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AnnZ [28]
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?
Physics
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 sourceE=12V

Lamp PowerP=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

6 0
3 years ago
An insulating sphere is 8.00 cm in diameter and carries a 6.50 µC 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

7 0
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]
<span>To answer the question above, if the day sky is clear it collects radiation. If air is dry snow sublimates faster. If both cases overlap it disappears faster where ever coldest. Thank you for posting your question here. I hope my answer helps. </span>
7 0
3 years ago
NEED HELP ASAP
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

6 0
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:

5 0
3 years ago
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