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OverLord2011 [107]
4 years ago
7

Choose the correct statement regarding energy production in the interior of a star and energy production in a power plant. A sta

r uses fusion as an energy source by building larger atoms from smaller atoms. A nuclear power plant uses fusion as an energy source by building larger atoms from smaller atoms. A nuclear power plant uses fusion as an energy source by breaking apart atoms. A star uses fusion as an energy source by breaking apart atoms.
Physics
2 answers:
sergij07 [2.7K]4 years ago
8 0

Answer:

A star uses fusion as an energy source by building larger atoms from smaller atoms.

Explanation:

Nuclear fission and fusion are two processes at which an atomic nucleus is changed to produce energy. Fission is the process splitting heavy atomic into lighter atomic nuclei.

So, fusion is the combination of smaller atoms to form larger atoms and star uses this as source of energy.

Fusion is the process at which light atomic nuclei are merged or fused together to form heavier nuclei.

The energy source for all stars is nuclear fusion. In a nuclear fusion reaction, the nuclei of two atoms combine to create a new atom. Most commonly, in the core of a star, two hydrogen atoms fuse to become a helium atom.

katrin2010 [14]4 years ago
7 0

Answer:

Option A) A star uses fusion as an energy source by building larger atoms from smaller ones

Explanation:

The energy source of stars is by nuclear fusion. Nuclear fusion is the process whereby two smaller nuclei combine to form a larger nucleus. The sun is an example of a star which is made up of mostly hydrogen and helium atoms. Two hydrogen atoms combine in the core of the star to form a helium atom.

Due to the fusion of hydrogen atoms that takes place in the sun to produce a helium atom, a great amount of heat is provided for the sun to utilize.

Nuclear power plants use heat produced during nuclear fission to heat water. In nuclear fission, atoms are split apart to form smaller atoms, thereby releasing energy.

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A 910 kg car is approaching a loop-the-loop. The loop has a diameter of 50 m. Determine the minimum speed the car must have at t
scoundrel [369]

Answer:

The minimum speed the car must have at the top of the loop to not fall = 35 m/s

Explanation:

Anywhere else on the loop, the speed needed to keep the car in the loop is obtained from the force that keeps the body in circular motion around the loop which has to just match the force of gravity on the car. (Given that frictional force = 0)

mv²/r = mg

v² = gr = 9.8 × 25 = 245

v = 15.65 m/s

But at the top, the change in kinetic energy of the car must match the potential energy at the very top of the loop-the-loop

Change in kinetic energy = potential energy at the top

Change in kinetic energy = (mv₂² - mv₁²)/2

v₁ = velocity required to stay in the loop anywhere else = 15.65 m/s

v₂ = minimum velocity the car must have at the top of the loop to not fall

And potential energy at the top of the loop = mgh (where h = the diameter of the loop)

(mv₂² - mv₁²)/2 = mgh

(v₂² - v₁²) = 2gh

(v₂² - (15.65)²) = 2×9.8×50

v₂² - 245 = 980

v₂² = 1225

v₂ = 35 m/s

Hence, the minimum speed the car must have at the top of the loop to not fall = 35 m/s

4 0
3 years ago
The object distance for a concave lens is 8.0 cm, and the image distance is 12.0 cm. The height of the object is 4.0 cm. What is
swat32
The answer for this question, If I am correct, should be answer "D".
3 0
3 years ago
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will the value of g be affected by the radius of the earth? consider the real shape of the earth. compare the acceleration due t
sp2606 [1]

Answer:

Yes, the value of g affected by the radius.

Explanation:

The formula for the force of gravity of 2 objects is

F_{gravity} = G\frac{m_{1}m_{2}}{r^2}, where m1 and m2 are the masses of the 2 objects, r is the radius, and G is the gravitational constant, which is approximately 6.67 \cdot 10^{-11}.

Therefore, as the radius if bigger, the force of gravity is going to be smaller exponentially.  

5 0
4 years ago
A satellite with mass 500 kg is placed in a circular orbit about Earth (Mass= 5.98 x 10^24 kg), radius = (6.4 x 10^6), a distanc
gizmo_the_mogwai [7]

Explanation:

a) F = GmM / r²

F = (6.67×10⁻¹¹) (500) (5.98×10²⁴) / (6.4×10⁶ + 1.5×10⁶)²

F = 3200 N

b) F = ma

3200 = 500a

a = 6.4 m/s²

c) a = v² / r

640 = v² / (6.4×10⁶ + 1.5×10⁶)

v = 7100 m/s

4 0
3 years ago
Astronomy question, <br> what keeps supernova explosions from being seen?
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a star that suddenly increases greatly in brightness because of a catastrophic explosion that ejects most of its mass.
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