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stepladder [879]
3 years ago
15

What is being despited in this picture

Physics
1 answer:
yan [13]3 years ago
8 0

Answer:

i am guessing for reflection but not so sure

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The voltage across the diode indicates the energy given to charge carriers (electrons and holes, but more about that later in th
Elis [28]

Answer:

the charge carriers have an energy 2.8 10⁻¹⁹ J

Explanation:

The energy in a diode is conserved so the energy supplied must be equal to the energy emitted in the form of photons.

The energy of a photon is given by the Planck expression

             E = h f

the speed of light, wavelength and frequency are related

             c = λ f

we substitute

             E = \frac{h \ c}{\lambda}

a red photon has a wavelength of lam = 700 nm = 700 10⁻⁹ m

we calculate the energy

              E = 6.626 10⁻³⁴  3 10⁸/700 10⁻⁹

              E = 2.8397 10⁻¹⁹J

therefore the charge carriers have an energy 2.8 10⁻¹⁹ J,

3 0
2 years ago
A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
Light of wavelength 597 nm falls on a double slit, and the first bright fringe of the interference pattern is seen at an angle o
Kazeer [188]

Answer:

2.2 µm

Explanation:

For constructive interference, the expression is:

d\times sin\theta=m\times \lambda

Where, m = 1, 2, .....

d is the distance between the slits.

Given wavelength = 597 nm

Angle, \theta  = 15.8°

First bright fringe means , m = 1

So,

d\times sin\ 15.8^0=1\times \597\ nm

d\times 0.2723=1\times \597\ nm

d=2192.43481\ nm

Also,

1 nm = 10⁻⁹ m

1 µm = 10⁻⁶ m

So,

1 nm = 10⁻³ nm

Thus,

<u>Distance between slits ≅ 2.2 µm</u>

8 0
3 years ago
Read 2 more answers
A woman pushes a 27 kg lawnmower at a steady speed. She exerts a 120 N force in a direction 35◦ below the horizontal. The accele
nikklg [1K]

The vertical force exerted on the lawn is 68.8 N downward

Explanation:

The vertical force exerted by the lawnmower on the lawn is equal to the vertical component of the force applied, therefore:

F_y = F sin \theta

where

F is the magnitude of the force applied

\theta is the angle between the direction of the force and the horizontal

In this problem:

F = 120 N

\theta=-35^{\circ}

Substituting,

F_y = (120)(sin 30)=-68.8 N

where the negative sign means the direction of the force is downward.

Learn more about vector components and forces here:

brainly.com/question/2678571

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

3 0
3 years ago
A 5000 kg African elephant has a resting metabolic rate of 2500 W. On a hot day, the elephant's environment is likely to be near
alina1380 [7]

Answer:

36 kg

Explanation:

To answer this question, a few assumptions have to be made:

  • That the temperature on the day is 35 °C
  • That all the heat from the elephant is goes to warming/evaporating the water on the surface of the elephant

Energy released per hour = 2500 J/s * 3600 s = 9 000 000 J

Q = mcΔT

9 000 000 J= m *4.186 J/g-K * (373K - 308K) + m*2260 J/g

m =  36 000 g = 36 kg

3 0
3 years ago
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