A girl whose weight is 220-N hangs from

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

levacccp [35]

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

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5 0

2 years ago

a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an

Mademuasel [1]

We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
$T_1 - W_p - W_s + T_2 =0$
where
$W_p = (90kg)(9.81 m/s^2)=883 N$ is the weight of the person
$W_s = (200kg)(9.81 m/s^2)=1962 N$ is the weight of the scaffold
Re-arranging, we can write the equation as
$T_1 = 2845 N-T_2$ (1)

2) Equilibrium of torques:
$T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0$
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using $W_p = 883 N$ and replacing T1 with (1), we find
$2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0$
from which we find
$T_2 = 1128 N$

And then, substituting T2 into (1), we find
$T_1 = 1717 N$

8 0

3 years ago

Compare and contrast climate and weather

Levart [38]

It’s 49% of its original gig hope this helps!

8 0

3 years ago

Read 2 more answers

One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing

Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

$m = m_{1}=m_{2}$

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Along X- axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}$

$v_{1}=u_{1}+u_{2}$

Put the value into the formula

$v_{1}=2.2i-0.80i$

$v_{1}=1.4i\ m/s$

Along Y-axis

$0=m_{1}v_{1}+m_{2}v_{2}$

$m_{1}v_{1}=-m_{2}v_{2}$

$v_{1}=-v_{2}$

Put the value into the formula

$v_{1}=-1.36j\ m/s$

Then the final speed of the first ball

$v_{1}=\sqrt{(1.4)^2+(1.36)^2}$

$v_{1}=1.95\ m/s$

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

$\tan\theta=\dfrac{v_{2}}{v_{1}}$

$\tan\theta=\dfrac{-1.36}{1.4}$

$\theta=\tan^{-1}\dfrac{-1.36}{1.4}$

$\theta=-44.16^{\circ}$

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0

3 years ago

Bromine, a liquid at room temperature, has a boiling point

lukranit [14]

Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.

3 0

3 years ago