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alina1380 [7]
3 years ago
6

A very long straight wire has charge per unit length 1.44×10-10C/m.

Physics
1 answer:
4vir4ik [10]3 years ago
3 0

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m            

Explanation:

We have given charge per unit length, that is liner charge density \lambda =1.44\times 10^{-10}C/m

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

E=\frac{\lambda }{2\pi \epsilon _0r}, here \lambda is linear charge density and r is distance of the point where we have to find the electric field

So 2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
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1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
3 years ago
Technician A says that side post terminals need to be removed to inspect them for corrosion. Technician B says that side post te
zlopas [31]

Answer: C

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Over tightening the terminal bolt can damage side post terminals.

The battery terminals and cable ends can corrode especially when the battery or car is not used for a long period of time. Corrosion limits a battery's lifespan and so should be prevented. To inspect the areas where corrosion occur on a side-post battery, you need to remove the terminals.

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4 0
3 years ago
Fill in the blank
aliina [53]

Answer:

hypothesis

Explanation:

8 0
3 years ago
According to the Belmont Report, the moral requirement that there be fair outcomes in the selection of research subjects, expres
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Answer:

The correct answer is: justice

Explanation:

The Belmont Report refers to a report that was published 25 year ago, focusing on the ethical treatment and protection of participants in medical and behavioral research. This report centers around 3 principles:

1. Beneficence- striving to maximize benefits for participants of the research study and minimizing any harms/ risks that might occur.

2. Justice- The fair selection of potential participants for a study. This ensures equitable and fair distribution of risks/ benefits to all potential participants of a research study. Subjects of a study must not be chosen merely out of convenience or easy access. The inclusion/ exclusion criteria should be chosen according to the nature of the study and steps/ treatments that it will involve.

3. Respect for persons- Each participant of a research study should be  able to provide informed consent prior to their participation, protected from controllable harm and treated with respect.

Therefore, moral requirement that there be fair outcomes in the selection of research subjects, expresses the principle of justice.

5 0
3 years ago
A battery having an emf of 9.76 V delivers 111 mA when connected to a 60.5 Ω load. Determine the internal resistance of the batt
kow [346]

Answer:

r = 27.4 \: ohm

Explanation:

Using

E = I(R + r)

9.76 = 111 \times  {10}^{ - 3} (60.5 + r) \\ 87.928 = 60.5 + r \\ r = 27.4 \: ohm

5 0
2 years ago
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