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givi [52]
2 years ago
15

Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the

charge density rho. (b) Find the total charge contained in a sphere of radius R, centered at the origin. (Do it two different ways.)
Physics
1 answer:
Assoli18 [71]2 years ago
7 0

Answer:

Part a)

\rho = 3\epsilon_0 k r^2

Part b)

Q = 4\pi \epsilon_0kR^5

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

E = kr^3 \hat r

now electric flux through a spherical surface of radius r is given as

\phi = E. A

\phi = kr^3(4\pi r^2)

now by Guass law we know that

E.A = \frac{q}{\epsilon_0}

q = 4\pi \epsilon_0kr^5

now volume charge density is given as

\rho = \frac{q}{\frac{4}{3}\pi r^3}

\rho = 3\epsilon_0 k r^2

Part b)

Total charge inside the radius R is given as

Q = 4\pi \epsilon_0kR^5

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Darina [25.2K]

Answer:

Q=1005 J

t= 0.67 sec

Explanation:

Lets take condition of room is 1 atm and 25°C.

Heat capacity ,c = 21 J /K.mol

If we assume that air is ideal gas that

P V = n R T

V= 5.5\times 6.5\times 3\ m^3

V=107.25\ m^3

V=107.25\times 1000 L

V= 107250 L

At STP number of moles given as

n=\dfrac{V}{V_{at\ S.T.P.}}

V=22.4 L at S.T.P.

n=\dfrac{107250}{22.4}\ moles

n=4787.94 moles

n= 4.784 Kmoles

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Q = n x c x ΔT

Q = 4.78794 x 21 x 10

Q=1004.64 J

Time t

t= Q/P

P= 1.5 KW

t = 1.004.64 /1.5

t= 0.66 sec

4 0
2 years ago
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A gas has an initial volume of 24.6 L at a pressure of 1.90 atm and a temperature of 335 K. The pressure of the gas increases to
Tatiana [17]

Answer:

the final temperature of the gas is 785.18 K

Explanation:

The computation of the final temperature of the gas is shown below:

Here we apply the gas law

= PV ÷ T

Given that

P1 = 1.9 atm

V1 = 24.6 L

T1 = 335 K  

P2 = 3.5 atm

V2 = 31.3 L  

T2 = ?

Now

P1V1 ÷ T1 = P2V2 ÷ T2

(1.9 × 24.6) ÷ 335 = (3.5 × 31.3)/T2

T2 = 785.18 K

hence, the final temperature of the gas is 785.18 K

8 0
2 years ago
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An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
2 years ago
What is the longest wavelength light capable of ionizing a hydrogen atom in the ground state?
Sindrei [870]

Answer:

9.12\cdot 10^{-8} m

Explanation:

The energy needed to ionize a hydrogen atom in the ground state is:

E=13.6 eV= 2.18\cdot 10^{-18}J

The energy of the photon is related to the wavelength by

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength

Solving the formula for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{2.18\cdot 10^{-18}J}=9.12\cdot 10^{-8} m

7 0
2 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.240 rev/s. The magnitude
bazaltina [42]

Explanation:

Given that,

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Diameter = 0.720 m

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\omega_{f}=\omega_{i}+\alpha t

Put the value in the equation

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\omega_{f}=0.426\ rev/s

The angular velocity is 0.426 rev/s.

(b). We need to calculate the tangential speed of the blade

Using formula of  tangential speed

v= r\omega

Put the value into the formula

v = \dfrac{0.720 }{2}\times0.426\times2\pi

v=0.963\ m/s

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(c). We need to calculate the magnitude at of the tangential acceleration

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a_{t}=r\alpha

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a_{t}=0.36\times0.917\times2\pi

a_{c}=2.074\ m/s^2

The tangential acceleration is 2.074 m/s².

Hence, This is required solution.

4 0
2 years ago
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