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oksano4ka [1.4K]
3 years ago
15

When temperature decreases but actual water vapor content remains the same, what happens to relative humidity?

Physics
2 answers:
ruslelena [56]3 years ago
8 0

Answer:

d. Relative humidity increases.

Explanation:

The expression of relative humidity in terms of absolute humidity, absolute pressure and saturation pression at measured temperature is:

\phi = \frac{\omega \cdot P}{(0.622+\omega)\cdot P_{sat}}

When temperature decreases, the saturation pressure decreases also and, consequently, relative humidity increases. Therefore, the right answer is option D.

vaieri [72.5K]3 years ago
3 0

Answer:

d. Relative humidity increases

Explanation:

Relative humidity is defined as the amount of water vapor which can be found in air.

When temperature decreases and the actual water vapor content remains the same the relative humidity increases.

When the temperature increases and the actual water vapor content remains the same the relative humidity decreases.

This is because air at a higher temperature needs more moisture for it to get saturated than air with a lower temperature.

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Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings
Hatshy [7]

Answer:0.316 rad/s

Explanation:

Given

mass of Person m=65 kg

velocity of person v=3.8 m/s

diameter of turntable d=6.5 m

moment of Inertia of the table I_0=1850 kg-m^2

Moment of inertia of Person I

I=mr^2=65\times (\frac{6.5}{2})^2

I=686.56 kg-m^2

initial angular velocity \omega _1=\frac{v}{r}

\omega _1=\frac{3.8}{3.25}=1.17 rad/s

Conserving Angular momentum

I\omega _1=(I+I_0)\omega _2 , where \omega _2=final\ angular\ velocity

686.56\times 1.17=(686.56+1850)\times \omega _2

\omega _2=\frac{686.56}{2536.56}\times 1.17

\omega _2=0.316 rad/s

4 0
3 years ago
A 1250-kg compact car is moving with velocity v1 =36.2i^+12.7j^m/s. It skids on a frictionless icy patch and collides with a 448
MA_775_DIABLO [31]

Momentum is conserved, so the sum of the separate momenta of the car and wagon is equal to the momentum of the combined system:

(1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s) = ((1250 + 448) kg) <em>v</em>

where <em>v</em> is the velocity of the system. Solve for <em>v</em> :

<em>v</em> = ((1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s)) / (1698 kg)

<em>v</em> ≈ (30.3 <em>i</em> + 12.0 <em>j</em> ) m/s

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3 years ago
Which one of the following explains how light energy helps us see all kinds of objects around us?
vampirchik [111]

Answer:

light reflects off the objects and enter our eyes

6 0
3 years ago
Why pilots need to study navigation
masya89 [10]
To know where they’re heading to
3 0
4 years ago
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Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assu
baherus [9]

Answer:

823.46 kgm/s

Explanation:

At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.

So, mgh = 1/2mv²

From here, his velocity just as he reaches the surface of the water is

v = √2gh

h = 9 m and g = 9.8 m/s²

v = √(2 × 9 × 9.8) m/s

v = √176.4 m/s

v₁ = 13.28 m/s

So his velocity just as he reaches the surface of the water is 13.28 m/s.

Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.

So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.

His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,

J = (62 × 0 + 62 × 13.28) kgm/s = 0 +  823.46 kgm/s = 823.46 kgm/s

So the magnitude of the impulse J of the water on him is 823.46 kgm/s

6 0
3 years ago
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