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3 years ago

What part of the electromagnetic spectrum do you think is most important and why?

Physics

2 answers:

vampirchik [111]3 years ago

6 0

Answer:

poooo

Explanation:

poooo

KonstantinChe [14]3 years ago

4 0

Radio waves, gamma-rays, visible light, and all the other parts of the electromagnetic spectrum are electromagnetic radiation. Electromagnetic radiation can be described in terms of a stream of mass-less particles, ...

The electromagnetic spectrum is a map of all the types of light that we can identify. It separates all the types of light by wavelength because that directly relates to how energetic the wave is. More energetic wave

For most of history, visible light was the only known part of the electromagnetic spectrum. The ancient Greeks recognized that light traveled in straight lines and studied some of its properties, including reflection

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Question :

riadik2000 [5.3K]

The second law of motion states that: the acceleration of an object is dependent upon two variables: - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object

<h3>Meaning of Motion</h3>

Motion can be defined as the process of changing position willingly or due to a force applied.

Motion can be seen in different forms and types depending on the object.

In conclusion, The second law of motion is used to deduce the formula for acceleration.

Learn more about second law of motion: brainly.com/question/2009830

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8 0

2 years ago

Read 2 more answers

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

3 years ago

A 60 kg pupil runs for 600m in 1 minute uniformly calculate kinetic energy

AURORKA [14]

velocity = traveled distance ÷ time of the traveled distance is seconds

velocity = 600 ÷ 60

velocity = 10 m/s

_________________________________

Kinetic Energy = 1/2 × mass × ( velocity )^2

KE = 1/2 × 60 × ( 10 )^2

KE = 30 × 100

KE = 3000 j

8 0

3 years ago

In practice, if a voltmeter was connected across any combination of the terminals, the potential difference would be less than w

VladimirAG [237]

Answer:

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

Explanation:

A voltmeter is built by a galvanometer and a resistance in series, this set is connected in parallel to the resistance where the voltage is to be measured, therefore the voltage is divided between the voltmeter and the element to be measured, consequently the measured voltage It is less than the calculated one, since for them the resistance of the voltmeter is assumed infinite.

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

8 0

2 years ago

Three identical capacitors are connected in parallel to a battery. if a total charge of q flows from the battery, how much charg

muminat

Each capacitor carry the same charge 'q'.

Discussion:
The voltage from the battery is distributed equally across all of the capacitors when they are linked in series. The three identical capacitors' combined voltage is computed as follows:

$V_{T}$ = V₁ +V₂ +V₃

This voltage may also be calculated using capacitance and charge;

V = Q/ C

$V_{T}$ = V₁ +V₂ +V₃

Provided that the total charge is 'q', hence the total voltage can be expressed as:

$V_{T}$ = (Q/C₁) + (Q/C₂) + (Q/C₃) = Q(1/C₁ +1/C₂ +1/C₃)

Therefore from the above explanation, it is concluded that each and every capacitor carry same charge 'q'.

Learn more about the capacitor here:

brainly.com/question/17176550

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7 0

1 year ago