Answer:
In a transverse wave the particles of the medium move perpendicular to the wave's direction of travel.
In a longitudinal wave, the particles of the medium move parallel to the wave's direction of travel.
Answer:
Explanation:
Given that,
Frequency f = 5 kHz = 5000Hz
Voltage V=45V
Current In inductor I = 65mA
I = 65 × 10^-3 = 0.065A
We want to find inductance L
We know that the reactive inductance cam be given as
XL = 2πFL
Where
XL is reactive inductance
F is frequency
L is inductance
Then,
L = XL/2πF
From ohms law
V = IR
We can calculate the receive reactance of the inductor
V = I•XL
Then, XL = V/I
XL = 45/0.065
XL = 692.31 ohms
Then,
L = XL/2πF
L = 692.31/(2π×5000)
L = 0.02204 H
Then, L = 22.04 mH
The inductance of the inductor is 22.04mH
Answer:
La Niña, because the jet stream is displaced northwards
Explanation:
La Nina is known to be a cool phase that affects the average ocean temperature. During this period the surface wind becomes stronger and there is lesser rainfall over the areas of the central tropical Pacific to which the Southern United States belongs.
Hence, when the southern United States experiences periodic occurrences of severe cold and drought conditions, the weather phenomenon most likely causes this and why is "La Niña, because the jet stream is displaced northwards."
Answer:
The current in the wire under the influence of the force is 216.033 A
Solution:
According to the question:
Length of the wire, l = 0.676 m
Magnetic field of the Earth, 
Forces experienced by the wire, 
Also, we know that the force in a magnetic field is given by:
I = 216.033 A
Answer: a) E= 6.63x10^-19J
E= 3.97×10^2KJ/mol
b) E = 3.31×10^-19J
E= 18.8×10^4 KJ/mol
C) E = 1.32×10^-33J
E= 8.01×10^-10KJ/mol
Explanation:
a) E = h ×f
h= planks constant= 6.626×10^-34
E=(6.626×10^-34)×(1.0×10^15)
E=6.63×10^-19J
1mole =6.02×10^23
E=( 6.63×10^-19)×(6.02×10^23)
E=3.97×10^2KJ/mol
b) E =(6.626×10^-34)/(1.0×10^15)
E=3.13×10^-19J
E= 3.13×10^-19) ×(6.02×10^23)
E= 18.8×10^3KJ/MOL
c) E= (6.626×10^-34) /0.5
E= 1.33×10^-33J
E= (1.33×10^-33) ×(6.02×10^23)
E= 8.01×10^-10KJ/mol
