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Crazy boy [7]
3 years ago
7

A solid conducting sphere of radius R carries a charge Q. Calculate the electric-field energy density at a point a distance r fr

om the center of the sphere for (a) rR. (c) Calculate the total electric field energy associated with the charged sphere. (d) How much work is required to assemble the charge Q on the sphere

Physics
1 answer:
SCORPION-xisa [38]3 years ago
8 0

Answer: electric field density at centre = zero.

Explanation:

Find the attached file for the remaining solutions

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Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir
yawa3891 [41]

Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

(c) E = 0.28 N/kg

Explanation:

Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s

Radius, r = 6.71 x 10^8 m

G = 6.67 x 10^-11 Nm^2/kg^2

(a) T=2\pi \sqrt{\frac{r^{3}}{GM}}

M=\frac{4\pi ^{2}r^{3}}{GT^{2}}

M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

v=\frac{2\pi r}{T}

v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}

v = 13739.5 m/s

v = 13.74 km/s

(b) The gravitational field E is given by

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

E = \frac{6.67\times10^{-11}\times 1.898\times 10^{27}}{6.71^{2}\times 10^{16}}

E = 0.28 N/kg

6 0
3 years ago
In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21
Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

z = 20km

And the velocity can be written as,

V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V = 980.55m/s

From the properties of standard atmosphere at altitude z = 20km temperature is

T = 216.66K

k = 1.4

R = 287 J/kg

Velocity of sound at this altitude is

a = \sqrt{kRT}

a = \sqrt{(1.4)(287)(216.66)}

a = 295.049m/s

Then the Mach number

Ma = \frac{V}{a}

Ma = \frac{980.55}{296.049}

Ma = 3.312

So front stagnation temperature

T_0 = T(1+\frac{k-1}{2}Ma^2)

T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)

T_0 = 689.87K

Therefore the temperature at its front stagnation point is 689.87K

6 0
3 years ago
Write a hypothesis about the effect a change in temperature will have on the volume of a gas.
Licemer1 [7]
The colder the more likely it is to become a liquid
7 0
3 years ago
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5L=__ML<br> Express your answer in scientific notation<br> 5 x 109<br> 5x 106<br> 5x 103
Sphinxa [80]

Answer:

5000ml is the answer

Explanation:

multiply 5 time 1000

7 0
3 years ago
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PLEASE HELP! 2ND TIME I ASKED THIS!!!
Goshia [24]

The train is moving at 50 m/s and Emma is walking down the aisle with 1 m/s speed in the same direction of train. The relative velocity of Emma with respect to other passengers pf the train would be 1 m/s. This is because, the train is not moving relative to them and only emma is moving at 1 m/s. If a person observes from outside, Emma would have (50 +1) m/s = 51 m/s velocity.

relative velocity when two objects are moving in same direction as oberved from outside observer:

v_{ab}=v_a+v_b

relative velocity when two objects are moving in the opposite directionas oberved from outside observer:

v_{ab}=v_a-v_b

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3 years ago
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