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3 years ago

Can u help me on the first one: a ship going out of sea goes out of sight

Physics

2 answers:

k0ka [10]3 years ago

7 0

A ship sailing away from you gradually goes out of sight,
starting from the bottom and moving up the sides of the ship.
The Earth is a sphere. As the ship gets farther away from
you, it's working its way around the curve of the sphere.
Since you can't see around the curve of the Earth, the
lowest point on the ship that you can see moves slowly
up the ship.

kow [346]3 years ago

7 0

The earth is a sphere

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The distance between two stations is 1995 Km. How much time will it take to cover the distance at an average speed of 19KM/hour

Flura [38]

105 hours or 4.375 days

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3 years ago

Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher

siniylev [52]

Answer:

$v = 4.44 \times 10^5 m/s$

Explanation:

By Einstein's Equation of photoelectric effect we know that

$h\nu = W + \frac{1}{2}mv^2$

here we know that

$h\nu$ = energy of the photons incident on the metal

$W$ = minimum energy required to remove photons from metal

$\frac{1}{2}mv^2$ = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

$W = \frac{hc}{351 nm}$

here we know that

$hc = 1242 eV-nm$

now we have

$W = \frac{1242}{351} = 3.54 eV$

now by energy equation above when photon of 303 nm incident on the surface

$\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2$

$4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2$

$(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2$

$8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2$

$v = 4.44 \times 10^5 m/s$

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3 years ago

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maria [59]

Answer:

15kg

Explanation:

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3 years ago

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scZoUnD [109]

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Redshift, or lower power

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2 years ago

A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vec

Phoenix [80]

Answer:

The particle path will follow

(d) a circular path

Explanation:

When a charged particle having charge of magnitude ' $q$ ' enters into a magnetic field such that its velocity vector ' $\vec{V}$ ' is perpendicular to the direction of the magnetic field ' $\vec{B}$ ', then it will experience a force, called Lorentz force ( $\vec{F_{L}}$ ), given by

$\vec{F_{L}} = \vec{V} \times \vec{B}$

As shown in the figure, the magnetic field is directed perpendicular to the plane and towards the plane (as shown by the circle and 'X'-sign) and the velocity vector is from left to right on the plane.

According to the property of cross-product, the Lorentz force ( $F_{L}$ ) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path,as shown in the figure.

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3 years ago