y axis:NandW and also f x axis:T and F T away from car.
The intensity of the electromagnetic wave which travels in an insulating magnetic material in a medium is 5.766×10⁻⁸ W/m².
The intensity of a wave is the total power delivered per unit area. It can be given as,
It can also be given as,
Here, () is relative permeability, () is physical constant, (k) is dielectric constant, (E) is the amplitude of electric field, and is the permittivity of free space.
Here, the electromagnetic wave with frequency 65.0hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency.
As the electric field has amplitude 7.20×10−3v/m. Thus, put the values in the above formula to find the intensity as,
Hence, the intensity of the electromagnetic wave which travels in an insulating magnetic material in a medium is 5.766×10⁻⁸ W/m².
Learn more about the intensity of the wave here;
Answer:
In parallel combination, the capacity of each capacitor is 11 F.
In series combination, the capacity of each capacitor is 44 F.
Explanation:
Let there are two capacitors each of capacitance C.
When they are connected in parallel:
In parallel combination, the effective capacitance is Cp.
Cp = C1 + C2 = C + C
22 = 2 C
C = 11 F
When they are connected in series:
In parallel combination, the effective capacitance is Cs.
1 / Cs = 1 / C1 + 1 / C2 = 1 / C + 1 / C = 2 / C
1 / 22 = 2 / C
C = 44 F
Answer: wavelength =3.52m
Explanation:
,λ=c/μ
where c=speed of the light,λ=wave length, μ=frequncy
c=3x10^8m/s
And
μ=83.5/MHz =85.3x10^6Hz==85.3x10^6Hz=
=85.3x10^6s-1
λ=c/μ
=3x10^8m/s/85.3x10^6s-1
=3.51699883
=3.52m
c) cubic equation
d) a higher order equation
Answer:
a) Linear equation
Explanation:
Definition of acceleration
if a=constant and we integrate the last equation
So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.
Answer:
U/U₀ = 2
(factor of 2 i.e U = 2U₀)
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected
Explanation:
Energy stored in a capacitor can be expressed as;
U = 0.5CV^2 = Q^2/2C
And
C = ε₀ A/d
Where
C = capacitance
V = potential difference
Q = charge
A = Area of plates
d = distance between plates
So
U = Q^2/2C = dQ^2/2ε₀ A
The initial energy of the capacitor at d = d₀ is
U₀ = Q^2/2C = d₀Q^2/2ε₀ A ....1
When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.
The final energy stored in the capacitor at d = 2d₀ is
U = 2d₀Q^2/2ε₀ A ...2
The factor U/U₀ can be derived by substituting equation 1 and 2
U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )
Simplifying we have;
U/U₀ = 2
U = 2U₀
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.
Answer:
5.05 m/s
Explanation:
The distance from the bottom of his feet to his center of mass is (when is hanging at rest) is 2.1 - 1.3 = 0.8 m. Assume he keeps the posture, as soon as his feet touches the ground, his center of mass is 0.8 m above the ground. This would mean that he has traveled a distance of 2.1 - 0.8 = 1.3 m vertically. Using the law of energy conservation for potential and kinetic energy, also let the ground be ground 0 for potential energy, we have the following mechanical conservation energy:
Since he was hanging at rest, his initial kinetic energy at H = 2.1m must be 0. Let g = 9.81m/s2 and m be his mass, we can calculate for his velocity v at h = 0.8 m. First start by dividing both sides by m