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lozanna [386]
3 years ago
11

Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.55 μC charge and flies due west at a speed of

685 m/s over the Earth's south magnetic pole, where the 8.00 x 10^-5 T magnetic field points straight up. What is the magnitude of the magnetic force on the plane, in newtons?
Physics
1 answer:
Illusion [34]3 years ago
6 0

Answer:

3.014 x 10⁻⁸ N

Explanation:

q = magnitude of charge on the supersonic jet = 0.55 μC = 0.55 x 10⁻⁶ C

v = speed of the jet = 685 m/s

B = magnitude of magnetic field in the region = 8 x 10⁻⁵ T

θ = angle between the magnetic field and direction of motion = 90

magnitude of the magnetic force is given as

F = q v B Sinθ

F = (0.55 x 10⁻⁶) (685) (8 x 10⁻⁵) Sin90

F = 3.014 x 10⁻⁸ N

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Dropping it on a hard surface. 
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Is there supposed to be a slit at the top of your adams apple?
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No, there isn't. Please consult your doctor if this is the case with yours or someone you know.
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3 years ago
An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength
Vladimir [108]

Answer:

3x10⁴v

Explanation:

Using

Wavelength= h/ √(2m.Ke)

880nm = 6.6E-34/√ 2.9.1E-31 x me

Ke= 6.6E-34/880nm x 18.2E -31.

5.6E-27/18.2E-31

= 3 x 10⁴ Volts

4 0
3 years ago
A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3
4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

8 0
3 years ago
A mine elevator is supported by a single steel cable 0.0125 m in diameter. The total mass of the elevator cage and occupants is
Sonja [21]

Answer:

0.8895m

Explanation:

Cable diameter = 0.0125m

Mass of elevator = 6450kg

Young Modulus(E) = 2.11*10¹¹N/m

∇l (change in length) =

L = 362m

A = Πr², but r = d / 2 = 0.0125 / 2 = 0.00625m

A = 3.142 * (0.00625)² = 1.227*10^-4m²

Young Modulus (E) = Tensile stress / Tensile strain

E = (F / A) / ∇l / L

F = mg = 6450 * 9.8 = 63210N

2.11*10¹¹ = (63210 / 1.22*10^-4) / (∇l / 362)

2.11*10¹¹ = 5.18*10⁸ / (∇l / 362)

2.11*10¹¹ = (5.18*10⁸ * 362) / ∇l

2.11*10¹¹ = 1.875*10¹¹ / ∇l

∇l = 1.875*10¹¹ / 2.11*10¹¹

∇l = 0.8895m

The change in length is 0.8895m

8 0
3 years ago
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