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3 years ago

6

Suppose the original segment of wire is stretched to 12 times its original length. How much charge must be added to the wire to

keep the linear charge density unchanged?

1 answer:

kondaur [170]3 years ago

4 0

Answer:

Explanation:

The linear charge density id defined as the charge per unit length.

Let q be the charge and L be the length of the wire.

So, the linear charge density is

λ = q / L

If the length is 12 times, then charge should also be 12 times to keep the linear charge density same.

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If you connected 10 of these 12V (10 W) lamps in parallel, to the 12V source, how much current would the source have to supply

Flauer [41]

Answer:

$T=8.33A$

Explanation:

From the question we are told that:

Number of battery $n=10$

Voltage source $E=12V$

Lamp Power $P=10W$

Generally the equation for Resistance is mathematically given by

$R=\frac{V^2}{P}$

$R=\frac{12^2}{10}$

$R=14.4ohms$

Therefore

$R_{eq}=\frac{14.4}{10}$

$R_{eq}=1.44$

Generally the equation for Current is mathematically given by

$T=\ffrac{V}{Req}$

$T=\frac{12}{1.44}$

$T=8.33A$

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3 years ago

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

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3 years ago

If you live in a very cold area, you may have seen the depth of a bank of snow shrink even though temperatures remain below the

Aleks04 [339]

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3 years ago

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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3 years ago

HELP Giving all the points I can. Physics virtual lab report: circuit design<br><br> Please help me.

mafiozo [28]

Answer:text me I can help

Explanation:

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3 years ago

Read 2 more answers