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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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At a given time of day, the ratio of the height of an object to the length of its shadow is the same for all objects. If a 4-f

katrin2010 [14]

Answer:35.2 ft

Explanation:

Given

height of stick =4 ft

shadow length =2.8 ft

Angle of elevation of sun is

$tan\theta =\frac{4}{2.8}$

let the height of tree be h

as $\theta$ will remain same thus

$tan\theta =\frac{h}{24.64}$

$\frac{4}{2.8}=\frac{h}{24.64}$

h=35.2 ft

8 0

3 years ago

To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.

ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current $i_1$ . In the case of resistance 2, it will cross a current $i_2$

Defined this we proceed to obtain our equations,

For $i_1$ ,

$\frac{di_1}{dt}+i_1R_1 = V$

$I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

For $i_2$ ,

$I_2R_2 =V$

$I_2 = \frac{V}{R_2}$

The current in the entire battery is equivalent to,

$i_t = I_1+I_2$

$i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

Our values are,

$V=1V$

$R_1 = 95\Omega$

$L= 1.5*10^{-2}H$

$R_2 =360\Omega$

Replacing in the current for t= 0.4m/s

$i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})$

$i= 0.0133A$

$i_1 = 0.01052A$

3 0

3 years ago

Three small masses are positioned as follows: 2.0 kg at (0.0 m, 0.0 m), 2.0 kg at (2.0 m, 0.0 m), and 4.0 kg at (2.0 m, 1.0 m).

melamori03 [73]

Refer to the diagram shown below.

The given data is

mass, kg Coordinates. m
------------- -----------------
2 (0, 0)
2 (2, 0)
4 (2, 1)

Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.

Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m

8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m

Answer: (1.5, 0.5) m