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Vilka [71]
3 years ago
12

What is the flow rate of the air in the pipeline if the cross sectional area of a pneumatic pipeline is 0.01 m2 and the speed of

the air in the pipeline is 20 m/s?
Mathematics
1 answer:
e-lub [12.9K]3 years ago
7 0

Answer:

0.2 m³/s

Step-by-step explanation:

Applying,

Q = vA.............. Equation 1

Where Q = Flow rate of  air in the pipeline, v = flow velocity of the air in the pipeline, A = cross sectional area of the pneumatic pipeline.

From the question,

Given: v = 20 m/s, A = 0.01 m²

Substitute these values into equation 1

Q = 20(0.01)

Q = 0.2 m³/s

Hence the flow rate of air in the pipeline is  0.2 m³/s

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Kylie is planning to provide online science and Chinese language tutoring sessions. The sciences
Ivahew [28]

Answer:

120 ≤ 20c + 40s

Step-by-step explanation:

(Assuming her name is Kylie, who is giving 40 minute science sessions and 20 minute Chinese sessions.)

The unit for time will be minutes.

Write an equation for time needed for science sessions

40s = t

Write an equation for time needed for Chinese sessions

20c = t

Combine the two equations for the total time.

20c + 40s = total time

She does not want the total time to be more than two hours.

Convert two hours to minutes. There are 60 minutes in an hour.

2h * 60mins = 120 mins

Therefore t ≤ 120. Include this into the equation.

120 ≤ 20c + 40s

8 0
3 years ago
Am i correct pls help
a_sh-v [17]

Answer:

yea ur correct

Step-by-step explanation:

use PhotoMath or something like that to check next time, it'll save a lot of time

6 0
2 years ago
Stacy decided that she would start doing sit ups every day. Her plan each day was to do twice as many sit ups as she had done th
Whitepunk [10]

Answer:

Day five= 32 sit ups

Step-by-step explanation:

Day 1= 2 sit ups

Day 2= 4 sit ups

Day 3= 8 sit ups

Day 4= 16 sit ups

Day 5= 32 sit ups

8 0
3 years ago
A random sample of 100 automobile owners in thestate of Virginia shows that an automobile is driven onaverage 23,500 kilometers
Anastaziya [24]

Answer:

a) 22497.7 < μ< 24502.3

b)  With 99% confidence the possible error will not exceed 1002.3

Step-by-step explanation:

Given that:

Mean (μ) = 23500 kilometers per year

Standard deviation (σ) = 3900 kilometers

Confidence level (c) = 99% = 0.99

number of samples (n) = 100

a) α = 1 - c = 1 - 0.99 = 0.01

\frac{\alpha }{2} =\frac{0.01}{2}=0.005\\ z_{\frac{\alpha }{2}}=z_{0.005}=2.57

Using normal distribution table, z_{0.005 is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.

The margin of error (e) is given as:

e= z_{0.005}\frac{\sigma}{\sqrt{n} }  = 2.57*\frac{3900}{\sqrt{100} } =1002.3

The 99% confidence interval = (μ - e, μ + e) = (23500 - 1002.3, 23500 + 1002.3) =  (22497.7, 24502.3)

Confidence interval = 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

5 0
3 years ago
Please help with answers A and B :)
Over [174]
A. Terms
b. 9
64 to 49 is -15, 49 to 36 is -13, 36 to 25 is -11, 25 to 16 is -9, so the next number is 16-7=9
5 0
2 years ago
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