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Alex_Xolod [135]
3 years ago
8

The total amount of energy in a closed system stays the same. t/f

Physics
1 answer:
zvonat [6]3 years ago
5 0
True. Law of Conservation of energy
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While planets are smaller than stars, planets are generally larger than which of the following?
Sauron [17]

the answer was B

hope that helps you

3 0
3 years ago
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Calculate the mean free path of air molecules at a pressure of 7.00×10^−13 atm and a temperature of 303 K . (This pressure is re
Grace [21]

Answer:

82.8986 km

Explanation:

Given:

Pressure = 7.00×10⁻¹³ atm

Since , 1 atm = 101325 Pa

So, Pressure = 7.00×10⁻¹³×101325 Pa = 7.09275×10⁻⁸ Pa

Radius = 2.00×10⁻¹⁰ m

Diameter = 4.00×10⁻¹⁰ m (2× Radius)

Temperature = 303 K

The expression for mean free path is:

\lambda (Mean\ free\ path)=\frac {K (Boltzmann\ Constant)\times Temperature}{\sqrt {2}\times \pi\times (Diameter)^2\times Pressure}

Boltzmann Constant = 1.38×10⁻²³ J/K

So,

\lambda (Mean\ free\ path)=\frac {1.38\times 10^{-23}\times 303}{\sqrt {2}\times \frac {22}{7}\times (4.00\times 10^{-10})^2\times 7.09275\times 10^{-8}}

<u>Mean free path = 82.8986×10³ m = 82.8986 km</u>

4 0
3 years ago
When an electric current flows through a long conductor, each free electron moves
nirvana33 [79]
<span>When an electric current flows through a long conductor, each free electron moves from one end of the other end. When an electric field is applied to a conductor (a wire) the free electrons of the conductor are subject to an electric force which will cause the electrons to move. Given that the electrons are negatively charged they will move counter-wise the field.. Each end of the wire is attached to one pole or end of a battery (or generator of electricity) then the electrons will move form the end joined to the negative pole toward the end attached to positive pole.</span><span />
8 0
3 years ago
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Can a conductor be given limitless charge
k0ka [10]

Answer:

No

Explanation:

You could try to give it enough to fill all valence electrons in all of the atoms in the conductor, but practically this could not be achieved.

5 0
2 years ago
A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec
kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, \lambda=\dfrac{\lambda_ox_o}{x}

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

dE=\dfrac{k\ dq}{x^2}..........(1)

The linear charge density is given by :

\lambda=\dfrac{dq}{dx}

dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx

Integrating equation (1) from x = x₀ to x = infinity

E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx

E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}

E=\dfrac{k\lambda_o}{2x_o}

Hence, this is the required solution.

5 0
3 years ago
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