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jonny [76]
3 years ago
11

Which of the following is most likely to be an observation made by a physiologist

Physics
1 answer:
Naya [18.7K]3 years ago
3 0
Do you have the answer choices ?
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How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

Read more about the flows rate

brainly.com/question/27880305

#SPJ1

5 0
2 years ago
The Celsius temperature scale is based on which of the following
Lapatulllka [165]
Freezing.................
6 0
3 years ago
Triangle ABC is similar to triangle A'B'C'. Which sequence of similar transformations could map △ABC onto △A'B'C'?
Olin [163]

Answer:

b.

Explanation:

Dilation and Translation

6 0
3 years ago
Read 2 more answers
A two-slit pattern is viewed on a screen 1.26 m from the slits. If the two fourth-order maxima are 53.6 cm apart, what is the to
Anettt [7]

Answer:

= 6.55cm

Explanation:

Given that,

distance = 1.26 m

distance between  two fourth-order maxima = 53.6 cm

distance between central bright fringe and fourth order maxima

y = Y / 2

  =  53.6cm / 2

  = 26.8 cm

  =0.268 m

tan θ = y / d

         = 0.268 m /  1.26 m

         = 0.2127

       θ = 12°

4th maxima

d sinθ = 4λ

d / λ = 4 / sinθ

d / λ = 4 / sin 12°

d / λ = 19.239

for first (minimum)

d sinθ = λ / 2

sinθ =  λ / 2d

       =  1 / 2(19.239)

       = 1 / 38.478

       = 0.02599

    θ =  1.489°

tan θ = y / d

y = d tan θ

  = 1.26 tan 1.489°

  = 0.03275

the total width of the central bright fringe  

Y = 2y

  = 2(0.03275)

  = 0.0655m

  = 6.55cm

4 0
3 years ago
An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 6.0 μC flows to the positive plate. The 4.0 V batte
Ugo [173]

Answer:

1 μC extra charge will be flow here

Explanation:

Given data

battery V1 = 4.0 V

flows Q1 =  6.0 μC

replace battery V2 = 7.0 V

to find out

what happen if we replace battery

solution

we apply here principal of capacitor

that is Q directly proportional voltage

so we say Q2/Q1 = V2/ V1

put all value here

Q2/Q1 = V2/ V1

Q2/6 = 7/ 6

Q2 = 7

so we see here 7 μC will be flow

and Q = Q2 - Q1 = 7 - 6 = 1 μC

so we also say that 1 μC extra charge will be flow here

5 0
3 years ago
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