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ivanzaharov [21]
3 years ago
10

Please help on this one?

Physics
1 answer:
scoray [572]3 years ago
8 0

option B open system

because in open system energy and mass can escape from the system or can be added to it.

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A new interstate highway is being built with a design speed of 120 km/h. For one of the horizontal curves, the radius (measured
nikklg [1K]

Answer:

28.79%

Explanation:

Given

Design Speed, V = 120km/h = 33.33m/s

Radius, R = 300m

Side Friction, Fs = 0.09

Gravitational Constant = 9.8m/s²

Using the following formula, we'll solve the required rate of superelevation.

e + Fs = V²/gR where e = rate

e = V²/gR - Fs

e = (33.33)²/(9.8 * 300) - 0.09

e = 0.287853367346938

e = 28.79%

Hence, the required rate of superelevation for the curve is calculated as 28.79%

4 0
3 years ago
A crude approximation for the x component of velocity in an incompressible laminar boundary layer is a linear variation from u =
slega [8]

Answer:

2.5 * 10^-3

Explanation:

<u>solution:</u>

The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:

<em>δu/δx+δv/δy=0</em>

so that:  

<em>δv/δy= -δu/δx</em>

Now, since u = Uy/δ, where δ = cx^1/2, we have that:

<em>u=U*y/cx^1/2</em>

and we obtain:  

<em>δv/δy=U*y/2cx^3/2</em>

The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):  

v=∫δv/δy(dy)=U*y/4cx^1/2

 =y/x*(U*y/4cx^1/2)

 =u*y/4x

which is exactly what we needed to demonstrate.  

Also, using u = U*y/δ in the last equation we can obtain:  

v/U=u*y/4*U*x

     =y^2/4*δ*x

which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:

(v/U)_max=δ^2/4δx

                =δ/4x

                =2.5 * 10^-3

7 0
3 years ago
HURRY! PLEASE HELP!!!!<br><br><br> 3. What methods are you using to test this (or each) hypothesis?
Pie

Answer:

How to Test Hypotheses

State the hypotheses. Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis. ...

Formulate an analysis plan. The analysis plan describes how to use sample data to accept or reject the null hypothesis. ...

Analyze sample data. ...

Interpret the results.

4 0
3 years ago
Indigenous people sometimes cooked in watertight baskets by placing enough hot rocks into the water to bring it to a boil. What
yaroslaw [1]

Answer:

The rock has a mass of 4.02 kg

Explanation:

<u>Step 1: </u>Data given

Mass of the rock = TO BE DETERMINED

Temperature of the rock = 500 °C

Mass of the water  =4.24 kg

⇒ loses 0.044kg as vapor

Initial temperature of the water = 29°C

Final temperature = 100°C

Specific heat of rock = 0.20 kcal/kg °C

Specific heat of water = 1kcal/kg°C

Latent heat of vaporization = 539 kcal/kg

<u>Step 2:</u> formules

Qlost,rock + Qgained,water = 0

Qtotal,water = Qwater +Qvapor

<u>Step 3: </u>Calculate Qvapor

Qvapor = mass of vapor * Latent heat of vapor

Qvapor = 0.044kg * 539 kcal/kg = 23.716 kcal

<u>Step 4: </u>Calculate Qwater

Qwater = mass of water * specific heat * Δtemperature

Qwater = 4.196 kg * 1kcal/kg°C *( 100-29)

Qwater = 297.916 kcal

<u>Step 5:</u> Calculate Qwater,total

Qwater,total = Qwater + Qvapor

Qwater,total = 23.716 kcal + 297.916 = 321.632 kcal

<u>Step 6</u>: Calculate Qrock

Qrock = mass of rock * specific heat rock * Δtemperature

Qrock = mass of rock * 0.20 kcal/kg°C * (100-500)

Qrock = mass of rock * -80 kcal/kg

<u>Step 7:</u> Calculate mass of rock

Qlost,rock + Qgained,water = 0

Qlost,rock = -Qgained,water

mass of rock * -80 kcal/kg = -321.632 kcal

mass of rock = 4.02 kg

The rock has a mass of 4.02 kg

7 0
3 years ago
The graph shows a heating curve for water. Between which points on the graph would condensation occur?​
Radda [10]
I would have to see the graph.. but by looking at one one online, they are between points D and E.

4 0
3 years ago
Read 2 more answers
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