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Lady bird [3.3K]
3 years ago
7

Simon is riding a bike at 12 km/h away from his friend Keesha.He throws a ball at 5 km/h back to Keesha, who is standing still o

n a sidewalk.
 
How fast would Keesha say the ball is traveling toward her?
Physics
1 answer:
Tems11 [23]3 years ago
5 0

I notice that even though we're working with frames of reference
here, you never said which frame the '5 km/hr' is measured in.

In fact !  You didn't even say which frame the '12 km/hr' of his
bike is measured in.

So there are several different ways this could go.  I'll do it the way
I THINK you meant it, but that doesn't guarantee anything.

-- Simon is riding his bike at 12 km/hr relative to the sidewalk,
away from Keesha.

-- He throws a ball at Keesha, at 5 km/hr relative to his own face.

-- Keesha sees the ball approaching her at (12 - 5) = 7 km/hr
relative to the ground and to her.

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A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate
Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is \bf{2.81 \times 10^{4}~n~C^{-1}}.

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm

Part(c):

If we apply Gauss' law of electrostatics, then

&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}

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Why is the answer C for this problem?
shusha [124]

Answer:

Explanation:

If you look closely, force 1 does not reach 0.2 until 0.4 force 2 reaches 0.2 at about 0.2 - hope that made sense :P

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