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3 years ago

state and explain any effect on sensitivity of liquid in a glass thermometer (i) reducing the diameter of capillary tube

Physics

1 answer:

Ad libitum [116K]3 years ago

4 0

Answer:

please give me brainlist and follow

Explanation:

The measuring sensitivity of liquid-in-glass thermometers increases with the amount of liquid in the thermometer. The more liquid there is, the more liquid will expand and rise in the glass tube. For this reason, liquid thermometers have a reservoir to increase the amount of liquid in the thermometer.

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A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin

Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

$\epsilon = \frac{NBA}{t}$

Where,

$\epsilon =$ electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

$R = R_c + R_d$

And that the current can be expressed as function of charge and time, then

$I = \frac{q}{t}$

Equation Faraday's law and Ohm's law we have,

$V = \epsilon$

$IR = \frac{NBA}{t}$

$(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}$

Re-arrange for Magnetic Field B, we have

$B = \frac{q(R_c+R_d)}{NA}$

Our values are given as,

$R_c = 58.7\Omega$

$R_d = 45.5\Omega$

$N = 120$

$q = 3.53*10^{-5}C$

$A = 3.21cm^2 = 3.21*10^{-4}m^2$

Replacing,

$B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}$

$B = 0.095T$

Therefore the magnetic field in the System is 0.095T

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3 years ago

Which is a property of light a: mass b: density c: wavelength d: pitch

Reptile [31]

Im pretty sure its C- Wavelength

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3 years ago

A solid ball with a mass of 4.25 kg and a

Liono4ka [1.6K]

7.20 units long yea

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3 years ago

Example 3 :

kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

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2 years ago

How far will Mr. Woodland’s pet coyote travel in 2 hours at an average speed of 28 km/he

Umnica [9.8K]

Explanation:

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3 years ago

state and explain any effect on sensitivity of liquid in a glass thermometer (i) reducing the diameter of capillary tube​

state and explain any effect on sensitivity of liquid in a glass thermometer (i) reducing the diameter of capillary tube