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NISA [10]
3 years ago
5

the charge density in an insulateed solid sphere of radius find the electric field at a distance of from the center of the solid

Physics
1 answer:
KatRina [158]3 years ago
6 0

Answer:

Assuming the charged density in the insulated solid sphere of radius 3.1m is 8.8e-9, the electric field at 5.2 meters is 73.1256 i.

Explanation:

The electric charge linear density is equal to 8.8 x10^{-9}

the radius of the sphere is 3.1m

The magnitude of the electric field at the radius of the sphere equal to 5.2 meters can be calculated with the formula ;

- E = λ / 4πε₀ [ r  / α ( α + r ) ] i

Solution:

E =  8.8 x10^{-9} / 4πε₀ [ 3.1/ 5.2( 5.2 + 3.1) ] i

= 1018.0995 [0.07183] i

=  73.1256 i

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Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

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