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kow
1 month ago
5

Find the electric field inside a hollow plastic ball of radius R that has charge Q uniformly distributed on its outer surface. G

ive your answer as a multiple of Q/ε0.Express your answer in terms of some or all of the variables R, r and the constant π.
Physics
1 answer:
Sav [3.1K]1 month ago
6 0
Inside the hollow plastic ball, there is no electric field present. According to Gauss's law, the electric field at a closed Gaussian surface is determined by the charges enclosed within it. To assess the electric field inside, we consider a spherical Gaussian surface inside the ball. Since all charge is located on the surface of the sphere, this Gaussian surface encloses no charge at all; thus, per Gauss's law, the result is that the electric field remains zero.
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Consult Conceptual Example 9 in preparation for this problem. Interactive LearningWare 6.3 also provides useful background. The
Keith_Richards [3271]

Answer:

11.56066 m/s

Explanation:

m = Mass of individual

v = Velocity of individual = 13.4 m/s

g = Gravitational acceleration = 9.81 m/s²

v' = Velocity of the individual after dropping

At the surface, kinetic and potential energy will equalize

\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{13.4^2}{2\times 9.81}\\\Rightarrow h=9.15188\ m

The cliff's height is 9.15188 m

Define fall height as h' = 2.34 m

\dfrac{1}{2}mv'^2+mgh'=mgh\\\Rightarrow v'=\sqrt{2g(h-h')}\\\Rightarrow v'=\sqrt{2\times 9.81(9.15188-2.34)}\\\Rightarrow v'=11.56066\ m/s

The person's speed is 11.56066 m/s

3 0
2 months ago
The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of
kicyunya [3294]

Answer:

6.6*10^{27}e/m^3

Explanation:

When calculating Hall voltage, it is crucial to have the current, magnetic field strength, length, area, and number of charge carriers available. The Hall voltage can be expressed using the equation:

V_h = \frac{iB}{neL}

Where:

i= the current

B= the magnetic field strength

L = the length

n = the number of charge carriers

e= charge of an electron

We need to replace values and solve for n:

n= \frac{iB}{V_h e L}

n= \frac{2*1.2}{4.5*10^{-6}*5^10^{-3}*1.6*10^{-19}}

n= 6.6*10^{27}electron.m^{-3}

As a result, the charge carrier density is 6.6*10^{27}e/m^3

5 0
2 months ago
In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
Yuliya22 [3333]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as w_{h}.

We aim to determine the heart's weight for a human

Using the provided information

w_{h}=0.5\times w

Where, h = heart weight

w = human weight

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

The final weight of a human heart is 0.93 lbs.

8 0
3 months ago
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