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user100 [1]
3 years ago
6

Could an average star, such as our sun, become a neutron star? Explain your answer.

Physics
1 answer:
larisa86 [58]3 years ago
5 0
<span>No. Neutron stars are the remnants of very large stars that have supernova'd. Anything below 1.44 solar masses becomes a dwarf, anything above 5 solar masses becomes a black hole. Everything in between becomes a neutron star (or quark star, but it's not proven).</span>
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Fill in the blanks below:<br><br> Urgently need help!!!
miskamm [114]

Answer:

In a circuit ,<u> VOLTAGE </u>can be said to be the "source" or the "push of electrons". This push then creates what is known as a <u> CURRENT , </u>which is the flow of electric charge through the circuit. This flow can the slowed down or restricted by <u>RESISTOR </u>, and this is also what can be harnessed in order to use electric <u>ENERGY </u>.

Explanation:

Voltage:

It is the 'push' that causes charges to move in a wire or other electrical conductor, also it is a Source input to the electric circuit.

Measured in Volts.

Current:

An electric current is the rate of flow of electric charge from a point or through a region.

Measured in Ampere.

Resistor:

Resistor is used to resist the flow of charge or to resist the current called as Resistance.

Measured in Ohms.

Electric Energy:

Electrical energy is a form of energy resulting from the flow of electric charge.

Measured in Joules.

In a circuit , voltage can be said to be the "source" or the "push of electrons". This push then creates what is known as a current, which is the flow of electric charge through the circuit. This flow can the slowed down or restricted by resistor, and this is also what can be harnessed in order to use electric energy.

8 0
3 years ago
A jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction
Elis [28]

Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

v=\frac{10}{2}=5m/s\;\cdots(i)

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, d=vt

From equation (i)

\Rightarrow d=5t\;\cdots(ii)

As the jogger starts from origin, so, the distance, d, also represents the position of the jogger at the time t s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

d=5\times 5=25 m

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

d=5\times 15=75 m

So, the jogger is at a distance of 75 m from the origin.

8 0
3 years ago
2. Which statement is correct about the Krebs cycle?
IRINA_888 [86]

The correct option is


a. Acetyl-CoA combines with a pyruvic acid to make glucose in the Krebs cycle.


Explanation:

The Krebs citric acid cycle happens within the mitochondrial matrix and generates a pool of energy (ATP, NADH, and FADH2) from the oxidization of pyruvate, the tip product of metabolism. Pyruvate is transported into the mitochondria and loses dioxide to make acetyl-CoA, a 2-carbon molecule.

5 0
3 years ago
Read 2 more answers
Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th
Kipish [7]

Answer:

Probability of tunneling is 10^{- 1.17\times 10^{32}}

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J

Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

Thus

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = 1.52\times 10^{-35}\ m

Now,

We can calculate the tunneling probability as:

P(t) = e^{\frac{- 2t}{\eta}}

P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}

P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

6 0
3 years ago
Which stage follows the red giant stage of star development, if the star is a high-mass star? 
Anestetic [448]
I believe it's a supernova 
3 0
4 years ago
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