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Brrunno [24]
3 years ago
6

Two long, parallel wires each carry the same current I, but the two currents are anti-parallel. The two wires are a distance r a

part. What is the magnitude and direction of the magnetic field, B, at a point that is at the midpoint, P, between the two wires?
Physics
1 answer:
Georgia [21]3 years ago
6 0

Answer:

8 x 10⁻⁷ x  I / r

Explanation:

Two parallel long wires are carrying current I . Let the direction be towards the right in the farthest and towards the left in the nearest. Magnetic field due to current I  at a  distance d is given by the expression

B = μ₀ 2 I / 4π d

I the present case distance d = r/2  

Magnetic field due to one wire at point d = r/2  is

B₁ = μ₀ 2 I / (4π r / 2 )

= 10⁻⁷ x 4I / r

Magnetic field due to the other wire at point d = r/2  is

B₂ = μ₀ 2 I / (4π r / 2 )

= 10⁻⁷ x 4I / r

Direction of magnetic field due to both the wires at the mid  point P will be same . It will be in downward direction in the given scenario

So total magnetic field

B = B₁ + B₂

= 2 x  10⁻⁷ x 4I / r

= 8 x 10⁻⁷ x  I / r

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Lana71 [14]
C

Terminal velocity is the maximum velocity attainable by an object as it falls through a fluid (air is the most common example). It occurs when the sum of the drag force (Fd) and the buoyancy is equal to the downward force of gravity (FG) acting on the object.(Wikipedia)




3 0
3 years ago
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

6 0
3 years ago
A stone is dropped into a deep well and hits the water after 3.41 seconds. How deep is<br> The well?
TiliK225 [7]

Answer:

57.0 m

Explanation:

Given:

v₀ = 0 m/s

t = 3.41 s

a = 9.8 m/s²

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (3.41 s) + ½ (9.8 m/s²) (3.41 s)²

Δx = 57.0 m

6 0
3 years ago
A jet - powered car called the spirit of America required 9600meters to stop from its highest speed . If the car decelerated at
Mazyrski [523]
v^{2} =  u^{2}  +  2ar
0 = u^2 + 2*(-2)*9600
u^2 = 38400
u = 195.96 m/s
6 0
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The table below shows the height of some horse fossils.
choli [55]

Answer:

Y W Z X, B

Explanation:

It wants you to figure out the correct order starting from oldest to newest.

7 0
3 years ago
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