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siniylev [52]
4 years ago
7

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly throug

h its volume between its inner and outer surfaces. The volume charge density is?
Physics
1 answer:
Vaselesa [24]4 years ago
4 0

Answer:

1.57 * 10^{3} Q

Explanation:

The volume charge density is defined by ρ = \frac{Q}{V} (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are \frac{Coulombs}{m^{3} }, so we have to express the radius in meters:

inner radius = 4 cm * \frac{1 m}{100 cm} = 0.04m

outer radius = 6 cm * \frac{1m}{100cm}  = 0.06m

Now, we know that the volume of the sphere is calculated by the formula:

V = \frac{4}{3}\pi r^{3}, and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V = \frac{4}{3}\pi (r_{o} ^{3} - r_{i} ^{3}), where r_{o} is the outer radius and r_{i} is the inner radius.

Replacing the volume formula in the Equation A:

ρ = \frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}

ρ = \frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }

Replacing the values of the outer and inner radius whe have:

ρ = \frac{3Q}{4\pi (1.52 * 10^{-4})}

ρ = 1.57 * 10^{3} Q

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a block initially at rest has a mass m and sits on a plane incline at angle. it slides a distance d before hitting a spring and
KiRa [710]

Answer:

k = \frac{2\cdot m \cdot g \cdot (d+x_{f})\cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}{x_{f}^{2}}

Explanation:

Let assume that spring reaches its maximum compression at a height of zero. The system is modelled after the Principle of Energy Conservation and the Work-Energy Theorem:

U_{g,A}=U_{k,B} + W_{f}

m\cdot g \cdot (d + x_{f})\cdot \sin \theta = \frac{1}{2}\cdot k \cdot x_{f}^{2}+\mu_{k}\cdot m \cdot g \cdot (d+x_{f})\cdot \cos \theta

m\cdot g \cdot (d + x_{f})\cdot (\sin \theta-\mu_{k}\cdot \cos \theta) = \frac{1}{2}\cdot k \cdot x_{f}^{2}

The spring constant is cleared in the expression described above:

k = \frac{2\cdot m \cdot g \cdot (d+x_{f})\cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}{x_{f}^{2}}

6 0
3 years ago
What is the momentum of a baseball with a mass of 0.12 kg being thrown
Savatey [412]

Answer:

5.4kgm/s towards the hitter.

Explanation:

Given parameters:

Mass  = 0.12kg

Velocity  = 45m/s

Unknown:

Momentum of the baseball  = ?

Solution:

To solve this problem, we  must understand that the momentum of a body is the quantity of motion it possesses.

 Momentum  = mass x velocity

 Now insert the parameters and solve;

 Momentum  = 0.12 x 45  = 5.4kgm/s towards the hitter.

5 0
3 years ago
This solid layer of the earth is made of mostly iron and nickel.
Studentka2010 [4]

Answer

Explanation:

Yes, it's true that the solid layer of the earth is known as the most dense part as it is made up of the heavy metals like iron and nickel. Inner part is the hotter part due to the high pressure and temperature. It has the temperature of about 5,200°C and the pressure of 3.6 million atm but still the iron and nickel are present there in the solid form as they withstand such high temperature and pressure values.

5 0
3 years ago
A circuit has a resistance of 2.5 Ω and is powered by a 12.0 V battery.
Marizza181 [45]

Just apply Ohm's Law:


I=V/R=12/2.5=4.8A

5 0
3 years ago
A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 8.1 m from this surface, the potenti
igomit [66]

Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, V_s = 450 V

potential at radial distance, V_r = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}

450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r}  \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m

Therefore, the radius of the sphere is 4.05 m

4 0
3 years ago
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