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BigorU [14]
2 years ago
15

If the relative humidity is 25 nd the saturation mixing ratio is 24g/kg, what is the mixing ratio?

Physics
1 answer:
levacccp [35]2 years ago
5 0

The mixing ratio is 6.

To find the answer, we have to know about the mixing ratio.

<h3>What is mixing ratio?</h3>
  • The mixing ratio must be calculated in a complex manner.
  • A saturated vapor pressure (es) for values of air temperature and an actual vapor pressure (e) for values of dewpoint temperature must be determined in order to determine the mixing ratio.
  • The air temperature and/or dewpoint temperature must first be converted to degrees Celsius (°C) before the vapor pressures can be calculated.
  • The equation below can be used to determine the relative humidity (rh), as well as the actual mixing ratio and saturated mixing ratio,

                         Rh=\frac{w}{w_s} *100

where; w is the mixing ratio and w(s) is the saturation mixing ratio.

  • In our question, it is given that,

                                    Rh=25\\w_s=24

  • Thus, the mixing ratio will be,

                            w=\frac{Rh*w_s}{100} =\frac{25*24}{100}=6

Thus, we can conclude that, the mixing ratio is 6.

Learn more about mixing ratio here:

brainly.com/question/8791831

#SPJ4

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The formula v = √2.5r models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radiu
mel-nik [20]

Answer:

The maximum safe speed of the car is 30.82 m/s.

Explanation:

It is given that,

The formula that models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r r, is in feet is given by :

v=\sqrt{2.5\ r}.........(1)

A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet

Put the value of r in equation (1) as :

v=\sqrt{2.5\ \times 380}

v = 30.82 m/s

So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.

4 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s

v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s

<u>Time taken to reach 1180 m is 11.29 seconds</u>

v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0
3 years ago
A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,
muminat

Answer:

The speed of the plank relative to the ice is:

v_{p}=-0.33\: m/s

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

m_{g}v_{g}+m_{p}v_{p}=0 (1)

Where:

  • m(g) is the mass of the girl
  • m(p) is the mass of the plank
  • v(g) is the speed of the girl
  • v(p) is the speed of the plank

Now, as we have relative velocities, we have:

v_{g/b}=v_{g}-v_{p}=1.55 \: m/s (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

45v_{g}+168v_{p}=0

v_{g}-v_{p}=1.55

v_{p}=-0.33\: m/s

I hope it helps you!      

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Damping of the sound
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tiny-mole [99]
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8 0
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