1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Lyrx [107]
3 years ago
11

Need help coding this i’m using python.

Engineering
1 answer:
Mazyrski [523]3 years ago
4 0

Answer:

ill help'

Explanation:

You might be interested in
Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w
Dafna11 [192]

Answer:

\mathbf{h_2 =1021.9 \  mm}

Explanation:

Given that :

The initial volume of water V_1 = 1000.00 mL = 1000000 mm³

The initial temperature of the water  T_1 = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water T_2 = 70° C

The coefficient of thermal expansion for the glass is  ∝ = 3.8*10^{-6 } mm/mm  \ per ^oC

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature T_1 = 10 ^ 0C  the density of the water is \rho = 999.7 \ kg/m^3

At temperature T_2 = 70^0 C  the density of the water is \rho = 977.8 \ kg/m^3

The mass of the water is  \rho V = \rho _1 V_1 = \rho _2 V_2

Thus; we can say \rho _1 V_1 = \rho _2 V_2;

⇒ 999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2

V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }

V_2 = 1022.40 \ mL

v_2 = 1022400 \ mm^3

Thus, the volume of the water after heating to a required temperature of  70^0C is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

V_1 = A_1 *h_1

The area of the water before heating is:

A_1 = \dfrac{V_1}{h_1}

A_1 = \dfrac{1000000}{1000}

A_1 = 1000 \ mm^2

The area of the heated water is :

A_2 = A_1 (1  + \Delta t  \alpha )^2

A_2 = A_1 (1  + (T_2-T_1) \alpha )^2

A_2 = 1000 (1  + (70-10) 3.8*10^{-6} )^2

A_2 = 1000.5 \ mm^2

Finally, the depth of the heated hot water is:

h_2 = \dfrac{V_2}{A_2}

h_2 = \dfrac{1022400}{1000.5}

\mathbf{h_2 =1021.9 \  mm}

Hence the depth of the heated hot  water is \mathbf{h_2 =1021.9 \  mm}

4 0
3 years ago
A resistivity meter is measured in
Bingel [31]
Ohms ..................
3 0
2 years ago
Read 2 more answers
Partes de un transformador
BartSMP [9]
Está constituido por dos bobinas de material conductor, devanadas sobre un núcleo cerrado de material ferromagnético, pero aisladas entre sí eléctricamente. ... Las bobinas o devanados se denominan primario y secundario según correspondan a la entrada o salida del sistema en cuestión, respectivamente.
6 0
3 years ago
A Venturi meter (see below) uses the principles of the Bernoulli equation to measure the velocity of a flow in a reactor system.
LenaWriter [7]

Answer:

Q=0.000604 m³/s

Explanation:

Given that

d₁=5 cm

d₂=1 cm

P= 30 KPa

Density of water ,ρ=1000 kg/m³

As we know that volume flow rate Q given as

Q=A_1A_2\sqrt{\dfrac{\dfrac{2\Delta P}{\rho}}{A_1^2-A_2^2}}

A_1=\dfrac{\pi}{4}\times 0.05^2\ m^2

A₁=0.0019 m²

A_2\dfrac{\pi}{4}\times 0.01^2\ m^2

A₂=0.000078 m²

Q=0.0019 \times 0.000078 \sqrt{\dfrac{\dfrac{2\times 30\times 1000}{1000}}{0.0019^2-0.000078^2}}\ m^3/s

Q=0.000604 m³/s

7 0
3 years ago
A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall when the basketball
faltersainse [42]

Answer:

2.65 MPa

Explanation:

To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

\sigma = \frac{p*r}{2t}

<u>Where:</u>

p: is the gage pressure = 108 kPa

r: is the inner radius of the ball

t: is the thickness = 3 mm  

Hence, we need to find r, as follows:

r_{inner} = r_{outer} - t    

r_{inner} = \frac{d}{2} - t

<u>Where:</u>

d: is the outer diameter = 300 mm

r_{inner} = \frac{300 mm}{2} - 3 mm = 147 mm

Now, we can find the normal stress (σ) in the wall of the basketball:

\sigma = \frac{p*r}{2t} = \frac{108 kPa*147 mm}{2*3 mm} = 2646 kPa = 2.65 MPa

Therefore, the normal stress is 2.65 MPa.

I hope it helps you!

3 0
3 years ago
Other questions:
  • g The pump inlet is located 1 m above an arbitrary datum. The pressure and velocity at the inlet are 100 kPa and 2 m/s, respecti
    8·1 answer
  • A evolução da malha rodoviária do Brasil é um marco notável
    9·1 answer
  • CS3733: Homework/Practice 05 Suppose we would like to write a program called monitor which allows two other programs to communic
    5·1 answer
  • A tensile-testing machine is to test specimens of diameter 15 mm which have ultimate tensile strengths of up to 820 N/mm2. What
    10·1 answer
  • 1. Represent each of the following combinations of units in the correct SI form using the appropriate prefix: (a) μMN, (b) N/μm,
    6·1 answer
  • -0-1"<br> -0<br> -20<br> -15<br> -10<br> 0<br> -5
    9·1 answer
  • A well-established way of power generation involves the utilization of geothermal energy-the energy of hot water that exists nat
    9·1 answer
  • Sarah and Raj take/takes me to a baseball game every year.
    11·1 answer
  • un contenedor de 0.01m∧3 se llena con 2kg de nitrogeno a una presion de 15mpa ¿cual es la temperatura del nitrogeno?resolver uti
    8·1 answer
  • Whose responsibility is it to provide direction on correct ladder usage?<br> select the best option.
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!