Answer:
1.52 nm
Explanation:
Using the De Broglie wavelength equation,
λ = h/p where λ = wavelength associated with electron, h = Planck's constant = 6.63 × 10⁻³⁴ Js and p = momentum of electron = mv where m = mass of electron = 9.1 × 10⁻³¹ kg and v = velocity of electron = 4.8 × 10⁵ m/s
So, λ = h/p
λ = h/mv
substituting the values of the variables into the equation, we have
λ = h/mv
λ = 6.63 × 10⁻³⁴ Js/(9.1 × 10⁻³¹ kg × 4.8 × 10⁵ m/s)
λ = 6.63 × 10⁻³⁴ Js/(43.68 × 10⁻²⁶ kgm/s)
λ = 0.1518 × 10⁻⁸ m
λ = 1.518 × 10⁻⁹ m
λ = 1.518 nm
λ ≅ 1.52 nm
The meters per second
+1t a second / 2t
Answer:
1. t = 0.0819s
2. W = 0.25N
3. n = 36
4. y(x , t)= Acos[172x + 2730t]
Explanation:
1) The given equation is
The relationship between velocity and propagation constant is
v = 15.87m/s
Time taken, 
t = 0.0819s
2)
The velocity of transverse wave is given by
mass of string is calculated thus
mg = 0.0125N
m = 0.00128kg
0.25N
3)
The propagation constant k is
hence
0.036 m
No of wavelengths, n is
n = 36
4)
The equation of wave travelling down the string is
![y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]](https://tex.z-dn.net/?f=y%28x%2C%20t%29%3DAcos%5Bkx%20-wt%5D%5C%5C%5C%5Cbecomes%5C%5C%5C%5Cy%28x%20%2C%20t%29%3D%20Acos%5B%28172%20rad.m%29x%20%2B%20%282730%20rad.s%29t%5D)
![without, unit\\\\y(x , t)= Acos[172x + 2730t]](https://tex.z-dn.net/?f=without%2C%20unit%5C%5C%5C%5Cy%28x%20%2C%20t%29%3D%20Acos%5B172x%20%2B%202730t%5D)
Answer:
Low Potential energy and High Kinetic energy
Explanation:
Hope this helps and have a good day! Apologies if it's wrong.<3
