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Novay_Z [31]
2 years ago
11

The natural enzyme reaction in food slowly breaks it down and eventually causes it to spoil. what 2 things halt this enzyme reac

tion and help preserve the food?
microwaving
cooking at high temperature
storing properly in the refrigerator
blanching and canning
Physics
2 answers:
ollegr [7]2 years ago
4 0

Answer: the answer is storing properly in refrigeration.

Explanation:

katrin2010 [14]2 years ago
3 0

Answer:  storing food in the fridge

Explanation:

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According to newton's first law, what is required to make an object slow down?
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I believe it is friction

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"A boat that can travel at 4.0 km/h in still water crosses a river with a current of 2.0 km/h. At what angle must the boat be po
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30 degrees

Explanation:

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Hey!!

here is your answer >>>

We know that, acids and strong bases are good conductors of electricity!. And gold conducts electricity , while the best is silver and niext comes gold!. And H20 which is water contains less ions , so it is not a good conductors , it is a conductors but not a good conductor!. Salt in it's solid for cannot produce ions, hence, it cannot conduct electricity , but when this salt is mixed with water or when it is in it's molten state it can conduct electricity!. So now,the answer for the question is ,

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3 years ago
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mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi
valina [46]

Answer:

The answer is

"x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs \frac{8}{3} feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\

The source of the hooks law is stable,

16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\

Number \frac{1}{2}  times the immediate speed, i.e .. Damping force

\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2}  \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\

The m^2+m+12=0 and m is an auxiliary equation,

m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \  m2 =\frac{-1 - \sqrt{47i}}{2}

Therefore, additional feature

x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]

Use the form of uncertain coefficients to find a particular solution.  

Assume that solution equation,

x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\

These values are replaced by equation ( 1):

\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\

Going to compare cos3 t and sin 3 t coefficients from both sides,  

The cost3 t is 3A + 3B= 20 coefficients  

The sin 3 t is 3B -3A = 0 coefficient  

The two equations solved:

3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\

Replace the very first equation with the meaning,

3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\

equation is

x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)

The ultimate plan for both the equation is therefore

x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.  

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\

x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t)  +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\

c_2=\frac{-64\sqrt{47}}{141}

x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))

5 0
3 years ago
The waste products of a nuclear fission powerplant can best be described as
marishachu [46]
The waste products of a nuclear fission power plant can best be described as radioactive waste. 
These are the by-products from the processes carried out that produce nuclear energy. This type of waste is highly dangerous. A lot of attention has to be paid to the collection and disposal of this waste as it must not reach any near by water bodies for example. It can be deadly for life.
7 0
3 years ago
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