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3 years ago

Two forces that act on very small distances (smaller than you can see ) are

Physics

1 answer:

BlackZzzverrR [31]3 years ago

7 0

There are three fundamental forces that act at a distance. They are gravitational, electromagnetic and nuclear forces.

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At 5.0 minutes, the temperature of the water reaches 100 °C. The volume of the water in the urn

serious [3.7K]

Answer:

fdvevddvevkejokef0jeovdlvkjeuiyv

Explanation:

ddf4edscd

4 0

3 years ago

The difference in potential between the accelerating plates of a TV set is about 24 kV. The distance between these plates is 1.5

ryzh [129]

Answer:

So electric field between the plates will be equal to $1600\times 10^3KN/C$

Explanation:

We have given potential difference between accelerating plates = 24 KV = 24000 volt

Distance between the plates d = 1.5 cm = 0.015 m

We know that potential difference is given by V = Ed, here E is electric field and d is distance between plates

So $24000=E\times 0.015$

E = 1600000 N/C = $1600\times 10^3KN/C$

So electric field between plates will be equal to $1600\times 10^3KN/C$

7 0

3 years ago

Use the claim to answer the question.

Pachacha [2.7K]

Scientific evidence such as reflection and photoelectric effect has proven the wave-particle model of electromagnetic radiation to be true.

<h3>What is electromagnet radiation?</h3>

Electromagnetic radiation is radiation produced as a result of the interactions of the electric and magnetic fields.

Some forms of electromagnetic radiation include:

radio waves
microwaves
ultraviolet radiation
visible light

Electromagnetic radiation can be described using a wave model or a particle model.

The wave property of electromagnetic radiation include:

diffraction
reflection
refraction

The particle property of electromagnetic radiation include:

photoelectric effect
Compton effect.

Therefore, the wave-particle model of electromagnetic radiation is correct because it can be backed up by experiments and evidence.

Learn more about wave-particle model at: brainly.com/question/20452331

6 0

2 years ago

Read 2 more answers

The rectangular plates in a parallel-plate capacitor are 0.063 m × 5.4 m. A distance of 3.5 × 10–5 m separates the plates. The p

djverab [1.8K]

Answer:

capacitance of the capacitor = 0.18 μ F

Explanation:

Area of the plate A = 0.063 m x 5.4 m = 0.3402 m²

distance between the plate d = 3.5 × 10–5 m

dielectric value for Teflon K = 2.1

capacitance of capacitor = ?

Formula for capacitance of parallel plate is as follows ,

$C= \frac{K\epsilon_0 A}{d}$ (Where $\epsilon_0 = 8.82 \times 10^-12\ \frac{F}{m}$ )

putting the values in the equation,

C = $\frac{2.1\times8.82\times10^-12\times0.3402 }{3.5\times10^-5}$ $=0.18\times 10^-6$ =0.18 μ F

4 0

3 years ago

The electric field strength is 5.50×10^4 N/C inside a parallel-plate capacitor with a 2.50 mm spacing. A proton is released from

valentina_108 [34]

Answer:

$v=1.6\times10^{5}m/s$

Explanation:

The equation that relates electric field strength and force that a charge experiments by it is F=qE.

Newton's 2nd Law states F=ma, which for our case will mean:

$a=\frac{F}{m}=\frac{qE}{m}$

We know from accelerated motion that $v^2=v_0^2+2ad$

In our case the proton is released from rest, so $v_0=0m/s$ and we get $v=\sqrt{2ad}$

Substituting, we get our final velocity equation:

$v=\sqrt{\frac{2qEd}{m}}$

For the <em>proton </em>we know that $q=1.6\times10^{-19}C$ and $m=1.67\times10^{-27}kg$ . Writing in S.I. d=0.0025m, we obtain:

$v=\sqrt{\frac{2(1.6\times10^{-19}C)(5.5\times10^{4}N/C)(0.0025m)}{(1.67\times10^{-27}kg)}}=162318.5m/s=1.6\times10^{5}m/s$

6 0

3 years ago