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a_sh-v [17]
3 years ago
13

To practice Problem-Solving Strategy 22.1: Gauss's Law.An infinite cylindrical rod has a uniform volume charge density rho(where

rho>0). The cross section of the rod has radius r0. Find the magnitude of the electric field E at a distance r from the axis of the rod. Assume that r
Physics
1 answer:
d1i1m1o1n [39]3 years ago
8 0

Answer:

Solution for 3 different cases are given in explanation.

Explanation:

Gauss's Law:

\int_S E.dA=Q_{encl}/\epsilon_0

for r :

E.2\pi rL=\frac{\rho\pi r^2L/\pi r_0^2}{\epsilon_0} \\\\E=\frac{\rho r}{2\pi\epsilon_0 r_0^2}

for r=r_0 :

E.2\pi r_0L=\frac{\rho\pi r_0^2L}{\epsilon_0} \\\\E=\frac{\rho r_0}{2\epsilon_0}

for r>r_0 :

E.2\pi rL=\frac{\rho\pi r_0^2L}{\epsilon_0} \\\\E=\frac{\rho r_0^2}{2\epsilon_0 r}

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A 2.0 kg wooden block is slid along a concrete surface (μk = 0.21) with an initial speed of 15 m/s. How far will the block slide
exis [7]

Answer:

The distance is 54.6 m

Explanation:

Given that,

Mass = 2.0 kg

Frictional coefficient = 0.21

Initial velocity = 15 m/s

We need to calculate the acceleration

Using formula of frictional force  

F = \mu mg

F=0.21\times2.0\times9.8

F = 4.12\ N

We need to calculate the acceleration

F = ma

a = \dfrac{F}{m}

a =\dfrac{4.12}{2.0}

a=2.06\ m/s^2

We need to calculate the initial velocity

Using equation of motion

v^2=u^2-2as

Put the value  

0=15^2-2\times2.06\times s

s = \dfrac{15^2}{2\times2.06}

s=54.6\ m

Hence, The distance will be 54.6 m.

3 0
3 years ago
In which situation can you be at rest and moving at the same time
sineoko [7]

Answer:Being at rest is a relative term. Relative to earth we are at rest when we are sitting down. However, you are moving because the earth is moving its orbit around the sun.

5 0
4 years ago
Two particles are fixed to an x axis: particle 1 of charge −8.00 ✕ 10⁻⁷ C at x = 6.00 cm, and particle 2 of charge +8.00 ✕ 10⁻⁷
victus00 [196]

Answer:

0 N/C

Explanation:

Parameters given:

q_1 = -8.00 * 10^{-7} C

x_1 = 6.00 cm

q_2 = +8.00 * 10^{-7} C

x_2 = 21 cm

The distance between q_1 and q_2 is

21 - 6 = 15cm

Electric field is given as

E = \frac{kq}{r^2}

r = 15/2 = 7.5cm = 0.075m

The electric field at their midpoint due to q_1 is:

E = \frac{9 * 10^9 * -8.0 * 10^{-7}}{0.075^2}

E_1 = -1.28 * 10^6 N/C

The electric field at the midpoint due to q_2 is:

E = \frac{9 * 10^9 * 8.0 * 10^{-7}}{0.075^2}

E_2 = 1.28 * 10^6 N/C

The net electric field will be:

E = E_1 + E_2

E = -1.28 * 10^6 + 1.28 * 10^6

E = 0 N/C

7 0
4 years ago
A copper ornament has a mass of 0.0693 Kg and changes from a temperature of 18 degrees celcius to 26 degrees Celsius. How much h
zhuklara [117]

Answer: 216.2 J

Explanation: The heat energy of copper is the expressed in Q=mc∆T. First make sure that the mass 0.0693 kg is converted into grams to cancel both units.

Q= 69.3 g x 390 J/g°C x 26 °C-18°C = 216.2J

5 0
4 years ago
Read 2 more answers
A 3.42 kg mass hanging vertically from a spring on the Earth (where g = 9.8 m/s2) undergoes simple oscillatory motion. If the sp
sineoko [7]

Answer:

Time period of oscillation on moon will be equal to 3.347 sec

Explanation:

We have given mass which is attached to the spring m = 3.42 kg

Spring constant K = 12 N/m

We have to find the period of oscillation

Period of oscillation is equal to T=2\pi \sqrt{\frac{m}{K}}, here m is mass and K is spring constant

So period of oscillation T=2\times 3.14\times \sqrt{\frac{3.42}{12}}

T=2\times 3.14\times 0.533=3.347sec

So time period of oscillation will be equal to 3.347 sec

As it is a spring mass system and from the relation we can see that time period is independent of g

So time period will be same on earth and moon

3 0
3 years ago
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