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3 years ago

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4. In a humidifier, air at a dry bulb temperature of 40 oC and relative humidity of 10% is humidified to a relative humidity of

40%. Determine the amount of moisture added in the humidifier per kg of dry air.

1 answer:

lesya692 [45]3 years ago

8 0

Answer:

0.0141 kg/kg.

Explanation:

So, we are given the following parameters or information from the problem above:

The relative humidity = 10% was humidified to 40%, the temperature = 40°C.

From the chart, at temperature = 40°C and relative humidity = 10%, the weight of dry air = a1 = 0.0046 kg/kg, volume = 0.89 m³/kg.

Also, from the chart at temperature = 40°C and relative humidity = 40%, the weight of dry air = a2 = 0.00187 kg/kg, volume = 0.913 m³/kg.

Therefore, the amount of moisture added in the humidifier per kg of dry air = 0.0046 kg/kg - 0.00187 kg/kg = 0.0141 kg/kg.

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The most recently discovered system close to Earth is a pair of brown dwarfs known as Luhman 16. It has a distance of 6.5 light-

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Determine the kinetic energy of a girl of mass 40kg running with a velocity of 3m/s

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Read 2 more answers

A ball rolls off a desk at a speed of 2 m/s and lands 0.50 seconds later.

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Answer:

(a) The ball lands <u>1 m</u> ahead of the desk's base.

(b) The height of the desk is <u>1.225 m.</u>

Explanation:

Given:

Initial velocity of the ball is, $u_{0}=2\ m/s$

Time of flight of the ball is, $t=0.50\ s$

(a)

Let the ball fall at a distance of $x$ from the base of the desk and let the height of the desk be $y$ .

The given motion of the ball is a projectile motion that can be divided into 2 mutually perpendicular directions.

Now, considering only the horizontal motion of the ball. The ball not under the influence of any force in the horizontal direction.

Therefore, the distance covered by the ball horizontally is given as:

$x=v_0\times t\\x=2\times 0.50=1\ m$

So, the ball lands 1 m ahead of the desk's base.

(b)

Now, consider only the vertical motion of the ball. The ball is under the influence of gravity.

So, vertical distance can be obtained using Newton's equations of motion.

We use the following equation of motion:

$y-y_0=u_{oy}t+\frac{1}{2}gt^2$

Where,

$y_{0}\rightarrow \textrm{initial vertical position of ball}\\u_{oy}\rightarrow \textrm{initial vertical velocity}\\g\rightarrow \textrm{acceleration due to gravity}\\y\rightarrow \textrm{final position of ball vertically}$

Here, as the ball leaves the desk horizontally, initial velocity is only in the horizontal direction and doesn't have any component in the vertical direction. So, $u_{oy}=0\ m/s$

Plug in $y_0=h,y=0,g=-9.8,u_{oy}=0,t=0.5$ . Solve for $h$ .

$0-h=0+\frac{1}{2}(-9.8)(0.5)^2\\-h=-4.9\times 0.25\\h=1.225\ m$

Therefore, the height of the desk is 1.225 m.

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3 years ago

A 70.0-kg skier is sliding at 4 m/s when they slide down a 2m high hill. At the bottom of the hill they run into a large 2800 N/

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Answer:

$\Delta x=245\ mm$

Explanation:

Given:

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height of the hill, $h=2\ m$

spring constant, $k=2800\ N.m^{-1}$

<u>final velocity of skier before coming in contact of spring:</u>

Using eq. of motion:

$v^2=u^2+gh$

$v^2=4^2+9.8\times 2$

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<u>Now the time taken by the skier to reach down:</u>

$v=u+gt$

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<u>Now we calculate force using Newton's second law:</u>

$F=\frac{dp}{dt}$

$F=\frac{m(v-u)}{t}$

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<u>∴Compression in spring before the skier momentarily comes to rest:</u>

$\Delta x=\frac{F}{k}$

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$\Delta x=0.245\ m$

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3 years ago