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4 years ago

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Create a document that includes a constructor function named Company in the document head area. Include four properties in the C

ompany object definition: name, products, motto, and numEmployees (for number of employees.) Initiate a new Company object in the document body , then assign values to each of the properties Print the name, products, motto, and the numEmployees properties to the screen using document.write methods. Combine each printed property with a descriptive string. For example, when you print the company name, it should read something like " The company name is MyComany."

1 answer:

Stella [2.4K]4 years ago

3 0

I have added the answer as a pic due to difficulties pasting the text here.

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Technician A says that a lack of lubrication on the back of the disc brake pads can cause brake noise. Technician B says that pa

sweet-ann [11.9K]

Answer:

Technician A and B both are correct.

Explanation:

According to Technician A, we can stop noise by applying lubricant (Break friction).

According to Technician A, we can stop noise by set-up right position for break pad.

7 0

3 years ago

Steam enters an adiabatic turbine at 10MPa and 500 C and leaves at 10 kPa with a quality of 90%. Neglecting the changes in kinet

Amanda [17]

Answer:

flow ( m ) = 4.852 kg/s

Explanation:

Given:

- Inlet of Turbine

P_1 = 10 MPa

T_1 = 500 C

- Outlet of Turbine

P_2 = 10 KPa

x = 0.9

- Power output of Turbine W_out = 5 MW

Find:

Determine the mass ow rate required

Solution:

- Use steam Table A.4 to determine specific enthalpy for inlet conditions:

P_1 = 10 MPa

T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg

- Use steam Table A.6 to determine specific enthalpy for outlet conditions:

P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg

x = 0.9 -------------> h_fg = 2392.1 KJ/kg

h_2 = h_f + x*h_fg

h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg

- The work produced by the turbine W_out is given by first Law of thermodynamics:

W_out = flow(m) * ( h_1 - h_2 )

flow ( m ) = W_out / ( h_1 - h_2 )

- Plug in values:

flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )

flow ( m ) = 4.852 kg/s

3 0

3 years ago

Triss [41]

Answer:

2.5 is the required details

8 0

3 years ago

A solid cylinder is concentric with a straight pipe. The cylinder is 0.5 m long and has an outside diameter of 8 cm. The pipe ha

poizon [28]

Answer :

The force needed to move the cylinder is 25.6 N

<h2>Further explanation </h2>

Given that,

Length of the cylinder, l = 0.5 m

Outer diameter of the cylinder, d = 8 cm = 0.08 m

Outer radius of the cylinder, $r=0.04\ m$

Inside diameter of the pipe, d = 8.5 cm = 0.085 m

Inside radius of the pipe, $r=0.0425\ m$

Specific gravity of the oil, $\rho=0.92$

Density of oil, $d=\rho\times \rho_w$

Kinematic viscosity of the oil, $v=5.57\times 10^{-4}\ m^2/s$

Velocity of the cylinder, u = 1 m/s

We need to find the force needed to move the cylinder. Let the force is F.

Specific gravity is defined as the ratio of the density of the substance to the density of water.

Kinematic viscosity is the acquired resistance of a fluid when there is no external force is acting except gravity. It is denoted by v.

Absolute viscosity is given by :

$v=\dfrac{\mu}{d}$

Where, $d$ = density of oil

And $d=\rho\times \rho_w$ (density of oil = specific gravity × density of water )

$d=0.92\times 10^3\ kg/m^3$

So,

$\mu=v\times d$ ..............(1)

$\mu=5.57\times 10^{-4}\ m^2/s\times 0.92\times 10^3\ kg/m^3$

$\mu=0.512\ Pa-s$

The separation between the cylinder and pipe is given by :

$dy=\dfrac{d_p-d_c}{2}=\dfrac{8.5-8}{2}=0.25\ cm=0.0025\ m$

$d_p\ and\ d_c$ are diameter of pipe and cylinder respectively.

The mathematical expression for the Newton's law of viscosity can be written as:

$\tau\propto\dfrac{du}{dy}$

$\tau=\mu\times \dfrac{du}{dy}$ ..........(2)

Where

$\tau$ = Shear stress, $\tau=\dfrac{F}{A}$ ............(3)

$\mu$ = viscosity

$\dfrac{du}{dy}$ = rate of shear deformation

On rearranging equation (1), (2) and (3) we get :

$\dfrac{F}{A}=v\times \rho\times \dfrac{du}{dy}$ ...............(4)

A is the area of the cylinder, $A=2\pi rl$

Equation (4) becomes :

$F=v\times \rho\times \dfrac{du}{dy}\times 2\pi rl$ ..............(5)

$A=\pi d\times l$

$A=\pi \times 0.08\ m\times 0.5\ m$

$A=0.125\ m^2$

Now, equation (5) becomes :

$F=(v\times \rho)\times \dfrac{du}{dy}\times 2\pi rl$

$F=(0.512\ Pa-s)\times (\dfrac{1}{0.0025\ m})\times \times 0.125\ m^2$

F = 25.6 N

<h2>Learn more </h2>

Kinematic viscosity : brainly.com/question/12947932

<h2>Keyword : </h2>

Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.

7 0

3 years ago

When something is plumb, it is

GrogVix [38]

Answer:

C. exactly vertical

Explanation:

"Plumb" means "exactly vertical."

5 0

3 years ago

Read 2 more answers