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3 years ago

________ occurs when the relative humidity is 100%.

Physics

2 answers:

kaheart [24]3 years ago

7 0

Saturation.............

ExtremeBDS [4]3 years ago

3 0

<u>"Saturation"</u> occurs when the relative humidity is 100%.

Humidity discloses to you the moisture content of the climate, or how much water vapor there is noticeable all around. At the point when the humidity is high it feels onerous outside in light of the fact that perspiration doesn't vanish and give cooling. At the point when the humidity is low you feel cooler, yet your skin dries out and you get dried out more effortlessly on the grounds that more dampness is being vanished from your body.

Relative humidity is given as a rate. This rate reveals to you how close the air is to being saturated. On the off chance that the relative humidity is 100%, the air is saturated.

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A tow rope pulls a 1450 kg truck, giving it an acceleration 1.25 m/s^2. What force does the rope exert?

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3 years ago

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3 years ago

Calculate the nuclear binding energy per nucleon for 136^Ba if its nuclear mass is 135.905 amu.

kompoz [17]

Answer:

1.312 x 10⁻¹² J/nucleon

Explanation:

mass of ¹³⁶Ba = 135.905 amu

¹³⁶Ba contain 56 proton and 80 neutron

mass of proton = 1.00728 amu

mass of neutron = 1.00867 amu

mass of ¹³⁶Ba = 56 x 1.00728 amu + 80 x 1.00867 amu

= 137.10128 amu

mass defect = 137.10128 - 135.905

= 1.19628 amu

mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg

= 1.9858 x 10⁻²⁷ Kg

speed of light = 3 x 10⁸ m/s

binding energy,

E = mass defect x c²

E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²

E = 17.87 x 10⁻¹¹ J/atom

now,

binding energy per nucleon = $\dfrac{17.87\times 10^{-11}}{136}$

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4 0

3 years ago

Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag fo

Sonbull [250]

Answer:

$2.267$

Explanation:

Drag force is given by

$F=\dfrac{1}{2}\rho Av^2C$

C = Drag coefficient is constant

A = Area is constant

$v_1$ = Velocity of the passenger jet = 1200 km/h = $\dfrac{1200}{3.6}\ \text{m/s}$

$v_2$ = Velocity of the prop plane = $\dfrac{1}{4}v_1$

$\rho_1$ = Density of the air where the jet was flying = $0.38\ \text{kg/m}^3$

$\rho_2$ = Density of the air where the prop plane was flying = $0.67\ \text{kg/m}^3$

$F\propto \rho v^2$

$\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267$

The ratio of the drag forces is $2.267$ .

5 0

3 years ago