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amid [387]
2 years ago
12

Which food must be cooked at to at least 145 f 63c cheese fries baked potatoes steamed broccoli for a buffet table scrambled egg

s for a cause
Physics
1 answer:
pantera1 [17]2 years ago
6 0

The dish that has to be cooked to at least 145°f is scrambled eggs. Option C is correct.

<h3>What is temperature?</h3>

Temperature directs to the hotness or coldness of a body. It is denoted by T.Measured mainly in the °C.

For good food safety beef, pig and eggs must be cooked at a temperature of 145°f. A minimum of 15 seconds must pass with the temperature held steady.

The temperature range in which the majority of bacteria thrive and multiply quickly is known as the temperature danger zone.

To keep away the eggs from the bacteria they must be cooked to at least 145°f. Scrambled eggs are the food that must be cooked to at least 145°f.

Hence, option C is correct.

To learn more about the temperature refer to the link;

brainly.com/question/7510619

#SPJ1

You might be interested in
A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.50 cm below t
bezimeni [28]

Answer:

the gauge pressure at the upper face of the block is 116 Pa

Explanation:

Given the data in the question;

A cubical block of wood, 10.0 cm on a side.

height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m

density ρ = 790 kg/m³

Using expression for the gauged pressure;

p-p₀ = ρgh

where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.

we know that, acceleration due to gravity g = 9.8 m/s²

so we substitute

p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m

= 116.13 ≈ 116 Pa

Therefore, the gauge pressure at the upper face of the block is 116 Pa

4 0
3 years ago
A load of 1 kW takes a current of 5 A from a 230 V supply. Calculate the power factor.
Pachacha [2.7K]

Answer:

Power factor = 0.87 (Approx)

Explanation:

Given:

Load = 1 Kw = 1000 watt

Current (I) = 5 A

Supply (V) = 230 V

Find:

Power factor.

Computation:

Power factor = watts / (V)(I)

Power factor = 1,000 / (230)(5)

Power factor = 1,000 / (1,150)

Power factor = 0.8695

Power factor = 0.87 (Approx)

6 0
3 years ago
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

3 0
3 years ago
Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0
3 years ago
Both high-side pressures and low-side pressures are low with the engine running and the selector set to the air-conditioning pos
elena-14-01-66 [18.8K]

Answer:

Technician A only

Explanation:

Both high-side pressures and low-side pressures are low with the engine running and the selector set to the air-conditioning position. Technician A says that the system is undercharged. Technician B says the cooling fan could be inoperative. Which technician is correct?

usually . An overcharged system will result in lower than normal low side pressures

An undercharged system will not enable  the compressor  to create pressure. As a result of the low amount of refrigerant, the cooling ability is reduced. When we say undercharged, we mean the refrigerant in the system is low, so the both the high side pressures and low side pressures will be low.

8 0
3 years ago
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