Answer: (a). E = 3.1656×10³⁴ √k/m
(b). f = 9.246 × 10¹² Hz
(c). Infrared region.
Explanation:
From Quantum Theory,
The energy of a proton is proportional to the frequency, from the equation;
E = hf
where E = energy in joules
h = planck's constant i.e. 6.626*10³⁴ Js
f = frequency
(a). from E = hf = 1 quanta
f = ω/2π
where ω = √k/m
consider 3 quanta of energy is lost;
E = 3hf = 3h/2π × √k/m
E = (3×6.626×10³⁴ / 2π) × √k/m
E = 3.1656×10³⁴ √k/m
(b). given from the question that K = 15 N/m
and mass M = 4 × 10⁻²⁶ kg
To get the frequency of the emitted photon,
Ephoton =hf = 3h/2π × √k/m (h cancels out)
f = 3h/2π × √k/m
f = 3h/2π × (√15 / 4 × 10⁻²⁶ )
f = 9.246 × 10¹² Hz
(c). The region of electromagnetic spectrum, the photon belongs to is the Infrared Spectrum because the frequency ranges from about 3 GHz to 400 THz in the electromagnetic spectrum.
Answer:
The velocity of flow is 10.0 m/s.
Explanation:
We shall use Manning's equation to calculate the velocity of flow
Velocity of flow by manning's equation is given by

where
n = manning's roughness coefficient
R = hydraulic radius
S = bed slope of the channel
We know that for an asphalt channel value of manning's roughness coefficient = 0.016
Applying values in the above equation we obtain velocity of flow as

Answer:
Tension in cable BE= 196.2 N
Reactions A and D both are 73.575 N
Explanation:
The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence
hence

Therefore, tension in the cable, 
Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then



Similarly,


Therefore, both reactions at A and D are 73.575 N
Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S = 
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M
Answer:
Explanation:
Run the code given in text file following instructions.