A certain article is manufactured in batches of 100. The direct material cost is 250,direct labour cost rs. 400 and factory overheads rs. 290. If the selling on cost is 40% of prime cost, what should be the selling price of each product to obtain a profit of 20%(i. E profit is 20% of the selling price)

The fact that Eli plans to set up a recycling program made him choose D. The material he chose is renewable and is much better for the environment.

It should be noted that recycling simply means the process of collecting and processing materials that can be turned into another product.

Therefore, the material he chose is renewable and is much better for the environment and thus influenced his decision.

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Since their blades are bent, wind turbines provide lift. Low air pressure is produced by the side of the blade-shaped aerofoil that has the largest curve, while high air pressure is produced by the side that is beneath it.

Due to the form of wind turbine blades, air molecules flowing across the downwind side of the blade move more quickly than those moving across the upwind side. An airfoil is a form that resembles a crooked teardrop.

Turbines can be made with a vertical or a horizontal axis. Their blades might be straight, curved, or rounded. An energy-producing turbine with a horizontal axis and three blades is most effective at producing power.

Most blades are constructed of epoxy or polyester reinforced with fibreglass. As reinforcement materials, we may alternatively utilise carbon fiber or aramid (Kevlar). These days, research is being done into the potential applications of wood compounds like wood-epoxy or wood-fiber-epoxy.

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The maximum shear stress is  R = 25 MPa

Since the Von Mises stress in this case is less than the yield stress

From the circle, we can read off the principal stresses  as the two intersections of the circle with the horizontal axis. In this case, the principal stresses are:

σ1 = σavg + R = 35 MPa

σ2 = σavg - R = -15 MPa

The maximum shear stress is given by the radius of the circle, which is equal to half of the difference between the two principal stresses:

τmax = R = 25 MPa

In order to determine if the plate will yield under uniaxial tension with a yield stress of 80 N/mm², we first need to convert the units of the stress values we obtained earlier from MPa to N/mm²:

σ1 = 35 MPa = 35 N/mm²

σ2 = -15 MPa = -15 N/mm²

Substituting these values into the Von Mises yield criterion equation, we get:

σVM = √(35² - 35(-15) + (-15)²) = 51.96 N/mm²

Since the Von Mises stress in this case is less than the yield stress

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The frequency deviation of the modulating signal is given by: Frequency deviation, Δf = βVm = 5 x 1.2 = 6 kHz.

Assuming that an FM system has a frequency deviation of 10 kHz when the modulating signal has an amplitude of 4 volts and a frequency of 2 kHz, let us determine the frequency deviation if the modulating signal is given below.

Therefore, we can say that the modulating signal is given by 1.2sin(2πft), where f is the frequency of the modulating signal. Let us first find the modulator sensitivity (km/Volt). Given, amplitude of modulating signal, Vm = 4 V. Frequency of modulating signal, fm = 2 kHz Frequency deviation, Δf = 10 kHzWe know that Modulation index, β = Δf/fm = 10/2 = 5 km/Volt

From the given question, we are given that the modulator's sensitivity (km/volt) is 5 km/Volt. Therefore, the answer is 6 kHz.

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Note that the magnitude of the resultant active thrust on the wall is 876.69 kN.

To calculate the magnitude of the resultant active thrust on the wall, we need to use Rankine's theory of earth pressure.

Let's assume that the wall height is also 5.4 meters, and the angle of wall friction is zero.

Then, the total active thrust (Q) is given by:

Q = Ka * H * gamma * H/2

Where,

Ka = Active earth pressure coefficient

H = height of the wall

gamma = unit weight of soil

The active earth pressure coefficient can be calculated using the following formula:

Ka = (1 - sin(phi)) / (1 + sin(phi))

Where, phi = angle of shearing resistance of soil

Substituting the given values, we get:

phi = 30 degrees

H = 5.4 meters

gamma = 19.8 kN/m^3

Ka = (1 - sin(30)) / (1 + sin(30)) = 1/3

Q = Ka * H * gamma * H/2 = (1/3) * 5.4 * 19.8 * 5.4/2 = 876.69 kN

Therefore, the magnitude of the resultant active thrust on the wall is 876.69 kN.

The line of action of the resultant active thrust on the wall will be at one-third of the height of the wall from the bottom. Therefore, the line of action of the active thrust will be at a height of 1.8 meters from the bottom of the wall.

