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fenix001 [56]
3 years ago
15

What determines if a star turns into a red giant or a red supergiant?

Physics
1 answer:
mixer [17]3 years ago
3 0

Answer:

Explanation:

As stars age, they evolve away from the main sequence to become red giants or supergiants. The core of a red giant is contracting, but the outer layers are expanding as a result of hydrogen fusion in a shell outside the core. The star gets larger, redder, and more luminous as it expands and cools.

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What is the basic building block of all matter
Natalka [10]
I think that is atom. 
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3 years ago
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If the pitched ball was traveling 77 mph before stanton's bat hit it and 120 mph after his bat hit it, by what amount did the sp
Mice21 [21]

here we will use the concept of Newton's III law

as per Newton's III law the impulse given to the ball is same as the impulse lost by the bat

So here we will say

impulse gain by the ball = impulse lost by the bat

m_1(v_f - v_i) = m_2(\Delta v)

given that

m_1 = 5 ounce

m_2 = 32 ounce

For ball the change in speed will be

v_f - v_i = (120 - 77)mph

now from above equation

5\times (120 - 77) = 32 \times \Delta v

\Delta v = 6.72 mph

so speed of bat will decrease by 6.72 mph

3 0
2 years ago
!!please help !!!!! I’ll give brainliest
Oxana [17]
The answer is 45 degrees
7 0
3 years ago
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A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
Stells [14]

Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

5 0
3 years ago
In a few sentences, compare and contrast pressurized water reactors and boiling water reactors.
kondaur [170]
For pressurized water reactors the coolant is not permitted to boil in the core of the PRW, however the coolant in boiling water reactors is permitted to do so in the core of BWR. Pressurized water reactors have an indirect cycle. Whereas, the boiling water reactors go through a direct cycle. Both are light water reactors. 
6 0
3 years ago
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