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3 years ago

When saturated air is cooled what happens to its dew point

1 answer:

6 0

When saturated air is cooled, it simply reaches its dew point. Dew point is simply the temperature at which dew begins to form.

Dew point of saturated air is already pre-determined by how much water vapor the air contains. A state of saturation exists when the air is holding the maximum amount of water vapor possible at the existing temperature and pressure. The higher the dew point, the higher the moisture content of the air. Cooling does not change the dew point of saturated air, rather its the level of saturation.

So if the air has more moisture, dew will form at a higher temperature and vice versa, but dew point is NEVER EVER GREATER than the air temperature.

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What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?

viktelen [127]

Answer: 27.21 V

Explanation:

The electric potential $V_{E}$ due to a point charge is expressed as:

$V_{E}=k\frac{q}{r}$

Where:

$k=9(10)^{9}\frac{Nm^{2}}{C^{2}}$ is the electric constant

$q=1.6(10)^{-19}C$ is the electric charge of the hydrogen nucleus, which is positive

$r=5.292(10)^{-11}m$ is the distance

Rewritting the equation with the known values:

$V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}$

Finally:

$V_{E}=27.21 V$

5 0

3 years ago

A cello string 0.75 m long has a 220 hz fundamental frequency. find the wave speed along the vibrating string. answer in units o

maxonik [38]

For fundamental frequency of a string to occur, the length of the string has to be half the wavelength. That is,

1/2y = L, where L = length of the string, y = wavelength.

Therefore,
y = 2L = 2*0.75 =1.5 m

Additionally,
y = v/f Where v = wave speed, and f = ferquncy

Then,
v = y*f = 1.5*220 = 330 m/s

4 0

3 years ago

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

$F = \frac{k|q_1||q_2|}{r^2}$

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

4 0

3 years ago

How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a spa

Zarrin [17]

Answer:

cutting the magnet in two parts each part has a North and South pole,

Explanation:

In magnetism the magnetic mono-poles are not found, this means that we do not have magnetic charges alone, therefore when cutting the magnet in two parts each part has a North and South pole, the magnetic lines go from the North pole to the South pole, see attached.

The density of the lines is approximately the intensity of the magnetic field.

4 0

2 years ago

Please answer help me

Andrews [41]

I believe the answer is x

7 0

3 years ago