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Answer:

L = 10.32 m

Explanation:

The resistance of a copper-nickel alloy wire is given by the following formula:

\(R = \frac{\rho L}{A}\)

where,

R = Resistance = 0.5 Ω

ρ = resistivity of copper-nickel alloy = 3.8 x 10⁻⁸ Ωm

L = Length of wire = ?

A = cross-sectional area of wire = \(\frac{\pi d^2}{4} = \frac{\pi(1\ x\ 10^{-3}\ m)^2}{4}\) = 7.85 x 10⁻⁷ m²

Therefore,

\(0.5\ \Omega = \frac{(3.8\ x\ 10^{-8}\ \Omega.m)L}{7.85\ x\ 10^{-7}\ m^2}\)

L = 10.32 m

Answer:

Plotters

Explanation:

https://en.wikipedia.org/wiki/Plotter

The decision tree is continuing support for judgment, that utilizes a tree-like decision model, and its possible effects, such as the chance event results, resources as well as utility costs. It is a way of displaying an automated system containing the only conditional statement of control, and the further discussion can be defind as follows:

Please find the diagram in the attachment file.

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Answer:

A) About \(9.273 \times 10^{-9}\) newtons

B) 76.518 newtons

C) 111.834 newtons

Explanation:

A) \(F_g=\dfrac{GM_1M_2}{r^2}\) , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:

\(F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}\)

B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.

C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.

Hope this helps!

Height of mirror is 0.075 m while width of mirror 0.325 m.

The distance between an object's two sides is considered to be its width. This is the smallest dimension. Height is the distance between the top and bottom of an object.

given data:

wide = 1.3 m

high = 0.30 m

driver’s eyes = 0.50 m

rear window = 1.50 m

we take here height / width of the mirror  is

height / width   =  h/w  .................1

and

height /width of the window is

height /width  = hw/ww    .................2

and

distance of eye / window by the mirror is

distance of eye / window = xe/xw      .................3

so here

θ = θi  = θr    ....................4

and  tanθ for vertical is

tanθ  =   h/xe

tanθ  =  hw/(xe+xw)       ....................5

so

h =     hw × xe/(xe+xw)    ....................6

put  here value and we get

h = 0.30 ×  0.5/(0.5+1.5)  

h = 0.075 m

and

when we take here tanθ for horizontal than it will be

tanθ =     w/xe

tanθ =   Ww/(xe+xw)     .......................7

so

w = Ww  × xe/(xe+xw)      ....................8

put here value and we get

w = 1.30 ×  0.5/(0.5+1.5)  

w = 0.325 m

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Answer:

By using a forward-inclined half-moon torch position, stopping momentarily at each end of the weld pool

Explanation:

By using a forward-inclined half-moon torch position, stopping momentarily at each end of the weld pool

Adjusting welding parameters, using appropriate electrode size, and maintaining proper torch angle can prevent undercutting in horizontal

welding of an inside corner.

We have,

When an inside corner is being welded in the horizontal position, undercutting can be decreased or prevented by adjusting the welding parameters, such as:

- Controlling Welding Parameters:

Adjusting the welding parameters is crucial in preventing undercutting.

It involves regulating the welding current, voltage, and travel speed.

- Selecting Appropriate Electrode Size:

Choosing the right electrode size is essential to control the heat generated during the welding process.

Smaller electrode diameters generally provide lower heat input, which can help prevent undercutting.

- Maintaining Proper Torch Angle and Manipulation Techniques:

Keeping the correct torch angle during welding is crucial to control the weld pool and metal deposition.

- reducing the welding current and travel speed

- using the appropriate electrode size

- maintaining proper torch angle and manipulation techniques.

Thus,

Adjusting welding parameters, using appropriate electrode size, and maintaining proper torch angle can prevent undercutting in horizontal welding of an inside corner.

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Answer:

the disadvantages can be people's thoughts about them and also a rival resort nearby which looks more posh pls mark brainliest i need 5 more for next rank thank you

Explanation:

Complete Question:

The cost to rent a scooter is $15.50 per hour and the cost to rent a watercraft

is $22.80 per hour. Use the expression 15.5s + 22.8w to determine how much it would cost to rent a scooter for 3 hours and a watercraft for 1 hours.

