Answer:
Electromagnetic wave are waves formed as a result of the oscillatory activities involving the electric and the magnetic field.
However in an Electromagnetic wave, the electric field and magnetic field carry equal amounts of energy and the magnitude of the electric field is directly proportional to the magnitude of the magnetic field. This direct proportionality gives rise to the speed of light being the constant between the two fields.
When the electric field is doubled then an equal action is to be set for the magnetic field so it doesn’t deviate from its main functions and characteristics.
Potential energy is the answer
Answer:5101.35v
Explanation:
Radius of gold nucleus=7.3×10-15m and a charge of +79e
Q= 79e
e=1.6×10^-19
q= +2e
The nucleus is considered as the point charge where the potential energy between the charges are
U = 1/(4×3.142×Eo)×(qQ)/r
Where r is distance between the charges and the nucleus
r=R+d
V=U/q
U= 1/(4×3.142×Eo)×Q/r
V= 1/(4×3.142×Eo)×Q/(r+d)
9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)
V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)
V= 9×10^9×(5.67×10^-14)= 5101.35v
Answer:
Explanation:
Physics gets involved in your daily life right from you wake up in the morning. The buzzing sound of an alarm clock helps you wake up in the morning as per your schedule. The sound is something that you can't see, but hear or experience. Physics studies the origin, propagation, and properties of sound
Answer:
The upper limit on the flow rate = 39.46 ft³/hr
Explanation:
Using Ergun Equation to calculate the pressure drop across packed bed;
we have:

where;
L = length of the bed
= viscosity
U = superficial velocity
= void fraction
dp = equivalent spherical diameter of bed material (m)
= liquid density (kg/m³)
However, since U ∝ Q and all parameters are constant ; we can write our equation to be :
ΔP = AQ + BQ²
where;
ΔP = pressure drop
Q = flow rate
Given that:
9.6 = A12 + B12²
Then
12A + 144B = 9.6 -------------- equation (1)
24A + 576B = 24.1 --------------- equation (2)
Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So
288 B = 4.9
B = 0.017014
From equation (1)
12A + 144B = 9.6
12A + 144(0.017014) = 9.6
12 A = 9.6 - 144(0.017014)

A = 0.5958
Thus;
ΔP = AQ + BQ²
Given that ΔP = 50 psi
Then
50 = 0.5958 Q + 0.017014 Q²
Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;
Q² + 35.02Q - 2938.8 = 0
Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;
Q = 39.46 ft³/hr