Answer:
length of the ladder is 13.47 feet
base of wall to latter distance 6.10 feet
angle between ladder and the wall is 26.95°
Explanation:
given data
height h = 12 feet
angle 63°
to find out
length of the ladder ( L) and length of wall to ladder ( A) and angle between ladder and the wall
solution
we consider here angle between base of wall and floor is right angle
we apply here trigonometry rule that is
sin63 = h/L
put here value
L = 12 / sin63
L = 13.47
so length of the ladder is 13.47 feet
and
we can say
tan 63 = h / A
put here value
A = 12 / tan63
A = 6.10
so base of wall to latter distance 6.10 feet
and
we say here
tanθ = 6.10 / 12
θ = 26.95°
so angle between ladder and the wall is 26.95°
Answer:
Technician A is correct
Explanation:
The best approach to solve the problem is that of technician A. using a fluorescent die is the easiest and most efficient way to trace leaks with unknown sources. The fluorescent die will simply illuminate the path to the leaking spot in the engine of the car, without any need for much speculations. This makes this method a sure approach.
However, Technician B's approach still has a lot of assumptions factored into the methodology, and would not work properly. It will still require the painstaking attempts trying to make guesses where the oil leak is coming from, which will lead to wastage of time and energy.
This makes Technician A have the right approach to solving the problem
Does it give you options?
Solar system is nested nearly 2/3 of the way from the center of the galaxy to the outskirt of the galactic disc.
Answer:
(a) False
(b) True
(c) True
(d) True
(e) True
(f) True
Explanation:
(a) Maxwell's equations not only applies to constant fields but it applies to both the fields, i.e., Time variant field as well as Time Invariant field.
(b) We make use of the Modified form of the Ampere's law and Faraday's Law to derive the wave equation.
(c) Electromagnetic waves contains both the electric and magnetic fields and these fields oscillates at an angle of 
to the direction of wave propagation.
(d) In free space both the electric and magnetic fields are in phase while considering electromagnetic waves.
(e) In free space or vacuum, the expression for the speed of light in terms of electric and magnetic field is given as:
Thus the ratio of the magnitudes of the electric and magnetic field vectors are equal to the speed of light in free space.
(f) In free space or in vacuum the energy density of the electromagnetic wave is divided equally in both the fields and hence are equal.
