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grandymaker [24]
3 years ago
13

Which of the following scientists won a noble prize of pioneering work in the study of the evolution of stars?

Physics
2 answers:
zhenek [66]3 years ago
7 0
Subrahmanyan chandrasekhar
Svet_ta [14]3 years ago
5 0

Subramanyan Chandrasekhar and William Alfred Fowler won a noble prize of pioneering work in the study of the evolution of stars.

The prize was divided equally between both scientists in 1983.

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A 320 g rubber ball is dropped from 2.0 m above the ground, bounces, and returns to a maximum height 1.2 m above the ground. If
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Answer:

0.71 J

Explanation:

320 g = 0.32 kg

According to law of energy conservation, the energy loss to external environment (air, ground) can be credited to the change in mechanical energy of the ball.

As the ball was dropped at H = 2 m above the ground then later reaches its maximum height at h = 1.2m, tts instant speed at those 2 points must be 0. So the kinetic energy at those points are 0 as well. The change in mechanical energy is the change in potential energy.

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\Delta E_p = mgH - mgh = mg(H - h) = 0.32*9.81*(2 - 1.2) = 2.51 J

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The particle model of light explains which light behaviors?
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3 years ago
Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o
KATRIN_1 [288]

Answer:

1.36\times 10^{-7} C

Explanation:

We are given that

Mass of charged tine spheres=m=1 g=\frac{1}{1000}=0.001 kg

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Let q be the charge on each sphere.

k=9\times 10^9Nm^2/C^2

The electric force between two charged particle

F=\frac{kq_1q_2}{r^2}

Using the formula

The force between two charged tiny spheres=F_e=\frac{kq^2}{(0.02)^2}

According to  Newton's second law , the net force

F=ma

F=F_e

0.001\times 414=\frac{9\times 10^9\times q^2}{(0.02)^2}

q^2=\frac{0.001\times 414\times (0.02)^2}{9\times 10^9}

q=\sqrt{\frac{0.001\times 414\times (0.02)^2}{9\times 10^9}}

q=1.36\times 10^{-7} C

Hence, the magnitude of charge on each tiny sphere=1.36\times 10^{-7} C

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