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3 years ago

Why is the nail driven into the wood? Explain.

2 answers:

6 0

The nail can penetrate the wood because it is sharp, applying high pressure to the surface area it contacts. The nail's head is relatively much wider, so it will not penetrate your hand as easily as the sharp end, but it is clearly not engineered to be pushohand.

NNADVOKAT [17]3 years ago

5 0

Answer:

The nail is driven into the wood so that it would hold into the wood because if the wood was driven to the nail it may not stay and it would take way to much effort.

Explanation:

You might be interested in

If you know the distance an object has traveled in a certain amount of time you can determine

Nutka1998 [239]

With that information, you can determine the object's speed.
Just divide the distance covered by the time to cover the distance.

If you also know the direction the object moved, then you can
determine its velocity. If you don't, then you can't.

5 0

4 years ago

Read 2 more answers

A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel

erastovalidia [21]

Answer:

Explanation:

To solve the problem it is necessary to apply the concepts of Destructive and constructive interference. The constructive interference in tin film is given by

$2t = (m+\frac{1}{2})\frac{\lambda}{n}$

Where,

t = thickness

$\lambda=$ Wavelenght

m= is an integer

n= film/refractive index

We use this equaton because phase change is only present for gasoline air interface, but not at the gasoline-water interface. <em>The minimum t only would be when the value of m=0 then</em>

$2nt = \frac{\lambda}{2}$

$t = \frac{560nm}{4*1-4}$

$t = 100nm$

Therefore the correct answer is D. The minimum thickness of the film to see ab right reflection is 100nm

4 0

4 years ago

What would you do if someone came up atta nowhere and stole your kneecaps, what do you do?

arsen [322]

Answer:

peat them with my dislocated legs

3 0

3 years ago

An electron moving to the left at 0.8c collides with a photon moving to the right. After the collision, the electron is moving t

SVETLANKA909090 [29]

Answer:

Wavelength = 2.91 x 10⁻¹² m, Energy = 6.8 x 10⁻¹⁴

Explanation:

In order to show that a free electron can’t completely absorb a photon, the equation for relativistic energy and momentum will be needed, along the equation for the energy and momentum of a photon. The conservation of energy and momentum will also be used.

E = y(u) mc²

Here c is the speed of light in vacuum and y(u) is the Lorentz factor

y(u) = 1/√[1-(u/c)²], where u is the velocity of the particle

The relativistic momentum p of an object of mass m and velocity u is given by

p = y(u)mu

Here y(u) being the Lorentz factor

The energy E of a photon of wavelength λ is

E = hc/λ, where h is the Planck’s constant 6.6 x 10⁻³⁴ J.s and c being the speed of light in vacuum 3 x 108m/s

The momentum p of a photon of wavelenght λ is,

P = h/λ

If the electron is moving, it will start the interaction with some momentum and energy already. Momentum of the electron and photon in the initial and final state is

p(pi) + p(ei) = p(pf) + p(ef), equation 1, where p refers to momentum and the e and p in the brackets refer to proton and electron respectively

The momentum of the photon in the initial state is,

p(pi) = h/λ(i)

The momentum of the electron in the initial state is,

p(ei) = y(i)mu(i)

The momentum of the electron in the final state is

p(ef) = y(f)mu(f)

Since the electron starts off going in the negative direction, that momentum will be negative, along with the photon’s momentum after the collision

Rearranging the equation 1 , we get

p(pi) – p(ei) = -p(pf) +p(ef)

Substitute h/λ(i) for p(pi) , h/λ(f) for p(pf) , y(i)mu(i) for p(ei), y(f)mu(f) for p(ef) in the equation 1 and solve

h/λ(i) – y(i)mu(i) = -h/λ(f) – y(f)mu(f), equation 2

Next write out the energy conservation equation and expand it

E(pi) + E(ei) = E(pf) + E(ei)

Kinetic energy of the electron and photon in the initial state is

E(p) + E(ei) = E(ef), equation 3

The energy of the electron in the initial state is

E(pi) = hc/λ(i)

