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vampirchik [111]
3 years ago
11

A proton has been accelerated from rest through a potential difference of -1350 V. What is the proton's kinetic energy, in elect

ron volts? What is the proton's kinetic energy, in joules? What is the proton's speed?
Physics
1 answer:
Luba_88 [7]3 years ago
8 0

Answer:

1 eV = 1.60 * 10^-19 J      work done in accelerating electron thru 1 V

KE (total energy) = 1350 ^ 1 eV     (note proton goes from +  to -)

KE = 1.60 * 10^-19 * 1350 = 2.16 * 10^-16 Joules

1/2 m v^2 = KE = 2.16 * 10^-16 J

v^2 = 4.32 * 10E-16 / 1.67 * 10-27 = 2.59 * 10^11

v = 5.09 * 10^5 m/s

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In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten
sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

V = iR

V = (2 mA)(10 kohm)

V = 20 volts

so voltage across the capacitor + voltage across resistor = V

V_c + 20 = 50

V_c = 30 V

Now we know that

U = \frac{q^2}{2C}

here rate of change in energy of the capacitor is given as

\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}

\frac{dU}{dt} = (30)(2 mA)

\frac{dU}{dt} = 0.06 W

3 0
3 years ago
A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A
Kruka [31]

Answer:

\omega_f = 585.37 \ rev/s

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)=  800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

I \omega_i = I' \omega_f

(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f

M\times 800 = ( M + m )\omega_f

3\times 800 = (3+1.1)\times \omega_f

2400 = (4.1)\times \omega_f

\omega_f = 585.37 \ rev/s

3 0
3 years ago
Metals combine easily with nonmetals. true or false
Ad libitum [116K]
The answer for that is True.
4 0
3 years ago
A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of
enot [183]

Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

\frac{n_1}{n_2}=\frac{V_1}{V_2}

In the old transformer the ratio of voltages was:

\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1

In the new transformer the ratio of voltages is:

\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1

If the old current output was 425 kV, the ratio of current was:

\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{425}= 0.028\\\\I_2=0.028I_1

Then, the ratio of the new output over the old output is:

\frac{I_{2new}}{I_{2old}} =\frac{0.024\cdot I_1}{0.028\cdot I_1}= 0.86

The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).

c) The power loss is expressed as:

P_L=I^2\cdot R

Then, the ratio of losses is (R is constant for both power losses):

\frac{P_n}{P_o} =\frac{I_n^2R}{I_o^2R} =(\frac{I_n}{I_o} )^2=0.86^2=0.74

The losses in the new line are 74% the losses in the old line.

7 0
3 years ago
What is the force of gravity between two 40.0kg masses that are separated by 3.00m?
Slav-nsk [51]

Answer:

f = g \times \frac{m1 \times m2}{ {d}^{2} }

f = 6.67 \times  {10}^{ - 11}  \times \frac{40 \times 40}{9}

<h2>F=1.2x 10^-8</h2>
3 0
3 years ago
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