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2 years ago

Temperature danger zone what is the biggest cause of foodborne illness

1 answer:

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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Static cling makes your clothes stick together. what causes this to happen? 1. external forces to the clothes 2. forces of natur

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2 years ago

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Which one of the following accurately describes the force of gravity?

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2 years ago

What does the 2 mean in the formula 5Mg3(PO4)2?

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2 years ago

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(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord ca

Aleks04 [339]

(a) Let $v$ be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

$a_{\rm rad} = \dfrac{v^2}R$

where $R$ is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

$F = (1.500\,\mathrm{kg}) a_{\rm rad}$

so that

$(1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N$

Solve for $v$ :

$v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}$

(b) The net force equation in part (a) leads us to the relation

$F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}$

so that $v$ is directly proportional to the square root of $R$ . As the radius $R$ increases, the maximum linear speed $v$ will also increase, so the cord is less likely to break if we keep up the same speed.

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1 year ago

A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un

KonstantinChe [14]

Answer:

32s

Explanation:

We must establish that by the time the police car catches up to the speeder, both have travelled a certain distance during the same amount of time. However, the police car experiences accelerated motion whereas the speeder travels at a constant velocity. Therefore we will establish two formulas for distance starting with the speeder's distance:

$x=vt=23.3\frac{m}{s}t$

and the police car distance:

$x=vt+\frac{at^{2}}{2}=0+\frac{2.75\frac{m}{s^{2}} t^{2}}{2}=0.73\frac{m}{s^{2}}$

Since they both travel the same distance x, we can equal both formulas and solve for t:

$0 = 0.73\frac{m}{s^{2}}t^{2}-23.3\frac{m}{s} t\\\\0=t(0.73\frac{m}{s^{2}}t-23.3\frac{m}{s} )\\\\$

Two solutions exist to the equation; the first one being $t=0$

The second solution will be:

$0.73\frac{m}{s^{2}}t=23.3\frac{m}{s}\\\\t=\frac{23.3\frac{m}{s}}{0.73\frac{m}{s^{2}}}=32s$

This result allows us to confirm that the police car will take 32s to catch up to the speeder

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2 years ago