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2 years ago

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An electron is released from rest at a distance of 9.00 cm from a proton. If the proton is held in place, how fast will the elec

tron be moving when it is 3.00 cm from the proton

1 answer:

inna [77]2 years ago

8 0

Answer:

v = 61.09m/s

Explanation:

In order to calculate the speed of the electron when it is 3.00cm from the proton, you first calculate the acceleration of the electron, produced by the electric force between the electron and the proton. By using the second Newton law you have:

$F=ma=k\frac{q^2}{r^2}$ (1)

m: mass of the electron = 9.1*10^-31kg

q: charge of electron and proton = 1.6*10^-19C

r: distance between electron and proton = 9.00cm = 0.09m

k: Coulomb's constant = 8.98*10^9Nm2/C^2

You solve the equation (1) for a, and replace the values of the other parameters:

$a=\frac{kq^2}{mr^2}=\frac{(8.98*10^9Nm^2/C^2)(1.6*10^{-19}C)^2}{(9.1*10^{-31}kg)(0.09m)^2}=3.11*10^4\frac{m}{s^2}$

Next, you use the following formula to calculate the final speed of the electron:

$v^2=v_o^2+2ax$ (2)

vo: initial speed of the electron = 0m/s

a: acceleration = 3.11*10^4m/s^2

x: distance traveled by the electron

When the electron is at 3.00cm from the proton the electron has traveled a distance of 9.00cm - 3.00cm = 6.00cm = 0.06m = x

You replace the values of the parameters in the equation (2):

$v=\sqrt{2ax}=\sqrt{2(3.11*10^4m/s)(0.06m)}=61.09\frac{m}{s}$

The speed of the electron is 61.09m/s

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Answer:

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Explanation:

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From the law of conservation of energy, KE=PE hence

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Substituting m2-m1 for m and 0.5l for h, $\frac {2v}{L}$ for $\omega$ we obtain

$0.5I(\frac {2v}{L})^{2}=0.5Lg(m2-m1)$

$(\frac {2v}{L})^{2}=\frac {gl(m2-m1)}{I}$ and making v the subject

$v^{2}=\frac {gl^{3}(m2-m1)}{4I}$

$v=\sqrt {\frac {gl^{3}(m2-m1)}{4I}}$

For the rod, moment of inertia $I=\frac {ML^{2}}{12}$ and for sphere $I=MR^{2}$ hence substituting 0.5L for R then $I=M(0.5L)^{2}$

For the sphere on the left hand side, moment of inertia I

$I=m1(0.5L)^{2}$ while for the sphere on right hand side, $I=m2(0.5L)^{2}$

The total moment of inertia is therefore given by adding

$I=\frac {ML^{2}}{12}+ m1(0.5L)^{2}+ m2(0.5L)^{2}=\frac {L^{2}(M+3m1+3m2)}{12}$

Substituting $\frac {L^{2}(M+3m1+3m2)}{12}$ for I in the equation $v=\sqrt {\frac {gL^{3}(m2-m1)}{4I}}$

Then we obtain

$v=\sqrt {\frac {gL^{3}(m2-m1)}{4(\frac {L^{2}(M+3m1+3m2)}{12})}}=\sqrt {\frac {3gL^{3}(m2-m1)}{L^{2}(M+3m1+3m2)}}$

This is the expression of linear speed. Substituting values given we get

$v=\sqrt {\frac {3*9.81*0.8^{3}(0.05-0.02)}{0.8^{2}(0.39+3(0.02)+3(0.05))}} \approx 1.08 m/s$

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Answer:

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Explanation:

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f = 15 rps

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What is the mass of an object weighing 63 N on Earth?

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Weight expressed in Newtons is expressed in the equation whereby Weight= the mass of an object * the force of gravity. The force of gravity on earth is a constant 9.8 meters per second squared. Therefore if weight (w) = 63 N and the force of gravity is 63 N then the mass must equal 6.43 kg. Because the equation for weight is w=mg so 63 N (w) = m * 9.8 m/s^2.

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2 years ago

Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.

Musya8 [376]

Answer:

A) V = 7.5 V

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Given:

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- The thickness of Mylar added t = 0.10 mm

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A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

E = V / k*d

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Q = C*V

Q = 7.5*(2.213)

Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

E = V / k*d

k = 3.13 E = 7.5 / (3.13)0.10*10^-3

E = 24,000 V/m

F) What is the capacitor's charge after the Mylar is inserted?

C = k*A*ε / d

k = 3.13 C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

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Q = C*V

Q = 7.5*(6.927)

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eimsori [14]

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The bag is moving at constant speed, so its acceleration is zero:

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Therefore the net force is zero as well:

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