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alina1380 [7]
3 years ago
12

A 3-m-diameter vertical cylindrical tank is filled with water to a depth of 11 m. The rest of the tank is filled with air at atm

ospheric pressure. What is the absolute pressure at the bottom of the tank
Physics
1 answer:
muminat3 years ago
5 0

We have that the absolute pressure at the <u>bottom</u> of the tank is mathematically given as

Pabs=209.21Kpa

<h3>Absolute Pressure at the bottom of the <em>tank</em></h3>

Question Parameters:

A 3-m-diameter vertical <em>cylindrical </em>tank is filled

with water to a depth of 11 m.

Generally the equation for the Absolute Pressure  is mathematically given as

Pabs=Patm+pgh

Therefore

Pabs=101.3e3+(1000*9.81*11)

Pabs=209.21Kpa

For more information on Pressure visit

brainly.com/question/25688500

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A person kicks a 4.0-kilogram door with a 48-newton force causing the door to accelerate at 12 meters per second squared. What i
inna [77]

Answer:

-48 N

Explanation:

mass of door (m) = 4 kg

acceleration of the door = 12 m/s^{2}

force exerted by the person = 48 N

From Newton's third law of motion, action and reaction are equal but opposite. Therefore the force exerted on the door by the person which is 48 N will be the same as the force exerted on the person by the door but opposite in its direction, and this would be - 48 N

7 0
3 years ago
The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on
sammy [17]

Answer: First, we determine the circumference of the Mars by the equation below.  

                                     C = 2πr

Substituting the known values,

                                     C = 2(π)(3,397 km) = 6794π km

To determine the tangential speed, we divide the circumference calculated above by the time it takes for Mars to complete one rotation and that is,

                   tangential speed = 6794π km / 24.6 hours = 867.64 km/h

6 0
3 years ago
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Please do number 25! Explain how you got your answer with detail to get Brainliest! Thank you!
Ad libitum [116K]
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.

The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds)  =  4,000 foot-pounds of work.

If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds)  =  645.2 foot-pounds per second.

The rate of doing work is called "power".

(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)

So back to our problem.

John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.

Well, 550 foot-pounds per second is called 1 "horsepower".

So as John runs up the steps to the balcony, he's doing the work
at the rate of

           (645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)

=  1.173 Horsepower.  GO JOHN !

(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________

Oh my gosh !  Look at #26 !  There are the metric units I was talking about.

Do you need #26 ?

I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.

a).  5
b).  750 Joules
c).  800 Joules
d).  93.75%

You're welcome.

And #27 is 0.667 m/s .
7 0
3 years ago
Given Vout = 17.33 vpp and R1 = 3 kΩ, find the value of RF required to provide Av = 4.33. (Round your answer to 2 decimal places
Olenka [21]

Answer:

The magnitude of V_{2} is 4 V and phase of input voltage is zero

Explanation:

Given:

Output voltage V_{out} = 17.33

Resistance R_{1} = 3 kΩ

Voltage gain A_{v} = 4.33

For finding feedback resistance we use gain equation

Gain equation for non inverting op-amp is given by,

     A_{v} = 1+\frac{R_{f} }{R_{1} }

   4.33 = 1+ \frac{R_{f} }{3 k }

     R_{f} ≅ 10 kΩ

For finding input voltage we use,

   A_{v} = \frac{V_{out} }{V_{2} }

    V_{2} = \frac{17.33}{4.33}

    V_{2} = 4 V

The Phase of V_{2} is zero because output voltage phase is 360°

Therefore, the magnitude of V_{2} is 4 V and phase of input voltage is zero

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3 years ago
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