For each statement, circle whether it is true or false.

How is cold front formation different from stationary front formation?

padilas [110]

Answer:

A

Explanation:

4 0

3 years ago

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

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5 0

2 years ago

A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. at the bottom, she is moving 20.3 m/s.how much work did f

zhuklara [117]

Answer:

5.791244495 KNm

Explanation:

The height h is given by, $h=42.6sin42.3^{o}$

Potential energy, PE is given by

PE=mgh where m is mass of the woman, g is acceleration due to gravity whose value is taken as $9.81 m/s^{2}$ and h is already given hence substituting 77 Kg for m we obtain

$PE=77*9.81*42.6sin42.3^{o}=21656.7095 Nm$

PE=21.6567095 KNm

We also know that Kinetic energy is given by $0.5mv^{2}$ where v is the velocity and substituting v for 20.3 we obtain

$KE=0.5*77*20.3^{2}=15865.465 Nm$

KE=15.865465 KNm

Friction work is the difference between PE and KE hence

Friction work=21.6567095 KNm-15865.465 Nm=5.791244495 KNm

8 0

3 years ago

Hii please help i’ll give brainliest if you give a correct answer please please hurry

fiasKO [112]

Answer:

B most likely

Explanation:

tell me if this is right or wrong

8 0

3 years ago

A golf ball is hit from the ground with an initial velocity of 208 feet per second. Assume the starting height of the ball is 0

kirill115 [55]

It will take 13 seconds for the golf ball to hit the ground. The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

7 0

3 years ago

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