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3 years ago

A 1 mF, a 2 mF, and a 3 mF capacitor are connected in parallel, the combination being

Physics

2 answers:

Tresset [83]3 years ago

7 0

Answer:

a) the 3mf capacitor has the greatest charge.

b) they all have equal voltage which is 9V

Explanation:

a) the charge on a capacitor is the product of the capacitance and the voltage across the capacitor. It can be written as;

Q = CV

And since they are connected in parallel, the voltage across each capacitor will be the same and equal to 9V

V1 = V2 = V3 = 9V

So,

For capacitors

Q1 = 1mF × 9V = 9mC

Q2 = 2mF × 9V = 18mC

Q3 = 3mF × 9V = 27mC

Therefore, the 3mF capacitor have the highest charge.

b) Since it is a parallel connection, the voltage across each capacitor is the same and equal to 9V.

Aleonysh [2.5K]3 years ago

6 0

Answer:

a) the greatest charge will be on C3=3 mF

b) Voltage will be same across each capacitors

Explanation:

Given: C1 = 1 mF, C2 =2 mF, C3 = 3 mf and V= 9 volts

To Find: greatest charge Q=? and which capacitor has the greatest voltage=?

Solution:

As the voltage across each will be same so (using Q = C V)

Q1 = 1 mF × V

Q2 = 2 mF × V

Q3 = 3 mF × V

so the greatest charge will be on C3=3 mF

b) Voltage will be same across each capacitors

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3 years ago

An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a

Blababa [14]

Answer:

$\frac{\Delta L}{L} =5.37\times 10^{-4}$

Explanation:

Given:

cross sectional area of the wire, $A=5.75\times 10^{-6}\ m^2$

density of aluminium wire, $\rho=2.7\times 10^3\ kg.m^{-3}$

young's modulus of the material, $E=7\times 10^{10}\ N.m^{-2}$

wave speed, $v=118\ m.s^{-1}$

<u>We have mathematical expression for strain as:</u>

$\frac{\Delta L}{L} =\frac{\sigma}{E}$ ...............................(1)

and since, $\sigma =\frac{T}{A}$

where, T = tension force in the wire

equation (1) becomes:

$\frac{\Delta L}{L} =\frac{T}{A.E}$ ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

$v=\sqrt{\frac{T}{\mu} }$ ...................................(3)

where: $\mu=$ linear mass density of the wire

$\therefore \mu=\rho \times A$

Now, equation (3) becomes

$v=\sqrt{\frac{T}{\rho \times A} }$

$T=v^2.\rho \times A$ ............................(4)

Using eq. (2) & (4) for tension T

$v^2.\rho \times A=A.E\times \frac{\Delta L}{L}$

$\frac{\Delta L}{L} =\frac{v^2.\rho}{E}$

putting the respective values

$\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}$

$\frac{\Delta L}{L} =5.37\times 10^{-4}$

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3 years ago

How are elements arranged in the modern periodic table?

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3 years ago

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A 4.0kg bowling ball sliding to the right at 8.0 m/s has an elastic head-on collision with another 4.0 kg bowling ball initially

Kisachek [45]

We use the formula :

m1v1i + m2v2i = m1v1f + m2v2f

Substituting the values in:

4.0kg*8.0m/s + 4.0kg*0m/s = 4.0kg*0m/s +4.0kg*v2f

Calculating this we get:

32.0kg*m/s + 0kg*m/s = 0kg*m/s + 4.0kg*v2f

Rearrange for v2f:

v2f = $\frac{32.0kg*m/s}{4.0kg}$

This gives us 8.0 m/s as the final velocity of the second ball.

Since the collision is assumed to be elastic it means that the kinetic energy must be equal before and after the collision.

This means we use the formula:

Ek = $\frac{1}{2} *m*v^{2}$ + $\frac{1}{2} *m*v^{2}$ = $\frac{1}{2} *m*v^{2}$ + $\frac{1}{2}*m*v^{2}$

Substituting in values:

Ek = 0.5*4.0kg*(8.0m/s)^2 + 0.5*4.0kg*(0m/s)^2 = 0.5*4.0kg*(0m/s)^2 + 0.5*4.0kg*(8.0m/s)^2

This simplifies to:

Ek= 128J + 0J = 0J + 128J

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3 years ago

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C. The net force on the car is greater than zero in the horizontal direction.

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4 years ago