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earnstyle [38]
4 years ago
12

You’re making a food safety mistake if you

Physics
2 answers:
Strike441 [17]4 years ago
4 0

The answer is; D

Cooking food kills bacteria. Putting it while still warm in the fridge gives a thermal shock to the remaining bacteria hence also acts towards positive food safety. Fridging food reduces reproduction rates of bacteria on the food hence their colonies cannot grow.  Allowing the food to cool before putting it in the fridge give time for bacteria colonies to settle on the food again.


creativ13 [48]4 years ago
3 0

Answer: Option D

Explanation: There are certain safety guidelines that should be maintained while preparing and storing food. These guidelines helps to keep the food healthy and prevents from spoilage.

One of the safety guidelines states that food should not be kept outside until it cools down and then putting it inside the refrigerator. The bacteria can grow in the food within two hours unless it is refrigerated.

So, it should be kept inside the refrigerator before two hours.

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15.0 kg mass is displaced 3.00 m south and then 4.00 m west by a 10.0 N force. What is the total work done on the object?
masha68 [24]
<h2>Answer: D. 70.0 J</h2>

Explanation:

3.00 m + 4.00 m = 7.00 m x 10.0 N = 70.0 J

5 0
3 years ago
device for heating a cup of water in a car connects to the car's battery, which has an emf E = 10.0 V and an internal resistance
tatuchka [14]

Answer:

R = 0.93 ohms

Explanation:

Let's start by calculating the resistance of the coil,

V = I ( R + r )

Make R the subject of the above equation.

Hence,

V = IR + Ir

V - Ir = IR

( V - Ir )/I = R

Given that V = 10V

r = 0.07ohms

I = 10A ( this is an assumption )

( 10 - 10×0.07 ) ÷ 10

R = 0.93 ohms

Note: r is the internal resistance, R is the resistance, I is the current and the emf is represented as V. Since current was supposed to be included in the question but it was not, hence it was assumed to be 10A . Meaning whatsoever value is given for current will be substituted into the equation. Thanks

6 0
3 years ago
The closely packed cones in the fovea of the eye have a diameter of about 2 μm. For the eye to discern two images on the fovea
Marina CMI [18]

Answer:

The distance between the object is l=0.0056\  cm

Explanation:

The free body diagram of this setup is on the first uploaded image

From the question

   The diameter of closely packed cones in the fovea of the eye is  =  2 \mu m

     The distance of separation by one cone(not excited ) is d = 4\mu m = 4*10^{-4}cm

     The distance between the two point-like object  is  l

     The diameter of the eye is D = 2 cm

     The distance of the two point-like object from the near point of the eye is A = 28 cm

 From the diagram we see that the light from the two point-like object form a triangle of similar base l and d  and height D and A

So for a triangle with similar base we have that

                \frac{l}{A} =\frac{d}{D}

                \frac{l}{28} = \frac{4*10^{-4}}{2}

making l the subject we have

            l = \frac{28 *4*10^{-4}}{2}

              l=0.0056\  cm

               

         

       

   

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3 years ago
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