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3 years ago

How can light energy solve our real life problem?

2 answers:

5 0

One big real-life problem is the inability to SEE stuff, like when it's dark all around you. When that happens and you try to move around, there's a big chance that you could bump into stuff, like walls, or step on stuff, like the dog. And honestly, driving the car is out, altogether.

Light energy can really help us solve this big problem, by bouncing off of the things around us and into our eyes. When light does this, our brain can figure out pretty quick WHERE all the things are, in what direction and how far from us. In that way, we can navigate ourself around, and we are prevented from bumping into something or stepping on it unless we want to. The light gives us our choice back.

kifflom [539]3 years ago

4 0

Answer:

It gives our light which we need for probably everything.

Explanation:

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2 years ago

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A laser beam is incident on two slits with a separation of 0.195 mm, and a screen is placed 5.30 m from the slits. If the bright

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Answer:

λ = 596 nm.

Explanation:

Fringe width = λ D / d

λ is wave length , D is screen distance and d is slit separation.

Putting the values

1.62 x 10⁻² =( λ x 5.3 ) / .195 x 10⁻³

$\lambda=\frac{1.62\times10^{-2}\times195\times10^{-6}}{5.3}$

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2 years ago

A parallel-plate capacitor in air has a plate separation of 1.76 cm and a plate area of

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Answer:

Explanation:

Plate separation, d = 1.76 cm = 0.0176 m

Area of plates, A = 25 cm^2 = 0.0025 m^2

V = 255 V

(a) Capacitance of capacitor

$C = \frac{\epsilon _0A}{d}$

$C = \frac{8.854\times 10^{-12}\times 0.0025}{0.0176}$

C = 1.258 x 10^-12 F

charge is same before and after immersion as the battery is disconnected

q = C V

q = 1.258 x 10^-12 x 255 = 3.2 x 10^-10 C

(b)

Capacitance before, C = 1.258 x 10^-12 C

capacitance after, C' = k x C = 80 x 1.258 x 10^-12 = 100.64 x 10^-12 C

Where, k is the dielectric constant of water = 80

Potential difference after immersion, V' = V / k = 255 / 80 = 3.1875 V

$U = \frac{q^{2}}{2C}$

$U = \frac{(3.2\times 10^{-10})^{2}}{2\times 1.258\times 10^{-12 }}=4.07\times 10^{-8}J$

Final energy

$U' = \frac{q^{2}}{2C'}$

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3 years ago

In space (no gravity or friction), you throw a ball with mass 0.1 kg at a target with mass 1 kg. You throw the ball at a speed o

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Answer:

Explanation:

We shall apply law of conservation of momentum in space to know the velocity of combination after the impact

m₁v₁ = m₂v₂

.1 x 4 = ( 1 + .1 ) v₂

v₂ = .3636 m /s

1 )

Kinetic energy of the combination

= 1/2 x 1.1 x ( .3636)²

= 7.3 x 10⁻² J

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= 1/2 x 0.1 x 4²

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This energy was converted into internal energy of the system .

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3 years ago

A boy swings a ball on a string at constant speed in a circle that has a circumference equal to 6 m. What is the work done on th

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Answer:

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Explanation:

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Also the angle θ adjacent to the perpendicular line = 90 °

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W = Fd cos θ

W = Fd cos 90°

W = Fd (0)

W = 0 J

Hence the force is perpendicular to the direction of displacement and the net work done in a circular motion in one complete revolution is = 0

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