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3 years ago

When your finger turns green from that new ring you bought ....what is happening?

Chemistry

2 answers:

Troyanec [42]3 years ago

6 0

Answer:

Your ring may be fake

Explanation:

xxMikexx [17]3 years ago

6 0

The dye on the ring could be projecting onto your skin. It could be that you’re having an allergic reaction or the ring could be too tight.

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The work function for metallic potassium is 2.3 eV. Calculate the velocity in km/s for electrons ejected from a metallic potassi

mote1985 [20]

Answer:

932.44 km/s

Explanation:

Given that:

The work function of the magnesium = 2.3 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = $2.3\times 1.6022\times 10^{-19}\ J=3.68506\times 10^{-19}\ J$

Using the equation for photoelectric effect as:

$E=\psi _0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

$\lambda$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

v is the velocity of electron

Given, $\lambda=260\ nm=260\times 10^{-9}\ m$

Thus, applying values as:

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{260\times 10^{-9}}=3.68506\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2$

$3.68506\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{260\times \:10^{-9}}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=7.64538\times 10^{-19}-3.68506\times 10^{-19}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=3.96032462\times 10^{-19}$

$v^2=0.869446\times 10^{12}$

v = 9.3244 × 10⁵ m/s

Also, 1 m = 0.001 km

5 0

3 years ago

A projectile is fired with speed v0 at an angle theta from the horizontal from the horizontal as shown in the figure.

irga5000 [103]

Answer:

v₀ = √(2gH/(sin²θ)) = (sin θ)√(2gH)

v₀ = √(gR/(sin2θ))

Explanation:

An image of the artillery officer, the hill and path of motionof the projectile is attached to this solution.

Given, R, H, g and θ (theta)

Using the equations of motion, we can get the initial velocity v₀

First of, we need to resolve this motion into the vertical and horizontal axis.

The horizontal component of the initial velocity, v₀ₓ = v₀ cos θ

Vertical component of the initial velocity, v₀ᵧ = v₀ sin θ

When the projectile reaches maximum height, Velocity at max height, vₕ = 0m/s

From equations of motion,

vₕ = v₀ᵧ - gt

0 = v₀ sinθ - gt

t = v₀ sinθ/g

This is the time taken to reach maximum height. The time take to comolete the toyal flight, T = 2t = (2v₀ sinθ)/g

The maximum height to be reached, H can be calculated from the equations of motion too

H = vₕt - 0.5gt² = 0 - 0.5g((v₀ sinθ)/g)²

H = (0.5g v₀² sin²θ)/g²

H = (v₀² sin²θ)/2g

The range, or horizontal distance to be covered by the projectile, R, will be calculated using the horizontal component of the initial Velocity, v₀ₓ = v₀ cos θ, this horizontal velocity is constant all through the motion, so, no acceleration in the horizontal direction.

R = v₀ₓT = (v₀ cos θ)((2v₀ sinθ)/g)

R = (v₀²(2cosθsinθ)/g)

2cosθsinθ = sin2θ

R = v₀²(sin2θ)/g

So, writing v₀ in terms of all the other parameters,

v₀ = √(2gH/(sin²θ)) = (sinθ)√(2gH

v₀ = √(gR/(sin2θ))