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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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What is a force, motion and energy??​
Leviafan [203]

Answer:Kinetic energy is the energy of motion. All moving objects have kinetic energy. When an object is in motion, it changes its position by moving in a direction: up, down, forward, or backward. 3.  A force is a push or pull that causes an object to move, change direction, change speed, or stop.

Explanation: Not sure if that's what you meant but that's the answer I can give you.

5 0
2 years ago
Read 2 more answers
Work is done on a locked door that remains closed while you try to pull it open. True False.
dsp73

Answer:False

Explanation:

Work is being done on a body when it causes displacement of body on the application of force

Work\ done=Force\times displacement

When we pull the door by a force it causes zero displacements of the door. So we can say that work done on it is zero.

Thus the above-given statement is false  

8 0
3 years ago
Pls Help Me with this problem
Naddika [18.5K]

Answer:

explanation

Explanation:

1 = C

2 = A

3 = D

4 = E

5 = B

8 0
2 years ago
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A tennis ball is dropped from a roof. If it takes 38.9 seconds to reach the ground, how fast is the ball moving just before it h
Ilya [14]

Answer:

The velocity of the ball before it hits the ground is 381.2 m/s

Explanation:

Given;

time taken to reach the ground, t = 38.9 s

The height of fall is given by;

h = ¹/₂gt²

h = ¹/₂(9.8)(38.9)²

h = 7414.73 m

The velocity of the ball before it hits the ground is given as;

v² = u² + 2gh

where;

u is the initial velocity of the on the root = 0

v is the final velocity of the ball before it hits the ground

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 7414.73 )

v = 381.2 m/s

Therefore, the velocity of the ball before it hits the ground is 381.2 m/s

5 0
2 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
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