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marta [7]
4 years ago
5

Two 2.0 cm * 2.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 V battery. a. What

are the charge on each electrode and the potential difference between them? The wires are disconnected, and insulated handles are used to pull the plates apart to a new spacing of 2.0 mm. b. What are the charge on each electrode and the potential difference between them?
Physics
1 answer:
NARA [144]4 years ago
5 0

Answer: a) 31.86 *10^-12 C=31.86 pC; b) 18 V

Explanation: In order to explain this problem we have to consider the expression a parallel plates capacitor,which is given by:

C=Q/V where C is equal to C=εo*A/d where A and D are the area and the separation between the plates.

also we have

Q=C*V=ε(o*A/d)*V=(8.85*10^-12*0.02*0.02/1*10^-3)*9=31.86*10^-12 C=31.86pC

Then if the plates apart to a new spacing of 2.0 mm the new capacitance is equal

Cnew=εo*A/2*d so Cnew =Cinitial/2

then Cnew =Q/Vnew (Q is constant after disconnection to the battery)

Finally Vnew= Q/(Cinitial/2)= 2*(Q/Cinitial)= 2*Vinitial= 2*9=18V

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What the kinetic energy quantities in calculation pls help me​
Rashid [163]

Answer:

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Explanation:

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At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
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