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WARRIOR [948]
3 years ago
15

Which would show an example of how physical changes are reversible?

Physics
2 answers:
kondor19780726 [428]3 years ago
8 0

Answer:

Melting tin and then cooling it into a mold

Explanation:

When you melt something, and when it cools it returns to its physical state, therefore the physical changes are reversible. For example, take chocolate. When chocolate melts its liquid, then when it's in a colder situation it becomes solid again and so on, the changes are reversible.

il63 [147K]3 years ago
8 0

Answer:

C

Explanation:

the physical change was the gold bar being cut and if it is put back together the physical change is reversed

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Loses electron as negative charge is lost therefore becomes overall positive.
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A wave is incident on the surface of a mirror at an angle of 41° with the normal. what can you say about its angle of reflection
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It's angle of reflection must be 41 degrees

we know, by the first law of reflection that angle of incidence is always equal to angle of reflection..........

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How is a theory different from a hypothesis?
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Answer:

A theory is a system of ideas intended to explain something, and a hypothesis is an educated guess.

Explanation: Hope this Helps! :)

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4 years ago
If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If
Rama09 [41]

Answer:

A) 0.120

B) 1.6s

C) 0.625 Hz

Explanation:

Here, X = 0.120 m

x = 0.120 m after t = 0.800 s

A- Amplitude = max displacement from equillibrium position = 0.120 m

B- Period = 2 * 0.800 = 1.6 s

C- Frequency = 1/peroid = 1/1.6

f = 0.625 Hz

3 0
3 years ago
A 25 {\rm pF} parallel-plate capacitor with an air gap between the plates is connected to a 100 {\rm V} battery. A Teflon slab i
jonny [76]

Answer:

0.275 nF

Explanation:

First of all, let's calculate the initial charge stored by the capactiro when it is in air. This is given by:

Q_0 = C_0 V

where

C_0 = 25 pF=25\cdot 10^{-12} F is the initial capacitance of the capacitor

V=100 V is the voltage of the battery

Substituting, we find

Q_0 = (25\cdot 10^{-12}F)(100 V)=25\cdot 10^{-10} F=0.25 nF

When the Teflon slab is inserted between the plates, the capacitance changes according to:

C'=kC_0

where

k=2.1 is the dielectric constant of Teflon. Therefore, the new charge stored on the capacitor will be:

Q'=C' V=(kC_0)V=(2.1)(25\cdot 10^{-12} F)(100 V)=0.525 nF

And so, the change in the charge on the capacitor is:

\Delta Q=Q'-Q=0.525 nF-0.25 nF=0.275 nF

7 0
3 years ago
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