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34kurt
3 years ago
5

a person who agrees to purchase the finished products or services presented with the prescribed or correct features​

Engineering
1 answer:
zzz [600]3 years ago
8 0

Answer:

Buyers - are Persons or organizations who agree to purchase the finished productsor services presented with the prescribed or correct features.

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Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows the pouring
umka2103 [35]

Answer:

the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

Explanation:

Given the data in the question ;

flowrate Q = 400 cm³/s

cross section of the sprue is round

Diameter of sprue at the top d_top = 3.4 cm

Height of sprue = 20 cm = 0.2 m³

the proper diameter at its base so as to maintain the same volume flow rate = ?

first we determine the velocity at the sprue base

V_base = √2gh = √( 2×9.81×0.2) = √3.924 = 1.980908 m = 198.0908 cm

so, diameter of the sprue at the bottom  will be

Q = AV = [ (( πd²_bottom)/4) × V_bottom ]

d_bottom =  √(4Q/πV_bottom)

we substitute

d_bottom =  √((4×400)/(π×198.0908 ))

d_bottom =  √( 1600/622.3206)

d_bottom =  √2.571022

d_bottom =  1.6034 cm

Therefore, the proper diameter at its base so as to maintain the same volume flow rate is 1.6034 cm

8 0
3 years ago
In contrast to the leading-trailing drum brake system, the duo-servo drum brake system will:
prohojiy [21]

Leading/trailing shoe type drum brake This is called the servo effect (self-boosting effect) which realizes the powerful braking forces of drum brakes. ... This is because drum brakes generate the same braking force in either direction. Generally, this type is used for the rear brakes of passenger cars.

6 0
4 years ago
An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after
V125BC [204]

Answer:

a) E = \frac{K}{e*d}  b) i.

Explanation:

Assuming no other forces acting on the electron, as the field is pushing the electron back till it comes to rest momentarily, applying the work-energy theorem, the loss in kinetic energy, must be equal to the work done by the field on the electron.

The force that does work on the electron is the one due to the electric field E, and by definition of electric field, can be expressed as follows:

W = q*E = -e*E (1)

This work must be equal to the change in Kinetic Energy:

ΔK = Kf -K₀ = 0 - K = -K (2)

From (1) and (2), and solving for E, we have:

E =\frac{K}{e*d}

b) As the electric field (by convention) has the direction that it would have a positive test charge, the electrons (being negative charges),when starting from rest),have the direction opposite to the field.

In this case, as the electron has an initial velocity, and the field is opposing to the electron movement, we conclude that the electric field is in the direction of the electron´s motion, due to the electron slows down, and then comes momentarily to an stop, before changing direction and move opposite to the field as it is its natural behavior.

3 0
3 years ago
One kilogram of water fills a 0.140 m^3 rigid container at an initial pressure of 1.8 MPa. The container is then cooled to 40°C.
Pavel [41]

Answer:

T1 = 299.18 °C

P2 = 0.00738443 MPa

Explanation:

From the data, we can get two properties for the initial condition. These are pressure and specific volume.

The pressure is 1.8 MPa and the specific volume, we can get it with the mass and volume of the container, since it’s filled this is also the volume of the water in it.

v=\frac{vol (m^{3})}{mass (kg)} = \frac{0.140 m^{3}}{1 kg} = 0.140 \frac{ m^{3}}{kg}

When we check in the thermodynamic tables, the conditions for saturation at 1.8 MPa we found the following:

P^{sat} = 1.8 MPa

T^{sat} = 207.12 C

v_{g} = 0.1103\frac {m^{3}}{kg} specific volume for the saturated vapor  

v_{l} = 0.001167 \frac{m^{3}}{kg} specific volume for the saturated liquid  

Since the specific volume in our condition is higher that the specific volume for the saturated vapor, we have a superheated steam.  

Looking in the thermodynamic tables for superheated steam we found that the temperature where the steam has a specific volume of 0.140 \frac{ m^{3}}{kg} at 1.8 MPa is 299.18 °C. This is the initial temperature in the container.

Since the only information that we have about the final condition is that the container was cooled. We can assume that it was cooled until a condition of saturation. So, the final pressure for the water will be the pressure of saturation for a temperature of 40°C. From thermodynamic tables we get:

P^{sat} at 40C = 0.00738443 MPa

7 0
3 years ago
The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 4840
Assoli18 [71]

Answer:

70,900

Explanation:

Given :

True stress (psi) _____ True strain (psi)

48400 ______________ 0.11

60400 ______________ 0.19

Using ratio simplification :

Let :

s = True stress ; t = true strain

s1 = 48400

s2 = 60400

t1 = 0.11

t2 = 0.19

True stress, s0 ; needed to produce a True plastic strain, tp = 0.26

(s0 - s1) / (s2 - s1) = (tp - t1) / (t2 - t1)

(s0 - 48400)/(60400 - 48400) = (0.26 - 0.11)/(0.19 - 0.11)

(s0 - 48400)/12000 = 0.15/0.08

Cross multiply :

0.08(s0 - 48400) = 0.15 * 12000

0.08s0 - 3872 = 1800

0.08s0 = 1800 + 3872

0.08s0 = 5672

s0 = 5762 / 0.08

s0 = 70,900

The true stress required to produce a true plastic strain of 0.26 is 70,900

5 0
3 years ago
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