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Roman55 [17]
3 years ago
8

Which best describes the relationship between humidity and air pressure?

Physics
2 answers:
irga5000 [103]3 years ago
7 0

Humid air is lighter, so it has lower pressure.

spayn [35]3 years ago
5 0
Answer:<span>Humid air is lighter, so it has lower pressure.

The reason is the molecules of water are H2O, whose molar mass is 18 g/mol.

These molecules displaces molecules of N2 and O2, whose molar masses are:

N2: 2*14g/mol = 28 g/mol, and
O2: 2*16g/mol = 32 g/mol.

Then molecules of 28g/mol and 32 g/mol are being replaced with molecules of 18g/mol, leading to a lower weight of the same volume of air, which results in lower pressure.
</span>
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PLEASE HELP ME WITH THIS ONE QUESTION
sdas [7]

Answer:

0.34 m

Explanation:

From the question,

v = λf................ Equation 1

Where v = speed of sound, f = frequency, λ = Wave length

Make λ the subject of the equation

λ = v/f............... Equation 2

Given: v = 340 m/s, f = 500 Hz.

Substitute these values into equation 2

λ = 340/500

λ = 0.68 m

But,  the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length

Therefore,

λ/2 = 0.68/2

λ/2 = 0.34 m

Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m

6 0
3 years ago
A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.
Airida [17]

Answer:

a) v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} ), b) \Delta K = 9.559\times 10^{-5}\,J

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v

The final velocity is:

v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )

b) The change in the kinetic energy of the 13.5 g coin is:

\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]

\Delta K = 9.559\times 10^{-5}\,J

4 0
3 years ago
Read 2 more answers
Can someone give me tips to passing AP Physics 1 with a solid A? Like, tips on how to memorize a lot of equations? Thank you!
Anna007 [38]
A good way for me to remember things is to study it, and to write it down! Say you want the formula for speed, I would write the formula multiple times on a piece of paper! 

Here's a video that I haven't actually watched, I just looked it up! It might help you out though: <span>https://www.youtube.com/watch?v=-Wqrw4G79Kc</span>
3 0
3 years ago
A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.
Maksim231197 [3]

Answer:

v=33.764m/s

Explanation:

Given data

Mass m=1.2 g=0.0012 kg

diameter d=0.76 m

Friction Force F=3.6 N

To find

Velocity v

Solution

From the Centripetal force we know that

F_{c}=\frac{mv^{2} }{r}

Where m is mass

v is velocity

r is radius

Substitute the given values to find velocity v

So

F_{c}=\frac{mv^{2} }{r}\\v^{2}=\frac{F_{c}(r)}{m}\\ v=\sqrt{\frac{F_{c}(r)}{m}}\\ v=\sqrt{\frac{F_{c}(diameter/2)}{m}}\\v=\sqrt{\frac{(3.6N)(0.76/2)m}{(0.0012kg)}}\\v=33.764m/s

4 0
3 years ago
Convert 2kg into SI unit of force​
Doss [256]

Answer:

19.6133 newton.

...?.......

3 0
3 years ago
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