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adoni [48]
3 years ago
9

In 1864, Louis Pasteur was asked to investigate diseases afflicting the wine in Arbois, France. He discovered that these disease

s were caused by micro-organisms that could be killed by heating the wine to 55 C for some time. What is this process called today that applies to milk?
A. Homeostasis

B. Replace this text with your second answer

C. Differentiation

D. Pasteurization
Physics
1 answer:
iragen [17]3 years ago
8 0

Answer:

D. Pasteurization

Explanation:

-Pasteurization is the process by which foods or liquids usually destined for long-term storage are heated the quickly cooled to eliminate pathogens or bacteria.

-The heating is mostly done at temperatures of around 212°F  for  atleast 30 minutes before the rapid coolind.

-Bacteria or pathogens cannot survive these extreme temperatures and will therefore die in the process thus extending the shelf life of a product.

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HELP ASAP PLEASE<br><br> Which of the following mechanical waves have the same energy?
serious [3.7K]

Answer:

1 and 4 is the answer

6 0
3 years ago
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If 90 grams of gold has a volume of 5 cm3, calculate the density (include the units and show your work or you get no credit
ehidna [41]

Answer:

The density of gold is of 18 grams per cm3.

Explanation:

The mass density of a homogeneous material expresses how much mass of that material is present in a given volume. Since the density of an object is obtained by dividing its mass by its volume, to obtain the density of gold, its 90 grams of mass must be divided by its 5 cm3 volume, performing the following calculation:

90/5 = X

18 = X

Thus, the density of gold is 18 grams per cm3.

6 0
3 years ago
Acceleration usually has the symbol a. It is a vector. What is the correct way
Ghella [55]

Explanation:

a straight line under the letter

3 0
2 years ago
The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail
Alik [6]

Answer:

1.34\cdot 10^{-16} C

Explanation:

The strength of the electric field produced by a charge Q is given by

E=k\frac{Q}{r^2}

where

Q is the charge

r is the distance from the charge

k is the Coulomb's constant

In this problem, the electric field that can be detected by the fish is

E=3.00 \mu N/C = 3.00\cdot 10^{-6}N/C

and the fish can detect the electric field at a distance of

r=63.5 cm = 0.635 m

Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^{-6} N/C)(0.635 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=1.34\cdot 10^{-16} C

4 0
2 years ago
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.
S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }

Where:

\omega - Angular frequency, measured in radians per second.

m - Mass of the physical pendulum, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

d - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

I_{O} - Moment of inertia with respect to pivot point, measured in kg\cdot m^{2}.

In addition, frequency and angular frequency are both related by the following formula:

\omega =2\pi\cdot f

Where:

f - Frequency, measured in hertz.

If f = 0.658\,hz, then angular frequency of the physical pendulum is:

\omega = 2\pi \cdot (0.658\,hz)

\omega = 4.134\,\frac{rad}{s}

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}

I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}

Given that m = 1.15\,kg, g = 9.807\,\frac{m}{s^{2}}, d = 0.425\,m and \omega = 4.134\,\frac{rad}{s}, the moment of inertia associated with the physical pendulum is:

I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}

I_{o} = 0.280\,kg\cdot m^{2}

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

8 0
2 years ago
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