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raketka [301]
1 year ago
11

The best way to cool soft and thick foods (such as beans, sauce or chili) when using the refrigerator is?

Physics
2 answers:
tigry1 [53]1 year ago
8 0
Answer choices are:

1. To use the big cooking pot from the stove and place directly into the refrigerator 
2. Placing the food into a shallow pan and stirring every few minutes 
3. To use a 5-gallon bucket and stirring every few minutes 
<span>4. By stacking several pans on top of the other cooled pans
</span>
Best answer choice:

2. Placing the food into a shallow pan and stirring every few minutes.

There is another way to cool soft and thick foods (such as beans, sauce or chili) when using the refrigerator is by having them to be placed in one container in which they are in a water bath, to be heated off. Hot food can be placed directly in the refrigerator or it can be rapidly chilled in an ice or cold water bath before refrigerating but putting hot food in the fridge right away is the safest. <span>Large amounts of </span>food<span> should be divided </span>into<span> small portions and </span>put in<span> shallow containers for quicker cooling </span>in the refrigerator because b<span>acteria </span>can<span> grow rapidly on </span>food<span> if it is left out at room temperature for more than 2 hours.</span>
AnnZ [28]1 year ago
3 0
<span>The best way to cool soft and thick foods when using the refrigerator is by having them to be placed and poured on a pan or another way is by having them to be placed in one container in which they are in a water bath, to be heated of.</span>
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A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th
son4ous [18]

Answer:

Part a)

E = 8436.7 N/C

Part b)

E = 8436.7 N/C

Explanation:

Part a)

Electric field due to large sheet is given as

E = \frac{\sigma}{2\epsilon_0}

\sigma = \frac{Q}{A}

Q = -3.8 nC

A = \pi(0.09)^2

A = 0.025 m^2

\sigma = \frac{-3.8\times 10^{-9}}{0.025}

\sigma = -1.5 \times 10^{-7} C/m^2

now the electric field is given as

E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}

E = 8436.7 N/C

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

E = 8436.7 N/C

5 0
2 years ago
Un cable está tendido sobre dos postes colocados con una separación de 10 m. A la mitad del cable se cuelga un letrero que provo
lisabon 2012 [21]

Answer:

El peso del cartel es 397,97 N

Explanation:

La tensión dada en cada segmento del cable = 2000 N

El desplazamiento vertical del cable = 50 cm = 0,5 m

La distancia entre los polos = 10 m

La posición del letrero en el cable = En el medio = 5

El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °

El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero

El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N

El peso del signo = 397,97 N.

8 0
2 years ago
If two waves with identical crests and troughs meet, what is happening? The wave is reflecting. Constructive interference is occ
Kazeer [188]
The correct answer would be that destructive interference is happening. In this interference, the crest of a wave meets a trough of another wave resulting to an amplitude that is lower. The opposite is called the constructive interference. Hope this answers the question.
7 0
1 year ago
Read 2 more answers
A 900 kg steel beam is supported by the two ropes shown in (Figure 1) . Calculate the tension in the rope.
Rzqust [24]
Let T1 and T2 be tension in ropes1 and 2 respectively. 
<span>since system is stationary (equilibrium), considering both ropes + beam as a system </span>

<span>for horizontal equilibrium (no movement in that direction, so resultant force must be zero horizontally) </span>
<span>T1sin(20) = T2sin(30) </span>
<span>=> T1 = T2sin(30) / sin(20) </span>

<span>for vertical equilibrium, (no movement in this direction, so resultant force must be zero vertically) </span>
<span>T1cos(20) + T2cos(30) = mg </span>

<span>m = 900kg, substituting for T1 </span>
<span>T2sin(30)*cos(20)/sin(20) + T2cos(30) = 900g </span>
<span>2.328*T2 = 900*9.8 </span>
<span>T2 = 3788.65N </span>
<span>so T1 from (1) </span>
<span>T1 = 5535.21N</span>
8 0
2 years ago
A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa
elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0
1 year ago
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