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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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vagabundo [1.1K]

Explanation:

Draw a free body diagram for each disc.

Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.

∑F = ma

86.5 N − T₁ − Wa = 0

Wa = 86.5 N − T₁

ma × 9.8 m/s² = 86.5 N − 55.6 N

ma = 3.2 kg

Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.

∑F = ma

T₁ − T₂ − Wb = 0

Wb = T₁ − T₂

mb × 9.8 m/s² = 55.6 N − 36.5 N

mb = 1.9 kg

Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.

∑F = ma

T₂ − T₃ − Wc = 0

Wc = T₂ − T₃

mc × 9.8 m/s² = 36.5 N − 9.6 N

mc = 2.7 kg

Disc D has two forces acting on it: T₃ up and Wd down.

∑F = ma

T₃ − Wd = 0

Wd = T₃

md × 9.8 m/s² = 9.6 N

md = 0.98 kg

3 0
3 years ago
Which of the following is not a synthesis reaction?
Nastasia [14]
C is the awnser because c is single replacement
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A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito
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voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

c =  \frac{c1 \times c2}{c1 + c2}

=  \frac{2 \times 4}{2 + 4}

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

= 1.33 \times 10 {}^{ - 6}  \times 8

= 10.64 \times 10 {}^{ - 6}

charge on 2.0μf capacitor is

\frac{Qeq}{2 \times 10 {}^{ - 6} }

=  \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} }

=5.32v

learn more about series capacitance from here: brainly.com/question/28166078

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1 year ago
What is a device that minimizes the force on a microscopic object during a collision?
liraira [26]
An impact which stops a moving object must do enough work to take away its kinetic energy, so extending the distance moved during the collision reduces the impact force.
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3 years ago
A horizontal force acts on an object on a fric- tionless horizontal surface. If the force is halved and the mass of the object i
guapka [62]

(a)  If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)  if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

The force on an object is determined by applying Newton's second law of motion;

F = ma

\frac{F_1}{m_1a_1} = \frac{F_2}{m_2a_2}

(a)

when the force is halved, F₂ = 0.5F₁,

mass is doubled, m₂ = 2m₁

\frac{F_1}{m_1a_1} = \frac{0.5F_1}{2m_1a_2} \\\\2m_1a_2F_1 = 0.5F_1 m_1a_1\\\\2a_2 = 0.5a_1\\\\a_2 = \frac{0.5a_1}{2} = \frac{a_1}{2 \times 2} = \frac{a_1}{4} \\\\a_2 = \frac{1}{4} (a_1)

Thus, If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)

when the force is doubled, F₂ = 2F₁,

mass is halved, m₂ = 0.5m₁

\frac{F_1}{m_1 a_1} = \frac{2F_1}{0.5m_1 a_2} \\\\0.5m_1a_2 F_1 = 2F_1m_1a_1\\\\0.5a_2 = 2a_1\\\\a_2 = \frac{2a_1}{0.5} \\\\a_2 = 4(a_1)

Thus, if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

Learn more here:brainly.com/question/19887955

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2 years ago
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