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Charra [1.4K]
2 years ago
15

Calculate a rate of cooling down of air from 80 C to 5C Show calculation. Give an answer in cubic meters per minute and cfm.

Physics
1 answer:
antoniya [11.8K]2 years ago
3 0

Explanation:

Given that,

Rate of cooling of air

Initial temperature= 80°C

Final temperature = 5°C

We need to calculate

Using newton's law of cooling

\dfrac{dT}{dt}=c(T-T_{0})

\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})

Where, dT=T_{1}-T_{2}

Here, T =\dfrac{T_{1}+T_{2}}{2}

T_{0} = 25°C  (surrounding temperature)

dt = 1 minute

\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})

Put the value into the formula

\dfrac{80-5}{1}=c(\dfrac{85}{2}-25)

c=\dfrac{75}{17.5}

c=4.285\ cubic\ meter/minute

Hence, This is the required answer.

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Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),
salantis [7]

Answer:

The  drag coefficient is  D_z  =  1.30512  

Explanation:

From the question we are told that

     The density of air is  \rho_a  = 1.21 \ kg/m^3

     The diameter of bottom part is  d = 0.15 \ m

The  power trend-line  equation is mathematically represented as

      F_{\alpha }  = 0.9226 * v^{0.5737}

let assume that the velocity is  20 m/s

Then

      F_{\alpha }  = 0.9226 * 20^{0.5737}

       F_{\alpha }  = 5.1453 \ N

The drag coefficient is mathematically represented as

      D_z  =  \frac{2 F_{\alpha } }{A \rho v^2 }

Where  

     F_{\alpha } is the drag force

      \rho is the density of the fluid

       v is the flow velocity

       A is the area which mathematically evaluated as

       A = \pi r^2 =  \pi  \frac{d^2}{4}

substituting values

     A =  3.142 *    \frac{(0.15)^2}{4}

     A = 0.0176 \  m^2

Then

   D_z  =  \frac{2 * 5.1453 }{0.0176 * 1.12 *  20^2 }

   D_z  =  1.30512  

3 0
2 years ago
A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f
zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s

v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

s = v_vt + gt^2/2

10 = 19.97t - 9.8t^2/2

4.9t^2 - 19.97t + 10 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}

t= \frac{19.9658877511823\pm14.24}{9.8}

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

s_1 = v_ht_1 = 15*0.58 = 8.8 m

s_2 = v_ht_2 = 15*3.49 = 52.5m

8 0
3 years ago
A lad, waiting for his friend walks in the sidewalk, in front of her house, from the front door, first, he moves towards the Pos
Andreas93 [3]

His total displacement from his original position is -1 m

We know that total displacement of an object from a position x to a position x', d = final position - initial position.

d = x' - x

If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.

Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.

His final position, x" after moving back 6 m is gotten from

x" - x' = -6 m

x" = -6 + x'

x" = -6 + 5

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Thus, his total displacement from his original position is

d = final position - initial position

d = x" - x

d = -1 m - 0 m

d = -1 m

So, his total displacement from his original position is -1 m

Learn more about displacement here:

brainly.com/question/17587058

3 0
3 years ago
A car, starting from rest, accelerates at
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Using V= vo +at with Vo = 0 and a= 4m/sec2.
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The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece
Brilliant_brown [7]

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m

or

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

8 0
2 years ago
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