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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

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If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :
gizmo_the_mogwai [7]

Let's see

Momentum be P

\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c

\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c

\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c

\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}

\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c

On comaparing

  • c=1

So

  • b-3c=1
  • b-3=1
  • b=1+3
  • b=4

and

  • -a-b=-1
  • -a-4=-1
  • -a=-1+4=3
  • a=-3

So the unit is

  • DV⁴/F³
5 0
3 years ago
Read 2 more answers
A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff
postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

F_c=\frac{mv^2}{R}

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

N=mg

Therefore, frictional force, f=\mu mg

Now, frictional force = centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0
4 years ago
How many seconds of space should you keep between you if you're driving a 100 foot truck 30 mph?
Marrrta [24]

Answer:

<em> The space in seconds that will be kept = 2.27 seconds</em>

Explanation:

S = d/t..................... Equation 1

making t the subject of formula in the equation above,

t = d/S.................... Equation 2

Where S = speed, d = distance, t = time.

<em>Conversion: (i)if 1 mph = 0.44704 m/s,</em>

<em>                 then, 30 mph = 30×0.44704    </em>

<em>                = 13.41 m/s</em>

<em>               (ii) If 1 foot = 0.3048 m</em>

<em>            then, 100 foot = 30.48 m.</em>

<em>Given: S = 30 mph = 13.41 m/s, d = 100 foot = 30.48 m</em>

<em>Substituting these values into equation 2</em>

<em>t = 30.48/13.41</em>

<em>t = 2.27 seconds.</em>

<em>Therefore the space in seconds that will be kept = 2.27 seconds</em>

7 0
4 years ago
Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

6 0
3 years ago
g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positi
Arisa [49]

Answer:

F_x = -\frac{6 C_6}{2^7}

Attractive

Explanation:

Data provided in the question

The potential energy of a pair of hydrogen atoms given by \frac{C_6}{X_6}

Based on the given information, the force that one atom exerts on the other is

Potential energy μ = \frac{C_6}{X_6}

Force exerted by one atom upon another

F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X}  (-\frac{C_6}{X^6})

or

F_x = \frac{\partial}{\partial X}  (\frac{C_6}{X^6})

or

F_x = -\frac{6 C_6}{2^7}

As we can see that the C_6 comes in positive and constant which represents that the force is negative that means the force is attractive in nature

5 0
3 years ago
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