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3 years ago

The Kelvin scale is the most common temperature scale used in what

Physics

2 answers:

dalvyx [7]3 years ago

7 0

It's mostly used in CHEMICAL PROCESSES.

Inessa [10]3 years ago

5 0

It is a chemical presses
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If i try to fail and succeed which one did I do

Novay_Z [31]

You've failed because you failing becomes a statement rather than it becoming fact or what actually happened.

7 0

2 years ago

Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second

vampirchik [111]

To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

$I' = \frac{I_0}{2} cos^2 \theta$

Where

$\theta =$ Angle From vertical of the axis of the polarizing filter

$I_0 =$ Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

$I = \frac{I_0}{2}$

Replacing we have that

$I = \frac{370}{2}$

$I = 185W/m^2$

Re-arrange the equation,

$I'= \frac{I_0}{2}cos^2\theta$

Re-arrange to find \theta

$cos^2\theta = \frac{2I'}{I_0}$

$cos^2\theta = \frac{2*138}{370}$

$\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})$

$\theta = 0.5282rad$

$\theta = 30.27\°$

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

4 0

2 years ago

What is the soil fabric?

inessss [21]

Answer:

The term soil fabric refers to the geometrical arrangement of individual particles in a soil, including the geometrical distribution of pore spaces. Soil fabric may be used to describe the particle arrangement in both cohesive soils such as clays and granular soils such as silts, sands, and gravels.

5 0

2 years ago

I’ve already figured out A. I just need help with B and C.

Damm [24]

Answer:

595433.00

1456543.00

Explanation:

7 0

3 years ago

Find the voltage change when: a. An electric field does 12 J of work on a 0.0001-C charge. b. The same electric field does 24 J

kondaur [170]

Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

$V=\dfrac{W}{q}$

$V=\dfrac{12}{0.0001}$

$V=12\times 10^4\ Volt$

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

$V=\dfrac{W}{q}$

$V=\dfrac{24}{0.0002}$

$V=12\times 10^4\ Volt$

Therefore, this is the required solution.

7 0

2 years ago