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emmainna [20.7K]
2 years ago
5

g he work functions of gold, aluminum, and cesium are 5.1 eV, 4.1 eV, and 2.1 eV, respectively. If light of a particular frequen

cy causes photoelectrons to be emitted when the light is incident on an aluminum surface, explain if we know whether this means that photoelectrons are emitted from a gold surface or a cesium surface when the light is incident on those surfaces.
Physics
1 answer:
faust18 [17]2 years ago
3 0

Answer:

Explanation:

Work function of aluminium  is  4.1 eV .

Therefor light of minimum energy of 4.1 eV will be required to cause the emission of electrons.

If we use this frequency of light or energy of light  to effect ejection of photoelectrons from cesium , it will be easily   ejected because energy requirement for this metal is lower ( work function is lower 2.1 eV ). than the energy of photon falling on it.

If we use this frequency of light or energy of light  to effect ejection of photoelectrons from aluminium  , it will be not  ejected because energy requirement for this metal is high ( work function is higher 5.1 eV ) . Even if we increase the intensity of light , electrons will not be ejected because energy of each photon depends upon the frequency .

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There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co
mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

q_1=q_2=2.06\mu C=2.06\times 10^{-6} C

q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force,F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2

F_1=F_3=F

F_{net}=\sqrt{F^2+F^2+0}-F_2

F_{net}=\sqrt 2F-F_2

F=\frac{kq^2}{d^2}

F_2=\frac{Kq^2}{2d^2}

F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

Where k=9\times 10^9

F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})

F_{net}=0.208 N

3 0
3 years ago
You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
3 years ago
Which rf characteristic best determines the range of a 2.4 ghz ism signal?
Rainbow [258]
<span>The radio frequency characteristic that best determines the range of a 2.4 GHz ism signal is the wavelength.

This frequency can be used in WiFi and can reach up to 46 meters when indoors and about 92 meters when outdoors.
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3 0
3 years ago
A car mass of 1.2 x 10 kilograms starts from rest and attains a speed of 20 meters/seconds in 5 seconds. What net force acted on
lorasvet [3.4K]

Net force on the car=F=4.8 x 10³ N

Explanation:

mass of car= 1.2 x 10³ Kg

initial velocity= Vi=0

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time = t= 5 s

Using kinematic equation,

Vf= Vi + at

20= 0 + a (5)

5 a=20

a= 20/5

a= 4 m/s²

Now force is given by F = ma

F= 1.2 x 10³ (4)

F=4.8 x 10³ N

7 0
3 years ago
A wave travels at a constant speed.How does the frequency change if the wavelength is reduced by a factor of 3 The frequency dec
nata0808 [166]

Answer:

The frequency increases by a factor of 3.

Explanation:

The relation between speed, wavelength and frequency of a wave is given by :

v=f\lambda

or

f\propto \dfrac{1}{\lambda}

A wave travels at a constant speed. If the wavelength is reduced by a factor of 3, it would mean that the frequency increases by a factor of 3 because there is an inverse relationship between wavelength and frequency.

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