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3 years ago

Is it realistic to build a tower with only 3 sheets of paper? why or why not

Engineering

2 answers:

lapo4ka [179]3 years ago

6 0

Answer:

For a tower made from paper, it would be rather difficult if you really worked hard.

Explanation:

I cannot say this is realistic or not and please dont use my words if it's a test that would get highly graded (I dont want you to get a bad grade due to the fact that this was not a full answer)

kupik [55]3 years ago

5 0

It is possible to do so but in my opinion I don’t think just plain sheets of paper will be realistic, maybe decorate them a little and it may look more realistic.

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The condensed Q formula may be used for operations in which the friction loss can be determined for:

yan [13]

The condensed Q formula may be used for operations in which the friction loss can be determined for a: 3, 4, or 5 inch hose.

<h3>What is a firehose friction loss?</h3>

A firehose friction loss can be defined as a measure of the effect of the resistance of water against the inner side of a firehose, which typically results in a pressure drop at the terminal end.

Generally, some of the factors that affect the resistance or friction in a firehose include:

Length of hose.

Age of hose.

Water flow (gpm)

Water turbulence

Gravity

Mathematically, the firehose friction loss can be calculated by using this formula:

FL = C × (Q/100)² × L/100.

Read more on friction loss here: brainly.com/question/17305262

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3 0

2 years ago

A well-insulated rigid tank contains 5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters of

vlabodo [156]

Answer:

$S_{gen} = 18.519\,\frac{kJ}{K}$

Explanation:

Given that rigid tank is a closed system, the following model is constructed after the First Law of Thermodynamics:

$W_{heater} + U_{sys,1} - U_{sys,2} = 0$

$W_{heater} = U_{sys,2} - U_{sys,1}$

$W_{heater} = m\cdot (u_{2}-u_{1})$

The entropy generation inside the rigid tank is determined by appropriate application of the Second Law of Thermodynamics:

$S_{sys,1} - S_{sys,2} + S_{gen} = 0$

$S_{gen} = S_{sys,2} - S_{sys,1}$

$S_{gen} = m\cdot (s_{2}-s_{1})$

The properties of the steam are obtained from steam tables:

Intial State

$P = 200\,kPa$

$T = 120.21\,^{\textdegree}C$

$\nu = 0.2222\,\frac{m^{3}}{kg}$

$u = 1010.7\,\frac{kJ}{kg}$

$s = 2.9294\,\frac{kJ}{kg\cdot K}$

$x = 0.25$

Final State

$P = 869.567\,kPa$

$T = 173.88\,^{\textdegree} C$

$\nu = 0.2222\,\frac{m^{3}}{kg}$

$u = 2578.6\,\frac{kJ}{kg}$

$s = 6.6332\,\frac{kJ}{kg\cdot K}$

$x = 1.00$

The entropy change of the steam during the process is:

$S_{gen} = (5\,kg)\cdot \left(6.6332\,\frac{kJ}{kg\cdot K} - 2.9294\,\frac{kJ}{kg\cdot K} \right)$

$S_{gen} = 18.519\,\frac{kJ}{K}$

8 0

3 years ago

A ½-hp motor requires 120 volts. What amperage is required?

Zolol [24]

Answer:

3.11 A

Explanation:

Use the given relations between power, horsepower, voltage, and current to find the current requirement for the motor.

P = hp(745.7 W) = (1/2)(745.7 W) = 372.85 W

I = P/V = (372.85 W)/(120 V) = 3.1071 A

The amperage required is about 3.11 A.

5 0

3 years ago

A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

3 years ago

Which of the following is the BEST example of statistical evidence?

alexandr1967 [171]

Well it would help if there was a picture but wild guess i’m going with B or the second choice.

5 0

3 years ago