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olchik [2.2K]
3 years ago
13

Zero gauge wire has a diameter of 0.32in and can carry a sustained current of 150A safely. What is the magnetic field a distance

5cm from the center of a long straight segment of zero gauge wire carrying a current of 150A? This would be the field experienced by a bird sitting on the wire. Select One of the Following:
(a) 1.2 × 10−2T
(b) 1.9 × 10−3T
(c) 3.0 × 10−4T
(d) 6.0 × 10−4T
(e) 1.0 × 10−6T
Physics
1 answer:
DanielleElmas [232]3 years ago
5 0

Answer: d) 6*10^-4 T

Explanation: In order to explain this problem we have to use the Ampere law to calculate the magnetic field, which is given by:

∫B*dl=μo*I  then

B=μo*I/(2*π*r)  where r and I are the distance to the wire and the current in the wire. μo is equal 4*π*10^-7 Wb/A*m

Finally, we have:

B=4*π*10^-7*150/(2*π*0.05)=6*10^-4 T

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A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge
Nesterboy [21]

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

\tau=RC

where

R=50,000 \Omega is the resistance

C=2.0\mu F=2.0\cdot 10^{-6}F is the capacitance

Substituting,

\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s

The charge on a charging capacitor is given by

Q(t)=Q_0 (1-e^{-t/\tau} ) (1)

where

Q_0 is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

Q(t)=0.90 Q_0

Substituting this into eq.(1) we find

0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s

4 0
3 years ago
the parking brake on a 1200kg automobile has broken, and the vehicle has reached a momentum of 7800.m/s. what is the velocity of
Softa [21]

Answer:

6.5 m/s

Explanation:

Momentum = mass × velocity

p = mv

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v = 6.5 m/s

4 0
3 years ago
Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound w
WINSTONCH [101]

Answer:

Explanation:

frequency of sound waves = 688 Hz

wavelength = 344 / 688 = .5 m

The problem is based on interference of sound waves

For the observer , path difference of sound waves reaching his ear

= 3.5 - 3.00

.5 m

= wavelength

Path difference is equal to wavelength so there will be constructive interference and hence louder sound will be heard by the listener than normal sound as sound waves interfere constructively.

6 0
3 years ago
A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant
balu736 [363]

Answer:

U_eq = 1.99 * 10^(-10) J

Explanation:

Given:

Plate Area = 10 cm^2

d = 0.01 m

k_dielectric = 3

k_air = 1

V = 15 V

e_o = 8.85 * 10 ^-12  C^2 / N .m

Equations used:

U = 0.5 C*V^2  .... Eq 1

C = e_o * k*A /d  .... Eq 2

U_i = 0.5 e_o * k_i*A_i*V^2 /d  ... Eq 3

For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:

U_electric = 0.5 e_o * k_1*A*V^2 /2*d

U_air = 0.5 e_o * k_2*A*V^2 /2*d

The total Energy is:

U_eq = U_electric + U_air

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U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d

Plug the given values:

U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01

U_eq = 1.99 * 10^(-10) J

3 0
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Darya [45]

Answer:i belive the answer is E = 1/2 mv^2

hope this helps plz mark me as braliest

8 0
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