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3 years ago

How do intrusive igneous rocks differ from extrusive igneous rocks?

1 answer:

Nutka1998 [239]3 years ago

6 0

Intrusive rocks are formed beneath the earth's surface and extrusive igneous rocks are formed beneath the earth's surface but cool quicken than intrusive igneous rocks

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Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

Learn more about pressure in a liquid:

brainly.com/question/9805263

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0 0

3 years ago

A 4 kg bowling bowl is sitting on a table 1 meter off the ground. How much potential energy does it have?

il63 [147K]

Answer:

$\huge\boxed{\sf P.E. = 39.2\ Joules}$

Explanation:

Given Data:

Mass = m = 4 kg

Acceleration due to gravity = g = 9.8 m/s²

Height = h = 1 m

Required:

Potential Energy = P.E. = ?

Formula:

P.E. = mgh

Solution:

P.E. = (4)(9.8)(1)

P.E. = 39.2 Joules

$\rule[225]{225}{2}$

Hope this helped!

5 0

3 years ago

Q 1 . How many significant figures are in the following measurement? 0.0009(1 point)

Crazy boy [7]

Here we have some questions about experimental errors.

Q1) We want to see how many significant figures have the measure:

0.0009

The number of significant figures is the number of known digits that are not the leading zeros.

Here we can see four leading zeros, and a single-digit different than zero, which is a 9.

Then we have only one significant figure, the 9.

Q2) Here we will use the measure that is the less exact, as the error of that measure may be larger than the smaller significant figures of the other measures.

Then:

31.2 lb + 38.02lb + 45 lb

The worst measure is 45lb, so the smallest significant figure that we should use is the first one at the left of the decimal point, then we need to round the other two measures to the next whole number, we will get:

31 lb + 38 lb + 45 lb = 114lbs

Q3) We know that the measure is 11.5 seconds and the uncertainty of 1.7%, then the uncertainty will be the 1.7% of the above measure:

(1.7%/100%)*11.5 s = 0.1955 s

Notice that our measure has one significant figure after the decimal point, so we need to round the error to the same significant figure.

0.1955 s ≈ 0.2s

Then the measure is:

11.5 s ± 0.20 s

Q4) We have the measure:

312.0 mph ± 3.9 mph.

The percent uncertainty will be the quotient between the error and the measure times 100%, or:

(3.9 mph/312.0 mph)*100% = 1.25%

This is a percent error, we do not need to round this.

If you want to learn more, you can read:

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5 0

2 years ago

Which of the following quantities are unknown?

Sauron [17]

Answer:

Charge of the alpha particle

Explanation:

6 0

3 years ago

The inventor of the photographic process in which a photograph produced without a negative by exposing objects to light on light

Nikolay [14]

A. Man Ray --------------------

7 0

3 years ago

Read 2 more answers