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astraxan [27]
3 years ago
14

Which option identifies a factor that would make the reasoning for a claim scientifically invalid?

Physics
1 answer:
lora16 [44]3 years ago
7 0

Answer:

C)a lack of supporting empirical evidence

Explanation:

A scientific reasoning can be rendered invalid if empirical evidences are lacking. Empirical evidences are usually a product of observations or experiments.

When these are lacking or cannot be gathered, scientific observations or sometimes reasoning can be rendered as invalid because there is no ground to prove them.

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The friction between a nonmoving object and the surface is resting on is called..?
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Answer:

Static friction exists between a stationary object and the surface on which it is resting. It prevents an object from moving against the surface. Example: Static friction prevents an object like a book from falling of the desk, even if the desk is slightly tilted

Explanation:

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Three different groups each measured the diameter of a CD three times (the actual diameter of a CD is 12.00cm). The class data i
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During nuclear decay, a new isotope is created. How is the nucleus of the new isotope different from the parent if the parent is
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Alpha decay will produce a daughter nucleus with more protons and beta decay will produce a daughter nucleus with fewer protons than the parent nucleus has.

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3 years ago
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Liquefaction _____.
gogolik [260]

Answer:

D).happens when soil is weakened by water and seismic waves              

Explanation:

Liquefaction is a phenomenon in which the strength of the soil weakens due to the earthquake or any other load. The solid behaves like a liquid. Liquefaction of soil below structures cause lot of damage during Earthquakes.The soil looses stiffness due to seismic waves and water.

Thus, the correct option is D.

3 0
3 years ago
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. If the surface charg
bekas [8.4K]

Answer:

5.3\times 10^3 N/C

Explanation:

We are given that

Distance between plates=d=2.2 cm=2.2\times 10^{-2} m

1 cm=10^{-2} m

\sigma=47nC/m^2=47\times 10^{-9}C/m^2

Using 1 nC=10^{-9} C

We have to find the magnitude of E in the region between the plates.

We know that the electric field for parallel plates

E=\frac{\sigma}{2\epsilon_0}

E_1=\frac{\sigma}{2\epsilon_0}

E_2=\frac{\sigma}{2\epsilon_0}

E=E_1+E_2

E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Where \epsilon_0=8.85\times 10^{-12}C^2/Nm^2

Substitute the values

E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}

E=5.3\times 10^3 N/C

Hence, the magnitude of E in the region between the plates=5.3\times 10^3 N/C

5 0
3 years ago
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