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blagie [28]
3 years ago
8

A spherical cavity is excised from the inside of the sphere. The cavity has radius a4 and is centered at position h⃗ , where |h⃗

|<34a, so that the entire cavity is contained within the larger sphere. Find the electric field inside the cavity.

Physics
1 answer:
kompoz [17]3 years ago
6 0

Answer:

Electric field =h/3Eo

Explanation:

Because of the intrisic notation and symbol necessary for the solution please refer to the attached file

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What is the most commonly used semiconducting material? Nickel Aluminum b. Cobalt d. Silicon​
UNO [17]

Answer:

i believe the answer is D

Explanation:

Hope it works

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Read 2 more answers
A 40-turn coil has a diameter of 11 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the
seropon [69]

Answer:

(a) emf = 0.507 V

(b) emf = 0.0507 V

(c) emf = 0.00234 V

Explanation:

Given;

number of turns of the coil, N = 40 turns

diameter of the coil, d = 11 cm

radius of the coil, r = 5.5 cm = 0.055 m

magnitude of the magnetic field, B = 0.4 T

The magnitude of the induced emf is calculated as;

emf = - N\frac{d\phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux= BA \\\\A \ is the \ area \ of \ the \ coil = \pi r^2 = \pi (0.055)^2 = 0.0095 \ m^2\\\\emf = - N \frac{dB.A}{dt} = -NA\frac{dB}{dt} \\\\emf = -NA\frac{(B_2 - B_1)}{t} \\\\emf = NA \frac{(B_1 - B_2)}{t} \\\\the \ final \ magnetic \ field \ is \ reduced \ to \ zero;\ B_2 = 0\\\\emf = \frac{NAB_1}{t}

(a) when the time, t = 0.3 s

emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{0.3} = 0.507 \ V

(b) when the time, t = 3.0 s

emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{3} = 0.0507 \ V

(c) when the time, t = 65 s

emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{65} = 0.00234 \ V

7 0
3 years ago
In which of the following situations would there be initial energy in the system?
pshichka [43]

Answer:

A spring is compressed and held at compression by a person before releasing it

Explanation:

All the other answers are showing things that are at rest and have no energy starting nor going through them. This answer show how the energy is starting and being released.

8 0
3 years ago
State Coulomb's law and use it to show that the force between any two charged particles
e-lub [12.9K]
The answer is b i hope this helps
8 0
3 years ago
A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n
satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

8 0
4 years ago
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