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3 years ago

What regulates labeling of food, drugs, and cosmetics?

Engineering

2 answers:

Flauer [41]3 years ago

8 0

Answer:

The FDA

Explanation:

The federal food and drug administration.

lana [24]3 years ago

6 0

FDA: the food and drug administration administrators the FPLA with respect to food, drugs, cosmetics, and medical devices

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Three single-phase, 10 kVA, 460/120 V, 60 Hz transformers are connected to form a three-phase 460/208 V transformer bank. The eq

evablogger [386]

Answer:

A) attached below

B) I₁ = 18.1 A , I₂ = 69.39 A

C) V( magnitude) = 454.5 ∠ 5.04° V , Voltage regulation = ≈ -1.2%

Explanation:

A) Schematic diagram attached below

attached below

B) magnitude of primary and secondary winding currents

I₂ ( secondary current ) = P / √3 * VL * cos∅ ---------- ( 1 )

VL = Line voltage = 208

cos∅ ( power factor ) = 0.8

P = 20 * 10^3 watts

insert values into equation 1

I₂ = 69.39 A

I₁ ( primary current ) = I₂V2 / V1

I₁ = ( 69.39 * 120 ) / 460 = 18.1 A

C ) Calculate the Primary voltage magnitude and the Voltage regulation

V(magnitude ) = Vp + ( I₁ ∠∅ ) Req ( 1 + j2 = 2.24 ∠63.43° )

= 460 + ( 18.1 * cos^-1 (0.8) ) ( 1 + j2 )

= 460 + 40.544 ∠ 100.3°

∴ V( magnitude) = 454.5 ∠ 5.04° V

Voltage regulation

= ((Vmag - V1) / V1 )) * 100

= (( 454.5 - 460 / 460 )) * 100

= -1.195 % ≈ -1.2%

8 0

3 years ago

A mass of air occupying a volume of 0.15m^3 at 3.5 bar and 150 °C is allowed [13] to expand isentropically to 1.05 bar. Its enth

e-lub [12.9K]

Answer:

Total work: -5.25 kJ

Total Heat: 52 kJ

Explanation:

V0 = 0.15

P0 = 350 kPa

t0 = 150 C = 423 K

P1 = 105 kPa (isentropical transformation)

Δh1-2 = 52 kJ (at constant pressure)

Ideal gas equation:

P * V = m * R * T

m = (R * T) / (P * V)

R is 0.287 kJ/kg for air

m = (0.287 * 423) / (350 * 0.15) = 2.25 kg

The specifiv volume is

v0 = V0/m = 0.15 / 2.25 = 0.067 m^3/kg

Now we calculate the parameters at point 1

T1/T0 = (P1/P0)^((k-1)/k)

k for air is 1.4

T1 = T0 * (P1/P0)^((k-1)/k)

T1 = 423 * (105/350)^((1.4-1)/1.4) = 300 K

The ideal gas equation:

P0 * v0 / T0 = P1 * v1 / T1

v1 = P0 * v0 * T1 / (T0 * P1)

v1 = 350 * 0.067 * 300 / (423 * 105) = 0.16 m^3/kg

V1 = m * v1 = 2.25 * 0.16 = 0.36 m^3

The work of this transformation is:

L1 = P1*V1 - P0*V0

L1 = 105*0.36 - 350*0.15 = -14.7 kJ/kg

Q1 = 0 because it is an isentropic process.

Then the second transformation. It is at constant pressure.

P2 = P1 = 105 kPa

The enthalpy is raised in 52 kJ

Cv * T1 + P1*v1 = Cv * T2 + P2*v2 + Δh

And the idal gas equation is:

P1 * v1 / T1 = P2 * v2 / T2

T2 = T1 * P2 * v2 / (P1 * v1)

Replacing:

Cv * T1 + P1*v1 + Δh = Cv * T1 * P2 * v2 / (P1 * v1) + P2*v2

Cv * T1 + P1*v1 + Δh = v2 * (Cv * T1 * P2 / (P1 * v1) + P2)

v2 = (Cv * T1 + P1*v1 + Δh) / (Cv * T1 * P2 / (P1 * v1) + P2)

The Cv of air is 0.7 kJ/kg

v2 = (0.7 * 300 + 105*0.16 + 52) / (0.7 * 300 * 105 / (105 * 0.16) + 105) = 0.2 m^3/kg

V2 = 2.25 * 0.2 = 0.45 m^3

T2 = 300 * 105 * 0.2 / (105 * 0.16) = 375 K

The heat exchanged is Q = Δh = 52 kJ

The work is:

L2 = P2*V2 - P1*V1

L2 = 105 * 0.45 - 105 * 0.36 = 9.45 kJ

The total work is

L = L1 + L2

L = -14.7 + 9.45 = -5.25 kJ

8 0

3 years ago

An escalator in a shopping center is designed to move 50 people, 75 kg each, at a constant speed of 0.6 m/s at 450 slope. Determ

REY [17]

Answer:

Power required to drive the escalator shall be equal to the rate at which the energies of the persons is increased.

$Energy=n\times mass\times g\times h\\\\\therefore Power=n\times mass\times g\times \frac{dh}{dt}\\\\Power=50\times 75\times 9.81\times 0.6sin(45)\\\\Power=15.607kW$

As we infer from the above equation If the velocity of the escalator is doubled then the Power required will also be doubled and become 31.215kW

3 0

4 years ago

The flowrate through a rectangular channel is 20 cfs. The upstream width of the channel is 10 ft, and the depth of the water in

Liula [17]

To solve this problem we will use the Froude number that relates the Forces of Inertia with the Forces of Gravity. There will be jump in the downstream only if Froude Number (Fr) is greater than 1 at upstream. Our values are given as,

$Q = 20cfs\\w= 10ft\\D= 1ft$

Then the velocity would be:

$V = \frac{Q}{wD}V = \frac{20}{10*1}V = 2ft/s$

The number of Froude is given as,

$Fr = \frac{V}{gD}^{1/2}$

Where,

V = Velocity

g = Gravity

D = Diameter

Replacing we have that

$Fr = \frac{2}{32.2}^{1/2}\\Fr = 0.352\\Fr$

There will be no Jump, correct answer is B.

5 0

3 years ago

Explain the difference, on the basis of the test results, between the ultimate strength and the "true" stress at fracture.

gayaneshka [121]

Answer / Explanation:

On the basis of the test result, Ultimate strength which is mostly known as the ultimate tensile strength is the strength attached to the ability or capacity of a structural element or material used in the test to withstand elongation forces or pull force applied to it.

WHILE,

True stress at fracture can be classified as stress or load associated to the point where yielding or fracture occurred divided by the cross-sectional area at the yield point.

5 0

3 years ago