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3 years ago

hi can someone please help me i can't figure this out and I tried to search it up on google please help me oyt

Physics

1 answer:

Bas_tet [7]3 years ago

6 0

The answer is C. Watson and Crick developed the Double Helix model seen in the diagram.

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What can happen overnight to soil?

UkoKoshka [18]

(GABS) Overnight, all of the particles settled down to the bottom , and the larger particles were on the bottom and the smaller particles were on the top. Therefore, clay was on top, hummus was in the middle, and soil was on the bottom.
Particles dissolve is an unique way

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3 years ago

What is -8,675,309.0 in scientific notation

Sindrei [870]

= 3.456 × 1011
(scientific notation)

= 3.456e11
(scientific e notation)

= 345.6 × 109
(engineering notation)
(billion; prefix giga- (G))

= 345600000000
(real number)

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3 years ago

How many atoms of Pb are there in the compound 3Pbl2

mash [69]

Answer: there are 6 atoms.

Explanation: 3 parts of Pb times 2 atoms per part = 6

good luck:)

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3 years ago

Read 2 more answers

Can someone please help me

Mars2501 [29]

The correct answer is letter b.

To find the answer follow the following steps.
1. 6524.96 x .25 = X

2. 1631.24 = X

This works for all of the given answers to find the correct answer.

5 0

3 years ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

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

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

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

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

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

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

Learn more about flow rate:

brainly.com/question/9805263

#LearnwithBrainly

7 0

3 years ago