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Eddi Din [679]
3 years ago
6

Which of the following scientists taught physics in the United States?

Physics
2 answers:
Sergio039 [100]3 years ago
6 0
<span>Subrahmanyan Chandrasekhar

</span>
GREYUIT [131]3 years ago
6 0

Answer : Option C) Subrahmanyan Chandrasekhar

Explanation : Scientist Subrahmanyan Chandrasekhar was a long-time professor in the University of Chicago. He has been awarded noble prize for Physics during his life time. He was an Indian by origin but later settled in US for his professional life.

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Describe Charle’s Law and Boyle’s Law.
ch4aika [34]

Explanation:

Boyle's law can be stated as the "volume of a fixed mass of a gas varies inversely as the pressure changes if the temperature is constant". It is mathematically expressed as;

    P1 V1  = P2 V2

P1 is the initial pressure

V1 is the initial volume

P2 is the final pressure

V2 is the final volume

Charles's law states that the volume of a fixed mass of a gas varies directly as its absolute temperature if the pressure is constant.

It is mathematically expressed as;

            \frac{V1}{T1}    = \frac{V2}{T2}  

V and T are volume and temperature respectively

1 and 2 are the initial and final states

8 0
3 years ago
What function do you think underground hyphae perform?
Helen [10]
Mushroom may be, didn't learn that it. Well, mushrooms' whole fungus is underground.

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4 years ago
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The aorta is approximately 25 mm in diameter. The mean pressure there is about 100 mmHg and the blood flows through the aorta at
Ksenya-84 [330]

Answer:

Explanation:

25 mm diameter

r₁  = 12.5 x 10⁻³ m radius.

cross sectional area =  a₁

Pressure P₁  = 100 x 10⁻³ x 13.6 x 9.8 Pa

a )

velocity of blood v₁ = .6 m /s

Cross sectional area at blockade = 3/4 a₁

Velocity at blockade area = v₂

As liquid is in-compressible

a₁v₁ = a₂v₂

a₁ x .6 m /s  = 3/4 a₁ v₂

v₂ = .8m/s

b )

Applying Bernauli's theorem formula

P₁ + 1/2 ρv₁² =  P₂ +  1/2 ρv₂²

100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ +  1/2x 1060 x .8²

13328 +190.8 = P₂ + 339.2

P₂ = 13179.6 Pa

= 13179 / 13.6 x 10³ x 9.8 m of Hg

P₂ =  .09888 m of Hg

98.88 mm of Hg

8 0
4 years ago
Scientists experimenting with two charged objects of 2.56 x 10-6 C and 3.34 x 10-7 C displayed a repulsion of 2.26 x 10-3 N. How
Pachacha [2.7K]

Answer:

The objects were 1.8m apart.

Explanation:

We will start stating the Coulomb's Law. It says that:

F_e=\frac{Kq_1q_2}{r^{2}}

Where F_e is the electric force between the objects, q_1 and q_2 are the magnitude of the charge of the objects, r is the distance between them and K is the Coulomb's constant (K=8.9*10^{9} \frac{Nm^{2} }{C^{2} } in vacuum). Solving for the distance r we have:

r=\sqrt{\frac{Kq_1q_2}{F_e} }

Plugging the given values into this equation, we obtain:

r=\sqrt{\frac{(8.9*10^{9}\frac{Nm^{2} }{C^{2} })(2.56*10^{-6}C)(3.34*10^{-7}C)}{2.26*10^{-3}N}}=1.8m

In words, the two charged objects were 1.8m apart.

6 0
3 years ago
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Which statement is not true about mass movements?
ololo11 [35]

B i think that the answer so

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3 years ago
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