, so the angular velocity must increase

Let's examine the answers

A) False. The opposite happens

B) False. Speed ​​changes

C) False. For this there must be an external force, which does not exist

D) True. You agree with the above

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A 300W hot plate produces 45,000 J of thermal energy while operating for 2 min. What is the efficiency of this devide? it need t
Gekata [30.6K]

Answer:

Explanation: P = 300 W   and t = 2 min = 120 s

Energy Q = Pt = 300 W · 120 s = 36 000 J.

Thus, plate can not produce 45 000 J heat.

5 0
3 years ago
An oil droplet is sprayed into a uniform electric field of adjustable magnitude. The 0.11 g droplet hovers
ohaa [14]

Answer:

The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.

F = N e E     where E is the value of the field and N e the charge Q

M g = N e E      and M g is the weight of the drop

N = M g / (e E)

N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13

.00011 kg is a very large drop

Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs

Check:     N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13   electrons

7 0
2 years ago
A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The d
blsea [12.9K]

The average power needed to produce this final speed is 1069.1 hp.

Mass of the dragster,  m = 500.0 kg,

Displacement travelled by the dragster,  s = 402 m,

Time taken in this travel,  t = 5.0 s,

Final velocity of the dragster,  v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation,  s = ut + (1/2)at^2.

402  =  u*5  + (1/2)*a*5^2

10*u + 25*a  = 804.      ...........(1)

Using kinematical equation, v = u +at.

130 = u + 5*a

5*u + 25*a = 650.       .............(2)

Solving (1)and (2), we get,

u =  30.8 m/s.

According to work-energy theorem,

Work done = change in kinetic energy

W  = (1/2)*m*(v^2 - u^2)

W = (1/2)*500*(130^2 - 30.8^2)

W  =  3987840. J

Therefore power rating of the dragster is given by,

P  ⇒  W/t. =  3987840/5 = 797568 watt.

P  ⇒ 797568/746 =  1069.1 hp.

Learn more about Power rating here brainly.com/question/20137708

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2 years ago
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The field-line representation of the e-field in a certain region in space is shown below. The dashed lines represent equipotenti
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You need to go on google an dlook this up

3 0
3 years ago
How long would it take an object to reach the ground from the top of a building that is 470 feet tall? Round to the nearest tent
Zinaida [17]

Answer:

It would take the object 5.4 s to reach the ground.

Explanation:

Hi there!

The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).

t = time

Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:

h = h0 + 1/2 · g · t²

We have to find the time at which h = 0:

0 = 470 ft - 1/2 · 32.2 ft/s² · t²

Solving for t:

-470 ft = -16.1 ft/s² · t²

-470 ft / -16.1 ft/s² = t²

t = 5.4 s

3 0
3 years ago
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