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juin [17]
3 years ago
7

Air at 20°c (1 atm) enters into a 5-mm-diameter and 10-cm long circular tube at an average velocity of 5.4 m/s. the tube wall is

maintained at a constant surface temperature of 160°c. determine the convection heat transfer coefficient and the outlet mean temperature. evaluate the air properties at 50°c. the properties of air at 50°c are: ρ = 1.092 kg/m3 μ = 1.963*10-5 kg/m∙s k = 0.02735 w/m∙k cp = 1007 j/kg∙k v = 1.798 *10−5 m2/s μs = 2.420*10−5 kg/m∙s pr = 0.7228
Physics
1 answer:
velikii [3]3 years ago
8 0
Im really sorry. pls dont report this comment. i need points. I have a question but I am bad at english and I cant answer peoples
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A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
Closed circuit
VMariaS [17]

A. the light bulb goes out once the circuit is open since it causes the flow of electricity to cut off. the light bulb dosent get the energy it needs to light up

Explanation:

B. a simple example of this in our every day life is a light switch. when you switch the light on then the circuit is closed and the energy transfers to the light bulb, when u switch the light off then you cut off the lights source of energy which causes the light to turn off.

5 0
2 years ago
What mass of water will fill a tank that is 100.0 cm long, 50.0 cm wide, and 30.0 cm high? Express the answer in grams.
Bond [772]
calculate\ mass\ from\ formula\ for\ density:\\\\
density=volume*mass\\
mass=\frac{density}{volume}\\\\
volume=width*length*height\\
volume=100*50*30=150000cm^3=0,15m^3\\\\
density\ of\ water=\ 999.9720 \frac{kg}{m^3} \\\\mass=0,15*999.9720=149,9958kg=149995,8grams\\\\ Answer\ is\ :\\water\ which\ weight\ is\ equal\ to\ 149995,8grams.

4 0
3 years ago
Brandon is flying to the Western United States. His plane manages to cover 700 miles in 2 hours.
Naddik [55]

thats cool for brandon

4 0
3 years ago
Read 2 more answers
A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a
photoshop1234 [79]

Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

f'=f(\frac{v-v_o}{v+v_s})    (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

v=\lambda f\\\\

v: speed of sound

λ: wavelength = 0.700 m

f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz

Next, you replace the values of all parameters in the equation (1):

f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

I=\frac{P}{4\pi r^2}

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}

hence, the intensity is 8.12*10^⁻4 W/m^2

3 0
3 years ago
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