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2 years ago

Question 2 of 10

Physics

1 answer:

makvit [3.9K]2 years ago

7 0

Answer:

D.Entropy tends to increase.

Explanation:

The second of thermodynamics states that the state of entropy of the entire universe,as an isolate system, will always increase over time

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My Notes An electron is released from rest on the axis of a uniform positively charged ring, 0.500 m from the ring's center. If

dsp73

Answer:

Velocity of the electron = v = 1.2\times 10^8\ m/s.

Explanation:

Given,

Mass of the electron = $m_e\ =\ 9\times 10^{-31}\ kg$
Charge on the electron = $q_e\ =\ 1.62\times 10^{-19}\ C$
Charge density of the ring = $\rho\ =\ +1.00\times 10^{-6}\ C/m$
Radius of the ring = R = 0.70 m
Distance between the electron ant the center or the ring = x = 0.5 m

Now total charge on the ring = $Q\ =\ \rho\times 2\pi R$

Potential energy due to the charged ring to the point on the x-axis is

$P.E.\ =\ \dfrac{KQq_e}{\sqrt{R^2\ +\ x^2}}\\$

Let v be the velocity of the electron at the center of the ring.

Total kinetic energy of the electron = $\dfrac{1}{2}m_ev^2\\$

Now, From the conservation of energy,

the total potential energy of the electron at initially is converted to the total kinetic energy of the electron at the center of the ring,

$\therefore P.E.\ =\ K.E.\\\Rightarrow \dfrac{KQ}{\sqrt{R^2\ +\ x^2}}\ =\ \dfrac{1}{2}m_ev^2\\\Rightarrow v\ =\ sqrt{\dfrac{2 KQq_e}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2Kq_e\rho \times 2\pi R}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2\times 9\times 10^9\times 1.0\times 10^{-6}\times 2\times 3.14\times 0.7\times 1.6\times 10^{-19}}{9\times 10^{-31}\times \sqrt{0.7^2\ +\ 0.5^2}}}\\\Rightarrow v\ =\ 1.2\times 10^8\ m/s.$

Hence the velocity of the electron on the center of the charged ring is $1.2\times 10^8\ m/s.$

5 0

3 years ago

A satellite is orbiting Earth with a distance R = 2REarth from Earth's center. If the satellite is moved to a distance R = 4REar

Maslowich

Answer:

Half

Explanation:

Given that:

radial distance of satellite from the earth, $R=2R_E$

Now, if the satellite is moved to a distance $R=4R_E$

We have the mathematical expression for the potential energy fue to gravitational field as:

$U=\frac{G.M.m}{R}$ ...................(1)

where:

$G = 6.67\times 10^{-11}\ m^3.kg^{-1}.s^{2}$

M = mass of earth

m = mass of satellite

R = radial distance of satellite

Now from eq. (1) initially we have:

$U=\frac{G.M.m}{2R_E}$

after the satellite is moved, we have:

$U'=\frac{G.M.m}{4R_E}$

$\Rightarrow U'=\frac{G.M.m}{2(2R_E)}$

$\Rightarrow U'=\frac{1}{2} \times U$

which is half of the initial condition.

3 0

3 years ago

Explain the relationship between air resistance and gravitational acceleration on a falling object

Allushta [10]

Air resistance, also called drag, acts upon a falling body by slowing the body down to thr point where it stops accelerating, and it falls at a constant speed, known as the terminal volocity of a falling object. Air resistance depends on the cross sectional area of the object, which is why the effect of air resistance on a large flat surfaced object is much greater than on a small, streamlined object.

7 0

3 years ago

What trends were seen in Mendeleev’s periodic table?

katrin [286]

He arranged his periodic table by each elements atomic mass

3 0

3 years ago

Read 2 more answers

What is the acceleration of a car that starts from rest and attains a final speed of 20 m

OlgaM077 [116]

Answer:

323.9