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3 years ago

A beam of ultraviolet radiation, with frequency

1 answer:

6 0

To develop this problem, we will apply Einstein's relationship which is in charge of the work done with the kinetic energy of the body versus the total energy of the system.

The energy can be calculated as

$E = hf$

Here,

h = Planck's Constant

f = Frequency

Our values are given as,

$f = 2.78*10^{15} Hz ,$

$W = 3.80 eV$

Therefore the Energy is

$E = hf$

$E = 6.625*10^{-34}J\cdot s(2.78*10^{15}Hz)$

$E = 1.84*10^{-18}J \rightarrow 1 J = 6.242*10^{18}eV$

Then,

$E = 11.48eV$

Applying the Einstein Relation we have that

$E = W+KE$

$KE = E -W$

$KE = 11.48eV-3.80 eV$

$KE = 7.68eV$

Therefore the maximum kinetic energy for an electron dislodged fromthe surface by the radiation is 7.68eV

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What property do the following elements have in common? Li, C, and F A) They are poor conductors of electricity. B) Each element

Aloiza [94]

Answer: Option (D) is the correct answer.

Explanation:

The given elements Li, C and F are all second period elements. So, when we move from left to right across a period then there occurs increase in number of valence electrons as there occurs increase in total number of electrons.

So, it means more electrons are added to the same energy level.

Thus, we can conclude that a property of valence electrons for each element is located in the same energy level is common in the given elements.

7 0

3 years ago

Read 2 more answers

The PV system is operating in a location where the annual average daily incident solar energy (the insolation) on the array equa

vlabodo [156]

The average amount of solar energy incident on the PV per day is 10000 kWh/day.

<h3>Equation</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Let the PV array has an area equal to 50 square meters. Hence:

Average amount of solar energy incident on the PV per day = 200 kWh/m²/day * 50 m² = 10000 kWh/day.

Find out more on Equation at: brainly.com/question/2972832

5 0

2 years ago

a ball is projected horizontally with a velocity of 5 m per second from the top of a building 19.6 m high how long will the ball

zepelin [54]

Answer:

1.98s

Explanation:

The time taken to hit the ground is given by

h=ut+ 1/2 at^2

but u =0

so we have

h=1/2at^2

making t the subject

t=√2h/g

√2×19.6/10

1.98s

8 0

3 years ago

If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m

AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

We are given $g,v_{i},h_{i},h_{f}$ .

We can now solve for the final velocity.

${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

${v_{f}=\sqrt{2*9.81*2.41$

${v_{f}=\sqrt{47.2842$

$v_f=6.88$

7 0

1 year ago

9) An object with a height of 18 cm is placed in front of a converging lens. The image has a

vovikov84 [41]

Answer:

Explanation:

a) Magnification = image height / object height = -9 / 18 = -0.5

b) Magnification = - image distance / object distance = -0.5

so image distance = 0.5 object distance

1/focal length = 1/image distance + 1/object distance

1/6 = 1/(0.5 object distance) + 1/object distance

object distance = 18.0 cm

c) Image appears behind the lens.

6 0

3 years ago

Read 2 more answers