Question 2

Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. Wha

devlian [24]

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

$K_{max}=\frac{hc}{\lambda}-W$

Here h is the Planck's constant, c is the speed of light, $\lambda$ is the wavelength of the light and W the work function of the element:

$W=\frac{hc}{\lambda}-K_{max}\\W=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{495*10^{-9}m}-0.5eV\\W=2.01eV$

Now, we calculate the wavelength for the new maximum kinetic energy:

$W+K_{max}=\frac{hc}{\lambda}\\\lambda=\frac{hc}{W+K_{max}}\\\lambda=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{2.01eV+1.5eV}\\\lambda=3.54*10^{-7}m=354*10^{-9}m=354nm$

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

8 0

4 years ago

Consider a block of mass m at rest on an inclined plane of angle theta. An acrobat of mass m_A is standing on the top corner of

shtirl [24]

Answer:

μ = tan θ

Explanation:

For this exercise let's use the translational equilibrium condition.

Let's set a datum with the x axis parallel to the plane and the y axis perpendicular to the plane.

Let's break down the weight of the block

sin θ = Wₓ / W

cos θ = W_y / W

Wₓ = W sin θ

W_y = W cos θ

The acrobat is vertically so his weight decomposition is

sin θ = = wₐₓ / wₐ

cos θ = wₐ_y / wₐ

wₐₓ = wₐ sin θ

wₐ_y = wₐ cos θ

let's write the equilibrium equations

Y axis

N- W_y - wₐ_y = 0

N = W cos θ + wₐ cos θ

X axis

Wₓ + wₐ_x - fr = 0

fr = W sin θ + wₐ sin θ

the friction force has the formula

fr = μ N

fr = μ (W cos θ + wₐ cos θ)

we substitute

μ (Mg cos θ + mg cos θ) = Mgsin θ + mg sin θ

μ = $\frac{(M +m) \ sin \ \theta }{(M +m) \ cos \ \theta }$

μ = tan θ

this is the minimum value of the coefficient of static friction for which the system is in equilibrium.

8 0

3 years ago

How much heat is released when 12.0 grams of helium gas condense at 2.17 K? The latent heat of vaporization of helium is 21 J/g.

uranmaximum [27]

To determine the heat released by the process of condensation, we simply multiply the amount of the gas that condensed to the latent heat of vaporization. We do as follows:

Heat released = 21 J/g (12.0 g ) = 252 J of heat released

8 0

3 years ago

Read 2 more answers

An 870 N firefighter, F_ff, stands on a ladder that is 8.00 m long and has a weight of 355 N, F_ladder. The weight of the ladder

castortr0y [4]

Answer:

Explanation:

Weight of the ladder is 355N

WL = 355N

The weight of the ladder acts at center of the ladder I.e at 4m from the bottom

Weight of firefighters is 870N

Wf = 870N

The fire fighter is at 6.3m from the bottom of the ladder.

N1 is the normal force exerted by the wall

N2 is the normal force exerted by the ground

Using Newton law

Check attachment

ΣFy = 0, since body is in equilibrium

N₂ - WL - Wf = 0

N₂ = WL + Wf

N₂ = 870 + 355

N₂ = 1225 N

This the normal force exerted by the ground on the wall.

Now to get N₁, let take moment about point A

So, before we take the moment we need to make sure that the forces are perpendicular to the plane(ladder), we need to resolve the weight of the ladder, firefighter and the normal of the wall to be perpendicular to the plane.

ΣMa = 0

Clockwise moment is equal to anti-clockwise moment

Moment Is the produce of force and perpendicular distance.

M = F×r

So,

WL•Cos50 × 4 + Wf•Cos50 × 6.3 —N₁•Sin50 × 8 = 0

355•Cos50 × 4 + 870•Cos50 × 6.3 =

N₁•Sin50 × 8

1825.52 + 3523.12 = 6.13N₁

6.13N₁ = 5348.64

N₁ = 5348.64/6.13

N₁ = 872.54 N.

The normal force exerted by the wall is 872.54N

5 0

3 years ago

Why is the surface salinity of the ocean higher in the subtropics than in equatorial regions?

larisa [96]

Evaporation is more contrasted with the rainfall which is less in the correlation of tropical locale. In tropical locale, protection is very little in light of mists. Alongside temperature, it is the main consideration in adding to changes in the thickness of seawater and consequently sea flow. Saltiness is the way to understanding the worldwide water cycle. 97% of the Earth's free water dwells in the seas. The water cycle is commanded by precipitation and evaporation.

7 0

3 years ago