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irga5000 [103]
3 years ago
12

what factors limit the accuracy of a potentiometer and what was the objective of reversing the terminals of the cell​

Physics
1 answer:
cricket20 [7]3 years ago
7 0

Explanation:

<em>"The accuracy of a potentiometer can be increased by decreasing the potential gradient across the potentiometer wire, and this can be achieved by increasing the length"</em>

<em />

<u>The factors that are affecting/limiting the accuracy of the potentiometer are: </u>

  1. The specific resistance of the material of the potentiometer wire.
  2. The potential gradient
  3. The current passing through the potentiometer wire.
  4. Area of a cross-section of the wire
  5. Internal temperature.

     

<u>The objective of reversing the terminals of the cell​</u>

If the jockey of the potentiometer is pressed for a long time, joule heating sets in, so that reversing the terminals of the potentiometer will prevent the resistance due to joule heat from being added to the measured resistance, ultimately preventing unwanted resistance

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How does reducing the volume of a gas affect its pressure if the temperature of the gas and the number of particles are constant
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have no idea but I think the particles would be affected because they would move faster then normal and have the space to

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Which wave has a wavelengh that is most likely seen as red light?
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D.

Explanation:

Red light has wavelengths around 620 to 750 nm.

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Physics is killing me. Any help?
Anni [7]

Answer:

The simplified expression is  3.833 x 10⁷ g

Explanation:

Given expression;

3.88 x 10⁷ g  -  4.701 x 10⁵ g

The expression above is simplified as follows;

= (3.88 x 100 x 10⁵ )g  -  ( 4.701 x 10⁵) g

= (388 x 10⁵ )g -  ( 4.701 x 10⁵) g

= (388  - 4.701 ) x (10⁵ )g

= 383.299 x 10⁵ g

In standard form, the simplified expression can be expressed as;

= (3.83299 x 100 x 10⁵) g

= 3.83299 x 10⁷ g

= 3.833 x 10⁷ g

Therefore, the simplified expression is  3.833 x 10⁷ g

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3 years ago
Planet-X has a mass of 3.42×1024 kg and a radius of 8450 km. What is the First Cosmic Speed i.e. the speed of a satellite on a l
tino4ka555 [31]

Answer:

First cosmic speed = 5195.74m/s

Second cosmic speed = 7346.05m/s

The raduis of the synchronous 0rbit of satellite is 2.80×10^7m

Explanation:

The first cosmic speed Is determined using the Orbital speed equation which is given by:

V = Sqrt(GM/r)

Where G = gravitational constant = 6.67 ×10^-11

M = Mass of planet

r = radius of the planet

V = Sqrt (6.67×10^-11)(3.42×10^24)/(8450×10^3)

V = Sqrt (2.28×10^14)/(8450×10^3)

V = Sqrt ( 26995739.64)

V = 5195.75m/s

The second cosmic speed is given by :

V = Sqrt(2 × GM)/r

V = Sqrt (2 × (6.67×10^-11)(3.42×10^24)/(8450×10^3)

V = Sqrt( 4.5×10^14)/ (8450×10^3)

V = Sqrt(53964497.04)

V = 7346.05m/s

The raduis of the synchronous orbit if the satellite around the planet is given by:

r = Cuberoot(T^2GM/4 pi r^2where T is the period of rotation of the planet in second

Given :

T = 17.1 hours converting to seconds

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r = Cuberoot ([(61560)^2×(6.67×10^-11)(3.42×10^24)/ (4 ×3.142×r^2)]

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5 0
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Bailey was driving down the street on her motorcycle and reduced her velocity from 25
Sav [38]

Answer:

a = - 25 m/s²

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The magnitude of acceleration or deceleration of an object can be found by using the third equation of motion as follows:

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2a(8 m) = (15 m/s)² - (25 m/s)²

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<u>a = - 25 m/s²</u>

<u>Here negative sign indicates deceleration</u>

5 0
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