What do most composites and plastics have in common?

Light with a wavelength range of 141–295 nm shines on a silicon surface in a photoelectric effect apparatus, and a reversing pot

Ksju [112]

Answer:

a) the longest wavelength of the light that will eject electrons from the silicon surface is 258.7891 nm

b) maximum kinetic energy will electrons reach the anode is 0.5098 eV

Explanation:

Given:

Wavelength range = 141-295 nm

Potential of 3.5 V

For the silicon, the work function is Φ = 4.8 eV = 7.68x10⁻¹⁹J

Questions:

a) What is the longest wavelength of the light that will eject electrons from the silicon surface, λ = ?

b) With what maximum kinetic energy will electrons reach the anode,

a) The longest wavelength that will eject electrons:

$\lambda =\frac{hc}{\phi }$

Here

h = Planck's constant = 6.625x10⁻³⁴J s

c = speed of light = 3x10⁸m/s

Substituting values:

$\lambda =\frac{6.625x10^{-34}*3x10^{8} }{7.68x10^{-19} } =2.588x10^{-7} m=258.7891nm$

b) The maximum kinetic energy (one electron):

$K=\frac{hc}{\lambda } -\phi =\frac{6.625x10^{-34}*3x10^{8} }{141x10^{-9} } -7.68x10^{-19} =6.416x10^{-19} J=4.0098eV$

Now, you need to calculate the potential difference:

$K'=eV$

Here

e = charge of electron = 1.6x10⁻¹⁹C

Substituting:

$K'=1.6x10^{-19} *3.5=5.6x10^{-19} J=3.5eV$

Now, the maximum kinetic energy of the electrons:

Kmax = 4.0098 - 3.5 = 0.5098 eV

6 0

4 years ago

10. Define and give 2 examples of kinetic energy. Be sure to include the two factors that make an object’s kinetic energy increa

Mademuasel [1]

You can describe kinetic energy and the potential energy of motion to get your answer.

7 0

4 years ago

What frequency is received by the ambulance after reflecting from a wall near the person watching the oncoming ambulance (as in

ElenaW [278]

Answer:

f=896Hz

Explanation:

Given data

Vs(speed of the ambulance)={(104 km/h)*(1000m*(1 h/3600)}=28.9m/s

f(frequency of the ambulance siren)=821 Hz

v(speed of sound)=345 m/s

Vo(speed of the observer)=0 m/s

To find

f(The ambulance is approaching the person)

Solution

From Doppler effect

$f^{i}=(\frac{v+v_{o} }{v-v_{s} })f$

As the ambulance approaches the we assign a positive sign for speed "vs" of the ambulance

So

$f^{i}=(\frac{v+0}{v-(+v_{s}) } )f\\f^{i}=(\frac{v}{v-v_{s} } )f$

Substitute the values from given data

$f^{i}=(\frac{345m/s}{345m/s-28.9m/s} )821Hz\\f^{i}=896Hz$

5 0

3 years ago

Beginning 145 miles directly south of the city of Hartville, a car travels due west. If the car is travelling at a speed of 42 m

ziro4ka [17]

Answer:

The rate of change of the distance is 14.89.

Explanation:

Given that,

Distance = 145 miles

Speed of car = 42 miles/hr

Distance covered by car = 55 miles

We need to calculate the the rate of change of the distance

According to figure,

Let OA is x, and AB is y.

Now, using Pythagorean theorem

$x^2=y^2+145^2$

On differentiating

$2x\dfrac{dx}{dt}=2y\dfrac{dy}{dt}$

$\dfrac{dx}{dt}=\dfrac{y}{x}\dfrac{dy}{dt}$

$\dfrac{dx}{dt}=\dfrac{55\times42}{\sqrt{55^2+145^2}}$

$\dfrac{dx}{dt}=14.89\ miles/hr$

Hence, The rate of change of the distance is 14.89.

8 0

4 years ago

On earth, you swing a simple pendulum in simple harmonic motion with a period of 1.6 seconds. what is the period of this same pe

polet [3.4K]

The period of a simple pendulum is given by
$T= 2 \pi \sqrt{ \frac{L}{g} }$
where
L is the pendulum length
g is the acceleration of gravity

If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
$T_e= 2 \pi \sqrt{ \frac{L}{g_e} }$
where $g_e$ is the acceleration of gravity on Earth, while the period of the pendulum on the Moon is
$T_m= 2 \pi \sqrt{ \frac{L}{g_m} }$
where $g_m$ is the acceleration of gravity on the Moon.

If we do the ratio of the two periods, we get
$\frac{T_m}{T_e} = \sqrt{ \frac{g_e}{g_m} }$
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write $g_e = 6 g_m$ and we can rewrite the previous ratio as
$\frac{T_m}{T_e} = \sqrt{ \frac{6 g_m}{g_m} }= \sqrt{6}$

so the period of the pendulum on the Moon is
$T_m = \sqrt{6} T_e = \sqrt{6} (1.6 s)=3.9 s$

8 0

4 years ago