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Bingel [31]
3 years ago
11

What uses electricity

Physics
1 answer:
Rus_ich [418]3 years ago
5 0
Our phones, lighting ect
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The photons of different light waves:
LuckyWell [14K]

Answer: contain different amounts of energy

Explanation:

The energy E of a photon is given by:

E=h\nu

Where:

h=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

\nu is the frequency of the light which is inversely related to the wavelength.

Now, if we have photons of different light waves, this means we have photons with different frequencies.

As the energy of the photon depends on its frequency:

Photons of different light waves <u>contain different amounts of energy.</u>

8 0
3 years ago
A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D
RideAnS [48]

Answer:

Explanation:

Initial kinetic energy of M = 1/2 M vi²

let final velocity be vf

v² = u² + 2a s

vf² =  vi² + 2 (F / M) x D

Kinetic energy

= 1/2 Mvf²

= 1/2 M ( vi² + 2 (F / M) x D

1/2 M vi² + FD

Ratio with initial value

1/2 M  vi² + FD) / 1/2 M  vi²

RK = 1 + FD / 2 M  vi²

4 0
3 years ago
It has been known for many years that ceramics can conduct electricity. In 2008, scientists found a ceramic material that can co
ollegr [7]

Answer:

The answer is the second option.

Explanation:

This is a higher temperature than Onnes's experiment, and it will allow for a broader use of superconductors.

8 0
3 years ago
Read 2 more answers
What minimum speed does a 200 g puck need to make it to the top of a frictionless ramp that is 4.1 m long and inclined at 22 ∘?
myrzilka [38]

Answer:

5.5 m/ sec

Explanation:

Because the inclined surface is frictionless so we can assume that total change of energy is zero

i-e ΔE = 0

Or we can say that difference between final and initial energy is zero i-e

Ef- Ei =0

Where,

Ef= final energy at the top of the ramp= KEf+PEf

Ei= Initial energy at the bottom of the ramp=KEi+PEi

So we have

(KEf+PEf)-(KEi+PEi)=0

==>KEf-KEi+PEf-PEi=0            -------------(1)

KEf = mgh = 200×9.8×h

Where h= Sin 22 = h/d= h/4.1

or

0.375×4.1=h

or h= 1.54 m

So, PEf= 200×9.8×1.54=3018.4 j

and KEf= 1/2 mVf^{2}= 0.5×200×0=0 j

PEi= mgh = 200×9.8×0=0 j

KEi= 1/2 mVi^{2}=0.5×200×Vi^{2}=100Vi^{2} j

Put these values in eq 1, we get;

0-100 Vi^{2}+3018.4-0=0

-100 Vi^{2}=-3018.4

==> Vi^{2}= \frac{3018.4}{100} = 30.184

==>  Vi = \sqrt{30.184}  = 5.5 m.sec

7 0
4 years ago
Helppppp pleaseeee 25 points please?????
Sedbober [7]

Answer:

Where is question 12, we need it to answer this question

Explanation:

6 0
3 years ago
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