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svetlana [45]
3 years ago
12

A charge of 0.14 C is moved from a position where the electric potential is 20 V to a position where the electric potential is 5

0 V. What is the change in potential energy of the charge associated with the change in position?
Physics
1 answer:
alukav5142 [94]3 years ago
7 0

The change in electric potential energy is given by:

ΔU = ΔVq

ΔU = change in PE, ΔV = potential difference, q = charge

Given values:

ΔV = 50V - 20V = 30V, q = 0.14C

Plug in and solve for ΔU:

ΔU = 30(0.14)

ΔU = 4.2J

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A circular hole in an aluminum plate is 2.739 cm in diameter at 0.000°C. What is the change in its diameter when the temperature
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Answer:

L = 2.746 cm

Explanation:

As we know that thermal expansion coefficient of aluminium is given as

\alpha = 24 \times 10^{-6} per ^oC

now we also know that after thermal expansion the final length is given as

L = L_o(1 + \alpha \Delta T)

here we know that

L_o = 2.739 cm

\alpha = 24 \times 10^{-6}

\Delta T = 108 - 0= 108^oC

now we will have

L = 2.739(1 + 24 \times 10^{-6} (108))

L = 2.746 cm

3 0
3 years ago
Una cuerda horizontal tiene una longitud de 5 m y masa de 0,00145 kg. Si sobre esta cuerda se da un pulso generando una longitud
FromTheMoon [43]

Answer:

Option (A) is correct.

Explanation:

A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is

mass of string, m = 0.00145 kg

Frequency, f = 120 Hz

wavelength = 0.6 m

Speed = frequency x wavelength

speed = 120 x 0.6 = 72 m/s

Let the tension is T.

Use the formula

v =\sqrt\frac{T L}{m}\\\\72 = \sqrt\frac{T\times 5}{0.00145}\\\\T = 1.5 N

Option (A) is correct.

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What is the scientific name given to potato​
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A 80 ohms resistor, 0.2 H inductance and 0.1 mF capacitor are connected in series across a generator (60 Hz, V rms=120 V). Deter
qaws [65]

Answer:

Impedance = 93.75 ohms

Current = 1.81 A

Explanation:

Resistance = R = 80 ohms

Inductance = L = 0.2 H

Inductive reactance = XL = X_{L}= = ωL = (2πf) L

= 2 (3.14) (60)(0.2) = 75.398 Ohms

Capacitive reactance = 1 / ωC = 1/(2πf)C = 1 / [(2π)(60)(0.1 × 10⁻3)]

= 26.526 Ohms  

Impedance = Z = \sqrt{R^{2} + (X_{L} - X_{C})^{^{2}}} =

= \sqrt{8788.511} = 93.747 ohms  

Voltage = \sqrt{2} × 120 = 169.7056 V

Current = I = V ÷ R = (169.7056) ÷ 93,747 = 1.81 A

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