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Nezavi [6.7K]
3 years ago
10

Technician A says that the freezing point of a substance is the temperature at which the substance changes state from a solid to

a liquid. Technician B says that the freezing point of a substance is the temperature at which the substance changes state from a liquid to a solid. Which technician is correct?
a) technician A
b) technician B
c) both technician A and B
d) neither technician A nor B
Physics
1 answer:
pashok25 [27]3 years ago
5 0

Answer:

Technician B is correct

Explanation:

Freezing is a method of conversion of substance in its liquid state to solid state. It is the process by which a liquid substance changes to a solid at a particular temperature.

Increasing the pressure and decreasing the temperature of a liquid increases its freezing point. For example, in other to freeze water i.e to change water to ice, it has to be kept in a fridge at a temperature lower than the temperature of the water. The essence of covering the fridge after placing the water in the fridge is to increase the pressure of the liquid hence increasing its freezing rate.

Based on the above explanation, it can be concluded that technician B is correct.

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Density is a physical property that relates the mass of a substance to its volume. A. Calculate the density, in g/mL , of a liqu
Angelina_Jolie [31]

Answer:

A) 0.660 g/ml

B) 1.297 ml

C) 0.272 g

Explanation:

Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes.  This is where density D appears as a  physical characteristic property of matter that establishes a relationship between the mass m of a body or substance and the volume V it occupies:

D=\frac{m}{V} (1)

Knowing this, let's begin with the answers:

<h2 /><h2>Answer A:</h2>

Here the mass is m=0.155g and th volume V=0.000235L=0.235mL

Solving (1) with these values:

D=\frac{0.155g}{0.235mL} (2)

D=0.660g/mL (3)

<h2>Answer B:</h2>

In this case the mass of a sample is m=4.71g and its density is D=3.63g/mL.

Isolating V from (1):

V=\frac{m}{D} (4)

V=\frac{4.71g}{3.63g/mL} (5)

V=1.297mL (5)

<h2>Answer C:</h2>

In this case the volume of a sample is V=0.293mL and its density is D=0.930g/mL.

Isolating m from (1):

m=D.V (6)

m=(0.930g/mL)(0.293mL) (7)

m=0.272g (8)

4 0
3 years ago
Read 2 more answers
Calculate the most probable speed of an ozone molecule in the stratosphere
Marysya12 [62]

Answer:

v_{mp}=305.83 m/s

Explanation:

The temperature in stratosphere is generally about 270 K

molecular weight of an ozone molecule = 48 gm/mole

now formula for most probable velocity

v_{mp}= \sqrt{\frac{2RT}{M} }

plugging the values we get

v_{mp}= \sqrt{\frac{2\8.314\times270}{48} }

v_{mp}=305.83 m/s

7 0
3 years ago
What is the definition of cell ?
vodomira [7]
<span>the smallest structural and functional unit of an organism, typically microscopic and consisting of cytoplasm and a nucleus enclosed in a membrane. Microscopic organisms typically consist of a single cell, which is either eukaryotic or prokaryotic.</span>
5 0
3 years ago
Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, in
olya-2409 [2.1K]

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

I=\Delta p = m\Delta v

where

m = 62.0 kg is the mass of the passenger

\Delta v is the change in velocity of the car (and the passenger), which is

\Delta v = 5.30 m/s - 0 = 5.30 m/s

So, the linear impulse experienced by the passenger is

I=(62.0 kg)(5.30 m/s)=328.6 kg m/s

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

I=F \Delta t

where in this case

I=328.6 kg m/s is the linear impulse

\Delta t = 0.812 s is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N

7 0
3 years ago
I need help, ASAP, I know what thermal energy is, but how does a skater use it when on the track?
Flauer [41]

Answer: Looked it up but

Explanation:

When the skater lands on the track, the vertical component of his kinetic energy is converted to thermal energy. You can do experiments where there is no loss to thermal energy (only PE and KE conversions) by turning friction off and by making sure the skater doesn't leave the track.

4 0
3 years ago
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