All categories

3 years ago

A given 60uc piont charge located at origion

2 answers:

7 0

For my first part it like that as we all know that flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2. NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2 calculate required region area with formula =double integral(r^2sintheta dtheta dphi) we get =.106m^2 now multiply D*Required region we get 7.5uc M I RIGHT PLZ tell me

Morgarella [4.7K]3 years ago

6 0

Answer:

flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2. NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2 calculate required region area with formula =double integral(r^2sintheta dtheta dphi) we get =.106m^2 now multiply D*Required region we get 7.5uc

Explanation:

You might be interested in

What is the potential energy for and object with mass of 15 Kg and 20 m above the ground?

vaieri [72.5K]

Answer:

3000J

Explanation:

The gravitational energy for any object is given by E=mgh

where m is mass, g is gravitational field strength anf h is height above ground

5 0

3 years ago

Two moles of helium are initially at a temperature of 21.0 ∘Cand occupy a volume of 3.30×10−2 m3 . The helium first expands at c

Anettt [7]

Answer:

(B) The total internal energy of the helium is 4888.6 Joules

(D) The final volume of the helium is 0.066 cubic meter

Explanation:

(B) ∆U = P(V2 - V1)

From ideal gas equation, PV = nRT

T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3

P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal

∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules

Assuming a closed system

(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules

6 0

3 years ago

A 5 kg rock is raised 28 m above the ground level. What is the change in its potential energy?

marin [14]

Let's assume that ground level is the height 0 meters. The change in potential energy is going to be gravitational potential energy, which is given by PE=mgh.
ΔPE=mgh-mgy
=mg(h-y)
=50(28-0)
=1400 J

3 0

3 years ago

What are the main activities involved in studying physics?

Firlakuza [10]

The main activity that is involved in studying of physics is the study of natural laws. The study of physics has to do with many aspects of the universe. Physics majorly looks into the natural laws that operate in the universe and describe how they affect matter in relation to time.

7 0

3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$