Physics: Kendra is flying a kite 35 feet in the air. The angle of depression from the kite to where Kendra holds the string is . If Kendra holds the string 3 feet off the ground, what is the distance from her hands to the kite? Round your answer to the nearest tenth.

The energy of individual photons of 450 nm light is 4.42 × 10⁻¹⁹ JouleFurther explanationThe term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:E = Energi of A Photon ( Joule )h = Planck s Constant ( 6.63 × 10⁻³⁴ Js )f = Frequency of Eletromagnetic Wave ( Hz )The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough...

Kendra is flying a kite 35 feet in the air. The

Ответ:

F00Dislife

19.01.2022

Physics

The distance from her hands to the kite is 60.4 feet.

Explanation:

Given that,

Kendra holds the string from the ground = 3 feet

Distance of kite from the ground = 35 feet

Suppose the angle is 32°.

According to the figure,

We need to calculate the distance from her hands to the kite

Using formula of angle

$\sin\theta=\dfrac{ED}{AE}$

Put the value into the formula

$\sin32=\dfrac{32}{H}$

$H=\dfrac{32}{\sin32}$

$H=60.4\ feet$

Hence, The distance from her hands to the kite is 60.4 feet.

Kendra is flying a kite 35 feet in the air. The angle of depression from the kite to where Kendra ho

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Ответ:

tifftiff22

03.02.2023

Physics

The energy of individual photons of 450 nm light is 4.42 × 10⁻¹⁹ Joule

Further explanation

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:

$\large {\boxed {E = h \times f}}$

E = Energi of A Photon ( Joule )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

f = Frequency of Eletromagnetic Wave ( Hz )

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

$\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}$

$\large {\boxed {E = qV + \Phi}}$

E = Energi of A Photon ( Joule )

m = Mass of an Electron ( kg )

v = Electron Release Speed ( m/s )

Ф = Work Function of Metal ( Joule )

q = Charge of an Electron ( Coulomb )

V = Stopping Potential ( Volt )

Let us now tackle the problem!

Given:

λ = 450 nm = 450 × 10⁻⁹ m

Unknown:

E = ?

Solution:

E = h f

$E = h \frac{c}{\lambda}$

$E = 6.63 \times 10^{-34} \frac{3 \times 10^8}{450 \times 10^{-9}}$

$\large {\boxed {E = 4.42 \times 10^{-19} ~ Joule } }$

Learn morePhotoelectric Effect : linkStatements about the Photoelectric Effect : linkRutherford model and Photoelecric Effect : linkPhotoelectric Threshold Wavelength : linkAnswer details

Grade: High School

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt

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Kendra is flying a kite 35 feet in the air. The angle of depression from the kite to where Kendra holds the string is . If Kendra holds the string 3 feet off the ground, what is the distance from her hands to the kite? Round your answer to the nearest tenth.

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