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Advocard [28]
3 years ago
6

In what way would a digital thermometer be preferable to those from a liquid-based thermometer?

Physics
1 answer:
garik1379 [7]3 years ago
5 0
Reading a digital thermometer is more preferable because you can just easily read the measurement from a screen, a number will just show in a screen as compared to a liquid-based thermometer where you still have to read the measurement by counting the markers in the thermometer. Hope this helps! Mark brainly please!
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(a) Calculate the self-inductance (in mH) of a 55.0 cm long, 10.0 cm diameter solenoid having 1000 loops.
DedPeter [7]

Explanation:

(a) We have,

Length of solenoid, l = 55 cm = 0.55 m

Diameter of the solenoid, d = 10 cm

Radius, r = 5 cm = 0.05 m

Number of loops in the solenoid is 1000.

(a) The self inductance in the solenoid is given by :

L=\dfrac{\mu_o N^2A}{l}

A is area

L=\dfrac{4\pi \times 10^{-7}\times (1000)^2\times \pi (0.05)^2}{0.55}\\\\L=0.0179\ H\\\\L=17.9\ mH

(b) The energy stored in the inductor is given by :

E=\dfrac{1}{2}LI^2\\\\E=\dfrac{1}{2}\times 0.0179\times (19.5)^2\\\\E=3.4\ J

Hence, this is the required solution.

4 0
3 years ago
1) Manipulate is to measure as
ValentinkaMS [17]

Answer: The variable that you manipulate is called the independent variable. The variable that you measure is called the dependent variable.

3 0
3 years ago
a specimen of oil having an initial volume of 5000cm³ is subjected to a pressure of 10⁴N/m² and the volume decreases by 0.20cm³.
inessss [21]

Answer:

 B = 2.5 10⁸ Pa

Explanation:

The volume modulus is defined by

           B = - \frac{P}{ \frac{\Delta V}{V} }

           

The negative fate is for the module to be positive since the volume change is negative

       

It is not necessary to reduce the volumes to the SI system, since they are both in the same units

             B = - \frac{10^4}{ \frac{-0.20}{5000} } = \frac{10^4}{4 \ 10^{-5} }

             B = 2.5 10⁸ Pa

4 0
3 years ago
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8 0
3 years ago
Monochromatic light with a wavelength of 384 nm passes through a single slit and falls on a screen 86 cm away. If the distance o
valkas [14]
This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
y= \frac{m\lambda D}{a}
from the center of the pattern. In the formula, m is the order of the minimum, \lambda the wavelenght, D the distance of the screen from the slit and a the width of the slit.

In our problem, the distance of the first-order band (m=1) is y=0.22 cm. The distance of the screen is D=86 cm while the wavelength is \lambda = 384 nm=384 \cdot 10^{-7}cm. Using these data and re-arranging the formula, we can find a, the width of the slit:
a= \frac{m \lambda D}{y}= \frac{1 \cdot 384 \cdot 10^{-7}cm \cdot 86 cm}{0.22 cm}=0.015 cm
3 0
3 years ago
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