The energy of the electron in the final state is

E(pf) = hc/λ(f)

Energy of the photon in the initial state is

E(ei) = y(i)mc2, where y(i) is the frequency of the photon int the initial state

Energy of the electron in the final state is

E(ef) = y(f)mc2

Substitute hc/λ(i) for E(pi), hc/λ(f) for E(pf), y(i)mc² for E(ei) and y(f)mc² for E(ef) in equation 3

Hc/λ(i) + y(i)mc² = hc/λ(f) + y(f)mc², equation 4

Solve the equation for h/λ(f)

h/λ(i) + y(i)mc = h/λ(f) + y(f)mc

h/λ(f) = h/lmda(i) + (y(i) – y(f)c)m

Substitute h/λ(i) + (y(i) – y(f)c)m for h/λ(f) in equation 2 and solve

h/λ(i) -y(i)mu(i) = -h/λ(f) + y(f)mu(f)

h/λ(i) -y(i)mu(i) = -h/λ(i) + (y(f) – y(i))mc + y(f)mu(f)

Rearrange to get all λ(i) terms on one side, we get

2h/λ(i) = m[y(i)u(i) +y(f)u(f) + (y(f) – y(i)c)]

λ(i) = 2h/[m{y(i)u(i) + y(f)u(f) + (y(f) – y(i))c}]

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

Calculate the Lorentz factor using u(i) = 0.8c for y(i) and u(i) = 0.6c for y(f)

y(i) = 1/[√[1 – (0.8c/c)²] = 5/3

y(f) = 1/√[1 – (0.6c/c)²] = 1.25

Substitute 6.63 x 10⁻³⁴ J.s for h, 0.511eV/c2 = 9.11 x 10⁻³¹ kg for m, 5/3 for y(i), 0.8c for u(i), 1.25 for y(f), 0.6c for u(f), and 3 x 10⁸ m/s for c in the equation derived for λ(i)

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

λ(i) = 2(6.63 x 10-34)/[(9.11 x 10-31)(3 x 108){(5/3)(0.8) + (1.25)(0.6) + ((1.25) – (5/3))}]

λ(i) = 2.91 x 10⁻¹² m

So, the initial wavelength of the photon was 2.91 x 10-12 m

Energy of the incoming photon is

E(pi) = hc/λ(i)

E(pi) = (6.63 x 10⁻³⁴)(3 x 10⁸)/(2.911 x 10⁻¹²) = 6.833 x 10⁻¹⁴ = 6.8 x 10⁻¹⁴

So the energy of the photon is 6.8 x 10⁻¹⁴ J

6 0

3 years ago

Place the items in order from the largest wavelength to the shortest wavelength.

Butoxors [25]

From largest to shortest wavelength:

Radio waves, Microwaves, Infrared radiation, Red visible, Orange visible, Yellow visible, Green visible, Blue visible, Violet visible, Ultraviolet, X-rays, Gamma rays

Explanation:

Electromagnetic waves are oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion the wave.

Electromagnetic waves are the only type of waves able to travel in a vacuum, and in a vacuum they always at the same speed, the speed of light, equal to:

$c=3.0\cdot 10^8 m/s$

Electromagnetic waves are classified into 7 different types, according to their wavelength/frequency. From slongest to shortest wavelength, they are ranked as follows:

Radio waves

Microwaves

Infrared radiation

Visible light

Ultraviolet

X-rays

Gamma rays

Visible light is the only part of the spectrum that the human eye is able to see. Depending on the wavelength of the visible light, we perceive the radiation as a different color. In order from longest to shortest wavelength, colors are:

Red

Orange

Yellow

Green

Blue

Indigo

Violet

Therefore, the correct order from largest to shortest wavelength in the given list is:

Radio waves

Microwaves

Infrared radiation

Red visible

Orange visible

Yellow visible

Green visible

Blue visible

Violet visible

Ultraviolet

X-rays

Gamma rays

Learn more about electromagnetic waves:

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brainly.com/question/12450147

#LearnwithBrainly

4 0

3 years ago