Answer:

$69.39

Explanation:

Given

15.5s + 22.8w

Required

Evaluate, when w = 1 and s = 3

To do this, we simply substitute 3 for a and 1 for w in the given expression.

15.5s + 22.8w becomes..

15.5 * 3 + 22.8 * 1

46.5 + 22.8

69.3

Hence, the cost is $69.30

Answer: This approach is allocated 32 seconds of effective green time. The cycle length is 100 seconds. Determine the average approach delay (using Eq. 7.27). A) 34.8 s.

Answer:

How does electricity produce magnetism?

Explanation:

When the electricity is switched on, the electric magnets create powerful magnetic fields. Coils of wire are mounted around the shaft. As the shaft with the magnets rotates, the coils of wire are exposed to changing magnetic fields, and an electric current is generated in the wires.

Answer:

Explanation:

I am pretty sure it depends but most of the time I think they produce direct current/voltage.

Based on the maximum capacity as a result of the pipe-laying operations, and the maximum capacity without obstruction to the four-lane dual carriageway, the mean speed of traffic in the bottleneck is -2.73km.

This can be found as:

= (Maximum restricted capacity - freeflowing rate) / (Kb - Ka)

= (1,100 - 1,500) / (209 - 62.5)

= -2.73 km/h

The rate that the queue outside the bottleneck grows is:

= Free flowing rate - (mean speed of traffic x ka)

= 1,500 - (-2.73 x 62.5)

= 1,670 veh/hour

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Answer:

Software reuse is the principal approach for constructing web-based systems, requirements for those systems cannot be completely specified in advance, User interfaces are constrained by the capabilities of web browsers.

Answer:

a) the safety factor according to the maximum normal-stress theory; n = 3

b) the safety factor according to the maximum-shear-stress theory; n = 1.714

c) the safety factor according to the maximum distortion-energy theory is; n = 1.97278

Explanation:

Given the data in the question;

a) What is the safety factor according to the maximum normal-stress theory;

According to the maximum normal-stress theory

n = S\(_y\) / σ\(_{max\)

since σ₁ = 20 ksi is greater than σ₂ = -15 ksi

σ\(_{max\) = 20 ksi and yield strengths in tension and compression S\(_y\) = 60 ksi

we substitute

n = 60 ksi / 20 ksi

n = 3

Therefore, the safety factor according to the maximum normal-stress theory; n = 3

b) What is the safety factor according to the maximum-shear-stress theory.

According to maximum-shear-stress theory;

τ\(_{max\) = [(σ₁ - σ₂) / 2]

= S\(_y\) / 2n

n = S\(_y\) / 2[(σ₁ - σ₂) / 2]

n = S\(_y\) / (σ₁ - σ₂)

we substitute

n = 60 ksi  / (20 ksi - (-15 ksi))

n = 60 ksi  / (20 ksi +15 ksi)

n = 60 ksi  / 35 ksi

n = 1.714

Therefore, the safety factor according to the maximum-shear-stress theory; n = 1.714

c) the safety factor according to the maximum distortion-energy theory?

By distortion energy theory

σ₁² + σ₂² - σ₁σ₂ = (S\(_y\)/n)²

we substitute

(20)² + (-15)² - ( 20 × -15 ) = ( 60 / n )²

400 + 225 + 300 = 3600 / n²

925 =  3600 / n²

n² = 3600 / 925

n = √( 3600 / 925 )

n = 1.97278

Therefore, the safety factor according to the maximum distortion-energy theory is; n = 1.97278

Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

\(z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}\)

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

\(50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)50 m + 30.704358 m + 20.4081633 m = 10.234786 m + \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

90.8777353 m = \(\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}\)

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

The required tank wall thickness is 87.7668 millimeters.

Detailed explanation:

Given:

V=1.5 m3 is the vessel's volume.

P=22 MPa compressed natural gas is contained in a spherical pressure vessel.

In the event that the tank is constructed of A516 grade 60 steel with a 150 MPa maximum normal stress.

ie omax =  150 Mpa

Solution :

The vessel's diameter is determined by,

d/2 =√3v/4\(\pi\)

d/2 =√3*.5/4*3.14

d/2=√0.358

d/2 =1.196m

Convert into mm,

d= 1.96*1000=1196.82mm

The wall tank's thickness is determined by,

t=pd/20

t=22*1196.82/2*150

t=26330.04/300

t=87.7668

Therefore, the required thickness of the tank wall is 87.7668 mm.

