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igomit [66]
4 years ago
6

Which statement is a correct expression of the Law of Conservation of Mass?

Physics
2 answers:
Mumz [18]4 years ago
8 0

Answer: B. The total mass remains the same during a chemical reaction.

Explanation:I Took The Test.

DiKsa [7]4 years ago
7 0
<span>The correct answer is B. The law of conservation of mass states that the mass of an isolated system is neither created nor destroy by chemical reaction or physical transformation. This means that the mass of subtances in an isolated system remains constant, it can not decrease and it can not increase.</span>
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if a person with a mass of 70kg is standing on scale in an elevator when the cable snaps, what will the reading on the scale be
dybincka [34]

Answer:

The reading will be the same.

Explanation:

Mass does not depend upon anything and it remains the same anywhere. What changes is the weight of the body because it depends upon gravity and is different at different places.

Giving me the brainest will be helpful.

3 0
3 years ago
A ball of 0.5kg slows down from 5m/s to 3m/s. Calculate the work done in the process
ivann1987 [24]
Answer: Cannot determine cause we need to know the change of time to calculate the work.

Explanation:

m = 0.5kg

V = 3m/s - 5m/s = -2m/s

P = W/t = Fv

F = ma

W = Fvt

W = (0.5)(9.8)t = 4.9t



4 0
2 years ago
Please any one need help on this
Romashka [77]

-- The wavelength and the amplitude were described in my answer to your  previous question.

-- A "compression" is a place where the wave is <em>compressed</em>.  It's the darker section of the wave in the picture, where the wavelength is temporarily shorter, so several waves are all bunched up (compressed) in a small time.

-- A "rarefaction" is exactly the opposite of a "compression".  It's a place where the wave gets more "<em>rare</em>" ... the wavelength temporarily gets longer, so that several waves get stretched out, and there are fewer of them in some  period of time.  The arrow in the picture points to a rarefaction.

3 0
3 years ago
A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of
levacccp [35]

\underline {\huge \boxed{ \sf \color{skyblue}Answer :  }}

<u>Given :</u>

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{D} = 5mA

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{DSS} = 10mA

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS(off)} = -6V

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS} =   {?}

\:  \:  \:

<u>Let's Slove :</u><u> </u>

  • \tt \large  I_{D} = I_{(DSS)}  (1 -   \frac {V_{GS}}{V_{GS(off)}} )^{2}

\:  \:  \:

  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times  V_{GS(off)}

\:  \:  \:

  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times  { - 6}

\:  \:

  • \underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}
3 0
2 years ago
The velocity of an object increases from –10 m/s to –15 m/s in 2.0 s. What is the average acceleration of the object? +2.5 m/s2
expeople1 [14]
Acceleration = cange in velocity ÷ change in time.
acceleration = (-15 - -10) ÷ 2
acceleration = -5 ÷ 2
ACCELERATION IS -2.5m/s2
5 0
4 years ago
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