Answer: hello question b is incomplete attached below is the missing question
a) attached below
b) V = 0.336 ft/s
Explanation:
Elongation ( Xo) = 16/ 7 feet
mass attached to 4-foot spring = 16 pounds
medium has 9/2 times instanteous velocity
<u>a) Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s</u>
The motion is an underdamped motion because the value of β < Wo
Wo = 3.741 s^-1
attached below is a detailed solution of the question
A) experimental because he isn’t sure and is testing out
Answer:
Light of a shorter wavelength should be used.
Explanation:
This is studied in the phenomenon called photoelectric effect, in which light is able to release electrons from a metal, said electrons are called photoelectrons .
The experiments that have been carried out show that <u>increasing or decreasing the intensity of the light will not cause the photoelectrons to be emitted</u>, what will cause the photoelectrons to be emitted is to increase the frequency of the incident light.
And a higher frequency corresponds to a shorter wavelength according to the equation:
(where 
is frequency, 
the speed of light, and 
the wavelength)
So the answer is that the wavelength of the light must be shortened to cause the emission of electrones.
Answer:
a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s
Explanation:
It is angular momentum given by
L = r x p
Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum
One of the easiest ways to make this vector product is with the use of determinants
![{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]](https://tex.z-dn.net/?f=%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%26z%5C%5Cpx%26py%26pz%5Cend%7Barray%7D%5Cright%5D)
Let's apply this relationship to our case
Let's start by breaking down the speed
v₀ₓ = v₀ cosn 45
voy =v₀ sin 45
v₀ₓ = 9 cos 45
voy = 9 without 45
v₀ₓ = 6.36 m / s
voy = 6.36 m / s
a) at launch point r = 0 whereby L = 0
. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero
vfy² = voy²- 2 g y
y = voy² / 2g
y = (6.36)²/2 9.8
y = 2.06 m
Let's calculate the angular momentum
L= ![\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%260%5C%5Cpx%260%260%5Cend%7Barray%7D%5Cright%5D)
L = -px y k ^
L = - (m vox) (2.06) k ^
L = - 20 6.36 2.06 k ^
L = 262 k ^ Kg m² / s
The angular momentum is on the z axis
c) At the point of impact, at this point the height is zero and the position on the x-axis is the range
R = vo² sin 2θ / g
R = 9² sin (2 45) /9.8
R = 8.26 m
L = ![\left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%260%260%5C%5Cpx%26py%260%5Cend%7Barray%7D%5Cright%5D)
L = - x py k ^
L = - x m voy
L = - 8.26 20 6.36 k ^
L = 1020.7 k^ kg m² /s
Answer:
electromagnetic wave i think
Explanation:
