Ask question

Ask question

All categories

3 years ago

9

What does a pshycometer measure

1 answer:

erma4kov [3.2K]3 years ago

5 0

It measures<span> the relative humidity in the atmosphere through the use of two thermometers. The first, a dry bulb thermometer, is used to </span>measure<span> the temperature by being exposed to the air. The second, a wet bulb thermometer,</span>measures<span> temperature by having the bulb dipped in a liquid.</span>

You might be interested in

Write with all the steps and formulas and drawing if needed.

lianna [129]

<u>The answer is not contained detail explanation, just a solution and the required values. </u>

All the details are in the pictures, the answers are marked with orange colour.

Note,

in the task no 20.:

$m_A - the \ mass \ of \ A; \ m_B-the \ mass \ of \ B \ balls.\\V_A \ and \ V_B-the \ velocities \ of \ the \ A&B \ balls \ before \ collision.\\V'_A \ and \ V'_B-the \ velocities \ of \ the \ A&B \ balls \ after \ collision.$

V - the velocity of the pair of the balls after collision.

in the task no 21:

m₁ - the mass of the copper ball; m₂ - the mass of the copper calorimeter; m₃ - the mass of the water; t₀ - the initial temperature of water in the copper calorimeter; θ - the final temperature in the calorimeter after the copper ball is transferred into a copper calorimeter; t₁ - the required initial temperature of the copper ball before it is transferred into the calorimeter.

7 0

3 years ago

In the diagram, q1= -2.60*10^-9 C and

Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

$E_{net}=90.37\: N/c$

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

$|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}$

Where:

k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
q1 is the first charge
d1 is the distance from q1 to P

$|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}$

$|E_{1}|=80.84\: N/C$

And E2 will be:

$|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}$

$|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}$

$|E_{2}|=40.39\: N/C$

Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.

$E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}$

$E_{net}=\sqrt{80.84^{2}+40.39^{2}}$

$E_{net}=90.37\: N/c$

I hope it helps you!

7 0

3 years ago

iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms

saw5 [17]

Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

$\frac{2}{a^{3} } = \frac{N_{A} }{V_{molar} }$ where $V_{molar} = \frac{A}{\rho }$ an and A is the atomic mass of iron.

Therefore:

$\frac{2}{a^{3} } = \frac{N_{A} * p }{A}$

This implies that:

$A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} } }$

= $\frac{4}{\sqrt{3} }r$

Assuming that there is no phase change gives:

$\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3} }$

= 8.60 g/m³

3 0

3 years ago

A reconnaissance plane flies 545 km away from

Nikolay [14]

Answer:

681.6/ms

Explanation:

A reconnaissance plane flies 545 km away from its base at 568 m/s. then flies back to its base at 852 m/s.

What is its average speed?

Answer in its of m/s

Avg speed of the round trip is

2*568*852/(568+852)= 681.6/ms

6 0

3 years ago

serious [3.7K]

I belive what your looking for is oxygen

5 0

3 years ago

Read 2 more answers