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3 years ago

21. How long can food that requires time-temperature control be left in the danger zone?

Engineering

2 answers:

dimulka [17.4K]3 years ago

8 0

Answer: A maximum of 1 hour

Explanation:

Read your lesson buddy!!

kati45 [8]3 years ago

4 0

Answer: D. A maximum of 4 hours. is the correct answer

Explanation:

You might be interested in

Who is using welding symbols to to communicate the detailed information necessary for welders to complete the weld?

muminat

Answer: An engineer or a craftsperson

Explanation:

Welding symbols allow the engineer or draftsperson to communicate detailed information about the weld to the welder. In most cases, the welding symbol has an arrow, which points to the location on the drawing where a weld is required. The arrow is attached to a leader line that intersects with a horizontal reference line. The arrow and leader line point to the joint in question, while the weld symbol tells the welder what type of weld to do. If the weld symbol is below the reference line, the weld should be made on the same side as the arrow.

3 0

3 years ago

A 450 MWt combined cycle plant has a Brayton cycle efficiency of 24% and a Rankine cycle efficiency of 29% with no heat augmenta

Romashka-Z-Leto [24]

Answer:

Output of plant=207.18 MW

Explanation:

Given input energy Q=450 MW

Brayton cycle efficiency $\eta_1=24$ %

Rankine cycle efficiency $\eta_2=29$ %

Combined efficiency for two cycle given as follow

$\eta _b=\eta _1+\eta _2-\eta _1\eta _2$

$\eta _b=0.24+0.29-0.24\times 0.29$

$\eta_b=0.4604$

We know that efficiency $\eta =\dfrac{out\ put}{in\ put}$

out put of plant= $450\times 0.4604$

So output of plant=207.18 MW

4 0

3 years ago

A 220-V electric heater has two heating coils that can be switched such that either coil can be used independently or the two ca

Nana76 [90]

Answer:

The resistances of both coils are 131.7 Ω and 29.64 Ω.

Explanation:

Since, there are two coils, they can be used independently or in series or parallel. The power is given as:

Power = P = VI

but, from Ohm's Law:

V = IR

I = V/R

therefore,

P = V²/R

R = V²/P

Hence, the resistance (R) and (P) are inversely proportional. Therefore, the maximum value of resistance will give minimum power, that is, 300 W. And the maximum resistance will be in series arrangement, as in series the total resistance gets higher than, any individual resistance.

Therefore,

Rmax = V²/Pmin = R1 + R2

R1 + R2 = (220 V)²/300 W

R1 + R2 = 161.333 Ω ______ en (1)

Similarly, the minimum resistance will give maximum power. And the minimum resistance will occur in parallel combination. Because equivalent resistance of parallel combination is less than any individual resistance.

Therefore,

(R1 R2)/(R1 + R2) = (220 V)²/2000 W

using eqn (1), we get:

(R1 R2) / 161.333 Ω = 24.2 Ω

R1 R2 = 3904.266 Ω²

R1 = 3904.266 Ω²/R2 _____ eqn (2)

Using this value of R1 in eqn (1), we get:

3904.266/R2 +R2 = 161.333

(R2)² - 161.333 R2 +3904.266 = 0

Solving this quadratic eqn we get two values of R2 as:

R2 = 131.7 Ω OR R2 = 29.64 Ω

when ,we substitute these values in eqn (1) to find R1, we get get the same two values as R2, alternatively. This means that the two coils have these resistance, and the order does not matter.

<u>Therefore, the resistance of both coils are found to be 131.7 Ω and 29.64 Ω</u>

7 0

3 years ago

A pressure gage connected to a tank reads 50 psi at a location where the barometric reading is 29.1 inches Hg. Determine the abs

Effectus [21]

Answer:

Absolute pressure , P(abs)= 433.31 KPa

Explanation:

Given that

Gauge pressure P(gauge)= 50 psi

We know that barometer reads atmospheric pressure

Atmospheric pressure P(atm) = 29.1 inches of Hg

We know that

1 psi = 6.89 KPa

So 50 psi = 6.89 x 50 KPa

P(gauge)= 50 psi =344.72 KPa

We know that

1 inch = 0.0254 m

29.1 inches = 0.739 m

Atmospheric pressure P(atm) = 0.739 m of Hg

We know that density of Hg = $13.6\times 10^3\ kg/m^3$

P = ρ g h

P(atm) = 13.6 x 1000 x 9.81 x 0.739 Pa

P(atm) = 13.6 x 9.81 x 0.739 KPa

P(atm) =98.54 KPa

Now

Absolute pressure = Gauge pressure + Atmospheric pressure

P(abs)=P(gauge) + P(atm)

P(abs)= 344.72 KPa + 98.54 KPa

P(abs)= 433.31 KPa

3 0

3 years ago

If something is 50fficient, how many joules of wasted energy will there be if 750j of energy is put in?’

Sedbober [7]

Efficiency is the minimum use of energy to accomplish the task. The wasted energy will be 375 J when 750 J of energy is given.

<h3>What is wasted energy?</h3>

Wasted energy is energy that is not useful when the transformation in the system occurs.

Total energy = 750 J

The efficiency of the system = 50 %

Output work (OW) is calculated as:

Efficiency = output work ÷ input work × 100%

750 × 50 = 100 OW

OW = 375 J

Wasted energy = Total energy - output work

= 750 - 375

= 375 J

Therefore, the machine is 50 % inefficient and has wasted energy of 375 J.

Learn more about wasted energy here:

brainly.com/question/16177264

#SPJ4

8 0

2 years ago