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g100num [7]
3 years ago
13

Which will most likely result in the revision of a theory?

Physics
2 answers:
viva [34]3 years ago
4 0
I would say A not 100℅ thou
N76 [4]3 years ago
3 0
<h3><u>Answer;</u></h3>

A) the development of new experimental methods

<h3><u>Explanation;</u></h3>
  • A scientific theory refers to a well-substantiated explanation of some aspect of natural world that is based on careful examination facts.
  • <em><u>Scientific theories can be tested and refined by additional research, and they allow scientists to make predictions.</u></em><em><u> Development of new experimental methods may lead to revision of a scientific theory.</u></em>
  • <em><u>Additionally, a scientific theory may be revised when new large amount of data refutes it.</u></em>
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MAXImum [283]

Answer:

true

Explanation:

Newton is the measure of the force with turns to be gravity multiplying the mass. Thus, the forces acts on the particles in the direction of the movement of the particles

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3 years ago
What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed
tekilochka [14]

Answer:

charge will be equal to 2.03\times 10^{-5}C  

Explanation:

We have given mass of the particle m = 1.45 gram = 0.00145 kg

Acceleration due to gravity g=9.8m/sec^2

Electric field E = 700 N/C

Electric force will be equal to F=qE, here q is charge and E is electric field

For particle to be stationary this force must be equal to force due to gravity , that is mg force

So qE = mg

q=\frac{mg}{E}=\frac{0.00145\times 9.8}{700}=2.03\times 10^{-5}C

So charge will be equal to 2.03\times 10^{-5}C

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2 years ago
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2 years ago
Which material would you take, steel or bricks to design a bridge? Why?
Vesna [10]

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4 0
2 years ago
Read 2 more answers
Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
netineya [11]

Answer:

Δu=1300kJ/kg  

Explanation:

Energy at the initial state

p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)

Is saturated vapor at initial pressure we have

p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)

Process 2-3 is a constant volume process

p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg

Δu=1300kJ/kg  

3 0
2 years ago
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