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2 years ago

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An adiabatic nozzle is used to accelerate 6000 kg/hour of CO2 to 450 m/s. CO2 enters the nozzle at 1000 kPa and 500 C. The inlet

area of the nozzle is 40 cm2. Find the inlet velocity and the exit temperature of CO2.

1 answer:

den301095 [7]2 years ago

7 0

Answer:

a. inlet velocity = 60.8m/s

b. exit temperature = 686k

Explanation:

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Which part of an atom has most of its mass? A. electrons B. neutrons C. nucleus D. protons

vredina [299]

<em>The </em><em>nucleus</em><em> has most of the atomic mass in an atom. The </em><em>nucleus</em><em> is made up of protons and neutrons.</em>

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3 years ago

Read 2 more answers

Whats the answer???????????

Leviafan [203]

Answer: Less than 4 ohms

Explanation:

We have three resistors with the following resistance:

$R_{1}=4\Omega$

$R_{2}=6\Omega$

$R_{3}=8\Omega$

Now, when the resistors are connected in parallel, the total resistance $R$ is calculated as follows:

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

Isolating $R$ :

$R=\frac{R_{1}R_{2}R_{3}}{R_{3}(R_{1}+R_{2})+R_{1}R_{2}}$

Rewriting with th known values:

$R=\frac{(4\Omega)(6\Omega)(8\Omega)}{8\Omega(4\Omega+6\Omega)+(4\Omega)(6\Omega)}$

Finally:

$R=1.84 \Omega$

Hence, the correct option is less than 4 ohms.

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2 years ago

A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?

Andru [333]

Answer:

F = 37.8 × 10^(6) N

Explanation:

The charges are 0.06 C and 0.07 C.

Thus;

Charge 1; q1 = 0.06 C

Charge 2; q2 = 0.07 C

Distance between them; r = 3 m

Formula for the force in between them is;

F = kq1•q2/r²

Where k is a constant = 9 × 10^(9) N.m²/C²

Thus;

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

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2 years ago

WILL GIVE BRAINLIEST PLZ HELP REALLY URGENT

ozzi

D.) 5kg

This is a trick question. The mass of an object does not change based in location. However the weight of an object does change, this is because Weight = Mass × Gravity. Also mass is measured in kilograms and so the answer is 5 kg. So if you ever want to lose weight just go to the moon!

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3 years ago

A far-sighted person has a near-point of 80 cm. To correct their vision so that they can see objects that are as close as 10 cm

bekas [8.4K]

Answer:

$f = 8.89 cm$

Explanation:

As we know that Far sighted person has near point shifted to 80 cm distance

so he is able to see the object 80 cm

now the distance of lens from eye is 2 cm

and the person want to see the objects at distance 10 cm

so here the image distance from lens is 80 cm and the object distance from lens is 8 cm

now from lens formula we have

$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$

$-\frac{1}{80} + \frac{1}{8} = \frac{1}{f}$

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3 years ago