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Ede4ka [16]
4 years ago
9

Two concentric conducting spherical shells produce a radially outward electric field of magnitude49,000 N/C at a point 4.10 m fr

om the center of the shells. The outer surface of the larger shell has aradius of 3.75 m. If the inner shell contains an excess charge of -5.30 μC, find the amount of chargeon the outer surface of the larger shell.
Physics
1 answer:
Korolek [52]4 years ago
6 0

To solve this problem it is necessary to apply the concepts related to the electric field according to the definition of Coulomb's law.

 The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point.

Mathematically this can be described as:

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

Where,

\epsilon_0 = permittivity of free space

r = Distance

q = Charge

E = Electric Field

Our values are given as,

E= 49.000N/C

r= 4.1m

\epsilon_0=8.854*10{-12}C^2N^{-1}m^{-2}

Replacing we have,

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

49000 = \frac{1}{4\pi (8.854*10{-12})}\frac{q}{(4.1)^2}

q= 9.16*10^{-5} C

q= 91.6\mu C

Therefore the amoun of charge on the outer surface of the larger shell is 91.6 \mu C

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A sprinter accelerates from rest to 9.00 in 1.38 sec.What is his acceleration in a. m/s2b. m/h2
dalvyx [7]

Answer:

a) The acceleration of the sprinter is 6.521 meters per square second, b) The acceleration of 84512.16 kilometers per square hour.

Explanation:

Note: Statement is incomplete, complete description of the problem is: <em>A sprinter accelerates from rest to </em>9.00\,\frac{m}{s}<em> in 1.38 seconds. What is his acceleration in </em><em>a.</em><em> </em>\frac{m}{s^{2}}<em> and </em><em>b.</em><em> </em>\frac{km}{h^{2}}<em>?</em>

a) We assume that sprinter accelerates uniformly, so that acceleration (a), measured in meters per square second, can be obtained from this kinematic expression as function of initial and final velocities (v_{o}, v), measured in meters per second, and time (t), measured in seconds, as well.

v = v_{o} + a\cdot t

a = \frac{v-v_{o}}{t}

If we know that v_{o} = 0\,\frac{m}{s}, v = 9\,\frac{m}{s} and t = 1.38\,s, the acceleration experimented by the sprinter is:

a = \frac{9\,\frac{m}{s}-0\,\frac{m}{s}  }{1.38\,s}

a = 6.521\,\frac{m}{s^{2}}

The acceleration of the sprinter is 6.521 meters per square second.

b) If we know that a hour is equivalent to 3600 seconds and a kilometer is equivalent to 1000 meters, the result of the previous item is converted by using two consecutive unit conversions:

a = \left(6.521\,\frac{m}{s^{2}} \right)\cdot \left(\frac{1}{1000}\,\frac{km}{m}  \right)\cdot \left(3600^{2}\,\frac{s^{2}}{h^{2}}  \right)

a = 84512.16\,\frac{km}{h^{2}}

The acceleration of 84512.16 kilometers per square hour.

6 0
4 years ago
(a ) A 14 turns circular coil is placed on a paper which lies in 1.2 T magnetic field pointing inwards to the paper. The coil's
bonufazy [111]

(i) The direction of the current will be in opposite direction to the magnetic field.

(ii) The magnitude of the emf induced in the circuit is 0.359 V.

(iii) The induced current if the circular coil is 0.048 A.

b(i) The magnetic flux linkage through the coil is 1.41 x 10⁻⁴ Tm².

b(ii) The radius of the coil is r√3 m.

<h3>Direction of the current</h3>

The direction of the current will be in opposite direction to the magnetic field.

<h3>Emf induced in the circuit</h3>

Initial area of the coil = (πd²)/4

Initial area of the coil = (π x 0.225²)/4 = 0.0398 m²

Final area of the coil = (π x 0.072²)/4 = 0.001296 m²

emf = \frac{NB(A_1 - A_2)}{t} \\\\emf = \frac{14 \times 1.2(0.0398 - 0.001296)}{1.8} \\\\emf = 0.359 \ V

<h3>Induced current</h3>

I = emf/R

I = 0.359/7.5

I = 0.048 A

<h3>Magnetic flux linkage through the coil</h3>

Ф = LI

Ф = 0.015 x 0.0094

Ф =  1.41 x 10⁻⁴ Tm²

<h3>Radius of the coil</h3>

L = \frac{\mu_o N^2\pi r^2}{l} \\\\r^2 = \frac{Ll}{\mu_o N^2\pi } \\\\r = \sqrt{\frac{Ll}{\mu_o N^2\pi } } \\\\r =  \sqrt{\frac{0.015(l)}{4\pi\times 10^{-7} \times (420)^2 \pi } } \\\\r = 3 \sqrt{l} \ m

where;

  • l is length of the coil

Learn more about inductance of coil here: brainly.com/question/17086348

3 0
3 years ago
The concentration of ozone in los angeles is 0.67 ppm on a summer day. what is the partial pressure of o3 if the total pressure
Katarina [22]
We use the Raoult's Law to answer this question.

Partial Pressure = Mole Fraction*Total Pressure

When concentration is expressed in ppm, it means 'parts per million'. So, there can be 0.67 mol of ozone per 10⁶ mol of mixture.

Mole fraction = 0.67/10⁶ = 0.67×10⁻⁶

Partial pressure = 0.67×10⁻⁶(735 torr) = <em>4.92×10⁻⁴ torr</em>
5 0
3 years ago
NEED AN ANSWER ASAP!
m_a_m_a [10]

Choice-D is the correct statement.

4 0
3 years ago
Based on this electric field diagram, which statement best compares the charge of A with B?
lilavasa [31]

Based on this electric field diagram, the statement which best compares the charge of A with B is "A is negatively charged and B is positively charged. The charge on A is greater than that on B".

<u>Answer:</u> Option A

<u>Explanation:</u>

The charge is quantized represented as elementary charge, about 1.602×10−19 coulombs. Their are two kinds of electric charging: positive and negative (usually transported, separately, by protons and electrons). Like charges repel each other, while attraction occurs among unlike charges. An entity without net charge is considered neutral. If a piece of matter comprises more electrons than protons, it has a negative charge, when there are fewer, it'll have a positive charge and when there are equal amounts, this will be neutral.

4 0
4 years ago
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