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3 years ago

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To model this process, assume two charged spherical conductors are connected by a long conducting wire and a 1.20-mC charge is p

laced on the combination. One sphere, representing the body of the airplane, has a radius of 6.00 cm; the other, representing the tip of the needle, has a radius of 2.00 cm. (a) What is the electric potential of each sphere? (b) What is the electric

1 answer:

il63 [147K]3 years ago

7 0

Answer:

Part a: The electric potential of each sphere is 1.35x10⁸V

Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

Explanation:

As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question

Let r_1 = 6 cm=0.06 m

r2 = 2 cm = 0.02 m

Q = 1.2 mC

Let q1 and q2 are the charges on each sphere.

q1 + q2 = 1.2 mC -------(1)

In the equilibrium, V1 = V2

k*q1/r1 = k*q2/r2

q1/0.06 = q2/0.02

q1/q2 = 0.06/0.02

q1/q2 = 3 ---------(2)

On solving equation 1 and 2

we get

q1 = 0.9 mC

q2 = 0.3 mC

So

V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts

V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts

So the electric potential of each sphere is 1.35x10⁸V

Part b

Now the electric potential is given as

E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C

E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C

So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

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Answer:

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Explanation:

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Width of slit d = 0.215\times 10⁻³ m

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Putting the values we get fringe width

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a) Width of central maxima = 2 times width of one fringe

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2 years ago

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Answer:

The ball doesn't strike the building because it strikes the ground at d=1.62 meters.

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An object is moving with the speed of light around the Earth. How much time will it take to complete one round trip along the eq

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Answer:

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Explanation:

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3 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

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Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

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1 kg =1000 g

Speed of bullet, v=357m/s

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Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

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Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

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$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

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$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

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