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deff fn [24]
3 years ago
9

Why does 100% relative humidity in the winter feel nothing like 100% relative humidity in the summer?

Physics
1 answer:
adell [148]3 years ago
4 0

The 100% relative humidity in the winter feel nothing like 100% in summer because it depends on "the saturation of the temperature".

<u>Explanation:</u>

Temperature really makes a big difference. Even once warm, a cold winter air produces much less humidity than summer heat. One cubic unit of air needs 0.001 ounces of water to saturate it, to elevate its ratio to one hundred per cent.

Nevertheless, it takes 0.022 ounces of water to saturate the one cubic unit of air once the temperature is eighty, which is twenty-two times that amount of water. Air with a humidity of one hundred percent at eighty degrees holds twenty-two times as much water as air at zero with humidity at one hundred percent.

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What fossil images, there are no pictures attached.
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A car of 10 kg is moving at a speed of 5 m/s collides with a
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Explanation:

M1U1+m2u2 = (m1+m2)vf

m1= 10kg

u1=5

m2=25

u2=0

vf=?

10×5+25×0=35 vf

50+0=35Vf

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What is Newton third law of motion?<br>​
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Answer:

Push and pull is the 3rd motion.

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4 years ago
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According to the definition of power, a powerful person is able to do which of the following?
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5 0
3 years ago
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,
eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

x(t) = A cos (\omega t + \phi) (1)

where

A is the amplitude

\omega is the angular frequency

\phi is the phase

t is the time

The displacement of the piston in the problem is given by

x(t) = (5.00 cm) cos (5t+\frac{\pi}{5}) (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

v(t) = x'(t) = -\omega A sin (\omega t + \phi) (3)

Differentiating eq.(2), we find

v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})

And substituting t=0, we find the velocity at time t=0:

v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)

Differentiating eq.(3), we find

a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})

And substituting t=0, we find the acceleration at time t=0:

a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

\omega=\frac{2\pi}{T}

Using \omega = 5.0 rad/s and solving for T, we find

T=\frac{2\pi}{5 rad/s}=1.26 s

4 0
3 years ago
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