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Step2247 [10]
2 years ago
13

Temperature danger zone what is the biggest cause of foodborne illness

Physics
1 answer:
Nina [5.8K]2 years ago
5 0
Foodborne illnesses are caused by eating contaminated food and drinking contaminated water. The contaminations are caused mostly by bacteria, virus, and parasites.
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Explain how microorganisms are friend and foe.​
Helga [31]

Answer:

Because they are beneficial

Explanation:

There are microorganisms in our large intestine that synthesis vitamins and allow them to be absorbed into the bloodstream. However a tiny minority are pathogens (disease-causing agents). These pathogens often called germs or bugs, are a threat to all life forms.

8 0
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The trade winds are located about ____ degrees north and south of the equator.
cupoosta [38]

The trade winds are located about 30 degrees north and south of the equator.

8 0
3 years ago
A small smooth object slides from rest down a smooth inclined plane inclined at 30degrees horizontal.What is the acceleration do
asambeis [7]
The acceleration is given as:

a = g sin(30°) where g is the gravitational acceleration

For g = 10 m/s^2, we get

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2 years ago
A large, massive object collides with a stationary, smaller object on an ice rink. If the large object transfers all of its mome
Hunter-Best [27]

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a ut will move faster than the large object was moving initially

5 0
2 years ago
A civil engineer plans to design a curved ramp such that a car may not have to rely on friction to round the curve without skidd
Delicious77 [7]

Answer:\theta =\tan ^{-1}(\frac{v^2}{gR})

Explanation:

let \theta be the inclination at which curve is tilted

v is the speed of car and R is the radius of curve

m is the mass of car

Suppose R is the reaction offered by road to car

Resolving R in x and y direction we get

R\cos \theta will balance weight and R\sin \theta will provide the necessary centripetal Force

thus R\sin \theta =\frac{mv^2}{R}  ------------1

R\cos \theta =m g  ----------------2

Divide 1 & 2 we get

\frac{R\sin \theta }{R\cos \theta }=\frac{mv^2}{mgR}

\tan \theta =\frac{v^2}{gR}

\theta =\tan ^{-1}(\frac{v^2}{gR})                                      

8 0
2 years ago
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