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xeze [42]
3 years ago
14

An eight-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm. The coil lies in the

plane of the page and carries a clockwise current of 6.20 A. If the coil is in a uniform magnetic field of 1.98 10-4 T directed toward the left of the page, what is the magnitude of the torque on the coil? Hint: The area of an ellipse is A = ?ab, where a and b are, respectively, the semimajor and semiminor axes of the ellipse.
Physics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

9.25 x 10^-4 Nm

Explanation:

number of turns, N = 8

major axis = 40 cm

semi major axis, a = 20 cm = 0.2 m

minor axis = 30 cm

semi minor axis, b = 15 cm = 0.15 m

current, i = 6.2 A

Magnetic field, B = 1.98 x 10^-4 T

Angle between the normal and the magnetic field is 90°.

Torque is given by

τ = N i A B SinФ

Where, A be the area of the coil.

Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²

τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°

τ = 9.25 x 10^-4 Nm

thus, the torque is 9.25 x 10^-4 Nm.

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Answer:

7.5 m

Explanation:

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3 years ago
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 12.9 rad/s in 2.98 s.
PtichkaEL [24]

Explanation:

(a) Find the magnitude of the angular acceleration of the wheel.

  • angular acceleration = angular speed /time
  • angular acceleration = 12.9/2.98 = 4.329rad/s²

(b) Find the angle in radians through which it rotates in this time interval.

  • angular speed = 2x3.14xf
  • 12.9rad = 2 x3.14

  • rad = 6.28/12.9
  • rad = 0.487

Now we convert rad to angle

  • 1 rad = 57.296°
  • 0.487 = unknown angle
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The angle in radians = 27.9°

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All BUT one is a function of the cell wall. That is___
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C the cell wall can’t transfer genetic material
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two point charges of magnitude 4.0 μc and -4.0 μc are situated along the x-axis at x1 = 2.0 m and x2 = -2.0 m, respectively. wha
user100 [1]

The electric potential at the origin of the xy coordinate system is negative infinity

<h3>What is the electric field due to the 4.0 μC charge?</h3>

The electric field due to the 4.0 μC charge is E = kq/r² where

  • k = electric constant = 9.0 × 10 Nm²/C²,
  • q = 4.0 μC = 4.0 × 10 C and
  • r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m

<h3>What is the electric field due to the -4.0 μC charge?</h3>

The electric field due to the -4.0 μC charge is E = kq'/r² where

  • k = electric constant = 9.0 × 10 Nm²/C²,
  • q' = -4.0 μC = -4.0 × 10 C and
  • r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m

Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is

E" = E + E'

= -2E

= -2kq/r²

<h3>What is the electric potential at the origin?</h3>

So, the electric potential at the origin is V = -∫₂⁰E".dr

= -∫₂⁰-2kq/r².dr

Since E and dr = dx are parallel and r = x, we have

= -∫₂⁰-2kqdxcos0/x²

= 2kq∫₂⁰dx/x²

= 2kq[-1/x]₂⁰

= -2kq[1/x]₂⁰

= -2kq[1/0 - 1/2]

= -2kq[∞ - 1/2]

= -2kq[∞]

= -∞

So, the electric potential at the origin of the xy coordinate system is negative infinity

Learn more about electric potential here:

brainly.com/question/26978411

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3 0
2 years ago
A jet taxiing down the runway receives word that it must return to the gate. The jet is traveling 37.6 m/s when the pilot receiv
IrinaVladis [17]

Answer:

7 m/s^2

Explanation:

Given that the jet is traveling 37.6 m/s when the pilot receives the message. 

And it takes the pilot 5.37 s to bring the plane to a halt.

Acceleration of the plane can be calculated by using first equation of motion

V = U - at

Since the plane is going to stop, the final velocity V = zero.

And the acceleration will be negative

Substitute all the parameters into the formula

0 = 37.6 - 5.37a

5.37a = 37.6

Make a the subject of formula

a = 37.6 / 5.37

a = 7.0 m/s^2

Therefore, the acceleration of the plane to bring the plane to a halt is 7 m/s^2

5 0
3 years ago
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