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nordsb [41]
3 years ago
11

I have attached the question

Physics
1 answer:
Licemer1 [7]3 years ago
7 0

Answer:

i don't know getting free points

Explanation:

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Series circuits are characterized by the fact that there is a single pathway by which charge can travel. True or false?
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Find the resistance of wire of<br>0.65m Radius 0.25<br>and<br>resistivity 3x10-6 OHM<br>​
densk [106]

Complete Question:

Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.

Answer:

Resistance = 9.95 Ohms

Explanation:

<u>Given the following data;</u>

Length = 0.65 m

Radius = 0.25 mm to meters = 0.00025 m

Resistivity = 3 * 10^{-6} ohm-metre.

To find the resistance of the wire;

Mathematically, resistance is given by the formula;

Resistance = P \frac {L}{A}

Where;

  • P is the resistivity of the material.
  • L is the length of the material.
  • A is the cross-sectional area of the material.

First of all, we would find the cross-sectional area of the wire.

Area of circle = πr²

Substituting into the equation, we have;

Area  = 3.142 * (0.00025)²

Area = 3.142 * 6.25 * 10^{-8}

Area = 1.96 * 10^{-7} m²

Now, to find the resistance of the wire;

Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}}

Resistance = 3 * 10^{-6} * 3316326.531

<em>Resistance = 9.95 Ohms </em>

5 0
3 years ago
A 7kg object is dropped on Earth,. Assuming it has not yet hit the ground, what is the velocity of the object after 2 seconds of
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The answer would be 9.8 meters/ sec^2.

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3 years ago
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A train's mass is 500 kg and it's acceleration is 5 m/s. What is the net force on the train
trapecia [35]
The answer is 2500 newtons. F = M * A, so 500 kg x 5 m/s = 2500 newtons.
7 0
4 years ago
The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov
son4ous [18]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  f_e  = 0.027 \ m

Explanation:

From the question we are told that

    The focal length is  f_o =  5.5 \ mm =  -0.0055 \ m

This negative sign shows the the microscope is diverging light

     The  angular magnification is m = 300

     The  distance between the objective and the eyepieces lenses is  Z =  19 \ cm  = 0.19 \ m

Generally the magnification is mathematically represented as

        m  =  [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]

Where f_e is the eyepiece focal length of the microscope

  Now  making f_e the subject  of the formula

         f_e  = \frac{Z}{1 - [\frac{M  *  f_o }{0.25}] }

substituting values

        f_e  = \frac{ 0.19 }{1 - [\frac{300  *  -0.0055 }{0.25}] }

         f_e  = 0.027 \ m

     

5 0
3 years ago
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