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Our current society has difficulty dealing with global population and material wealth due to systemic issues and a lack of prioritization.

The challenges in dealing with global population and material wealth stem from a combination of systemic factors and a lack of prioritization. Firstly, our society is built on a global economic model that emphasizes constant growth and consumption, which contributes to the unsustainable depletion of resources and widening wealth gaps.

Additionally, the issue of global population growth is complex and interconnected with various social, economic, and environmental factors, making it challenging to find comprehensive solutions.

Furthermore, addressing these difficulties requires a shift in mindset and a reevaluation of societal values. We need to recognize that the pursuit of infinite material wealth is not sustainable and does not lead to overall well-being. Instead, we must focus on creating a more equitable distribution of resources, promoting sustainable practices, and investing in education, healthcare, and social infrastructure.

By prioritizing these aspects, we can ensure a better quality of life for all individuals while minimizing the negative impacts of population growth and wealth disparities.

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To determine the requested values, we can use the heat exchanger equations and principles. Let's solve the problem step by step:

a. The overall heat transfer coefficient for the cold side can be calculated using the formula:

1/U_c = 1/h_o + R_fo + R_w + R_fi + 1/h_i

where h_o and h_i are the oil-side and water-side convective heat transfer coefficients, respectively. R_fo and R_fi are the fouling resistances on the oil and water sides, and R_w is the wall resistance.

Since no values are provided for h_o and R_fo, we can assume reasonable estimates. Let's assume h_o = 1500 W/m^2 K and R_fo = 0.0005 m^2 K/W (common values for oil-side heat exchangers). R_w can be calculated using the formula:

R_w = ln(r_o/r_i) / (2πLk)

where r_o and r_i are the outer and inner radii of the tube, L is the tube length, and k is the thermal conductivity of the tube material.

Substituting the given values into the formulas, we can calculate 1/U_c.

b. To find the outlet temperature of the engine oil, we can use the energy balance equation:

m_o * Cp_o * (T_o,out - T_o,in) = Q

where m_o is the mass flow rate of the oil, Cp_o is the specific heat capacity of the oil, T_o out is the outlet temperature of the oil, T_o in is the inlet temperature of the oil, and Q is the rate of heat transfer between the streams.

We can rearrange the equation to solve for T_o, out.

c. Similarly, for the outlet temperature of the water, we can use the energy balance equation:

m_w * Cp_w * (T_w,in - T_w,out) = Q

where m_w is the mass flow rate of the water, Cp_w is the specific heat capacity of the water, T_w in is the inlet temperature of the water, T_w, out is the outlet temperature of the water, and Q is the rate of heat transfer between the streams.

Rearranging the equation will allow us to solve for T_w, out.

d. The rate of heat transfer between the streams can be calculated using the formula:

Q = m_w * Cp_w * (T_w,in - T_w,out) = m_o * Cp_o * (T_o,out - T_o,in)

Substituting the known values, we can determine the rate of heat transfer.

By solving these equations using the given values and the assumptions made, we can obtain the desired results.

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Answer:

85 and 86 are the coil pins while 30, 87, and 87a are the switch pins. 87 and 87a are the two contacts to which 30 will connect. If the coil is not activated, 30 will always be connected to 87a. Think of this as the relay in the Normally Closed (OFF) position.

Explanation:

ᴄᴏᴍᴘʟᴇᴛᴇɴᴇꜱꜱ ᴍᴇᴀɴꜱ ᴛʜᴇ ꜱᴛᴀᴛᴇ ᴏꜰ ʙᴇɪɴɢ ᴄᴏᴍᴘʟᴇᴛᴇ ᴀɴᴅ ᴇɴᴛɪʀᴇ; ʜᴀᴠɪɴɢ ᴇᴠᴇʀʏᴛʜɪɴɢ ᴛʜᴀᴛ ɪꜱ ɴᴇᴇᴅᴇᴅ.

ʙᴇᴀᴜᴛʏ ᴍᴇᴀɴꜱ combination of qualities, such as shape, colour, or form, that pleases the aesthetic senses, especially the sight.

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... ​Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation: