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Tems11 [23]
3 years ago
5

A chef sanitized a thermometer probe and then checked the temperature of minestrone soup being held in a hot-holding unit. the t

emperature was 120 which did not meet the operation's critical limit of 135. the chef recorded the temperature in the log and reheated the soup to 165 for 15 seconds. which was the corrective action?
A) reheating the soup
B) checking the critical limit
C) sanitizing the thermometer probe
D) recording the temperature in the log
Physics
1 answer:
ss7ja [257]3 years ago
4 0

Answer: Option (A) is the correct answer.

Explanation:

A corrective action is defined as the action with the help of which a person can avoid a difficulty or problem that he/she was facing earlier.

For example, when the chef checked the temperature of soup using thermometer then it was 120 but his operation's critical limit was 135.

So, to avoid this problem he heated the soup to 165 at 15 seconds following which he got the result as desired.

Therefore, reheating the soup was his corrective action.

Thus, we can conclude that reheating the soup was the corrective action.

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As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p
Oksana_A [137]

Answer:

a.Attractive

2.31531\times 10^{-16}\ N

Explanation:

When it comes to charges, the charges which are alike repel each other and the charges which are different will attract each other.

Here, there is a proton and electron which are different particles hence, they will attract each other.

q_1=q_2 = Charge of electron and proton = 1.6\times 10^{-19}\ C

r = Distance between them = 997 nm

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(997\times 10^{-9})^2}\\\Rightarrow F=2.31531\times 10^{-16}\ N

The force of attraction between the particles will be 2.31531\times 10^{-16}\ N

8 0
3 years ago
Help please<br> It’s kinda urgent
user100 [1]

Answer:

a = - 50 [m/s²]

Explanation:

To solve this problem we simply have to replace the values supplied in the given equation.

Vf = final velocity = 0.5 [m/s]

Vi = initial velocity = 10 [m/s]

s = distance = 100 [m]

a = acceleration [m/s²]

Now replacing we have:

(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]

The negative sign of acceleration means that the ship slows down its velocity in order to land.

4 0
2 years ago
Which statement describes the magnetic field inside a bar magnet? It points from north to south. It points from south to north.
Levart [38]

Inside the bar magnet, the magnetic field points from north to south. Statement A is correct.

Magnetic Field:

It is defined as a vector field or the influence of the magnet on the electric current, charges and ferromagnetic substance.

The strength of magnetic field is depends up on the numbers of magnetic field lines per unit area.

  • Magnetic field lines emerge from the North pole and end in the South pole of a bar magnet.
  • Inside the magnet are also present inside the bar magnet and never intersect at any point.

Therefore, inside the bar magnet, the magnetic field points from north to south.

To know more about Magnetic Field:

brainly.com/question/19542022

4 0
2 years ago
Read 2 more answers
An object of mass 4kg is moving along a horizontal plane. If the coefficient of kinetic friction is 0.2 find the friction force
lana66690 [7]

Answer:

The friction force acting on the object is 7.84 N

Explanation:

Given;

mass of object, m = 4 kg

coefficient of kinetic friction, μk = 0.2

The friction force acting on the object is calculated as;

F = μkN

F = μkmg

where;

F is the frictional force

m is the mass of the object

g is the acceleration due to gravity

F = 0.2 x 4 x 9.8

F = 7.84 N

Therefore, the friction force acting on the object is 7.84 N

5 0
3 years ago
Compare the time period of two simple pendulums of length 4m and 16m at a place.
Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L)  is defined as:

T=2\,\pi\,\sqrt{ \frac{L}{g} }

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0
2 years ago
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