90.2 m/s. A rock dropped from rest after it has falling for 9.2 seconds will have a velocity of 90.2m/s.
The bodies left in free fall increase their speed (downwards) by 9.8 m/s² every second. The acceleration of gravity is the same for all objects and is independent of the masses of these. In the free fall the air resistance is not taken into account.
v = g*t
v = (9.8m/s²)(9.2s) = 90.2 m/s
Answer:
Explanation:
Area of square loop = L²
Flux Φ = area x magnetic field
= L²B
Frequency = f
angular velocity ω = 2πf
a )
Let at time t = 0 , the magnetic field is making 90 degree with the face of the loop
flux through loop = L²B
After time t , coil will turn by angle ω t = 2πft
Flux through the loop = L²B cosω t
Φ (t) = L²B cosω t
= L²B cos2πft
b )
emf induced e
= - d/dt [Φ (t)]
= - d/dt [ L²B cosω t]
= L²B ω sinω t
= L²B 2πf sin2πft
c )
current = e / R
(L²B ω/ R ) sinω t
Power delivered
P(t) = VI ,
VOLT X CURRENT
= AB ω sinω t X ( AB ω/ R ) sinω t
= L⁴B² 4π²f²/R sin²2πft
e )
torque = MB sinω t
τ(t) = i(L²B ) sinω t
= (L²B ω/ R ) sinω t x (L²B ) sinω t
= (L²B )²ω/ R sin²ω t
= (L²B )² 2πf/ R sin²2πft
As per the question the wavelength of the microwave is given as 3.52 mm.
we are asked to calculate the frequency of the wave.
we know that microwave is a electromagnetic wave.
As per Clark Maxwell's electromagnetic theory ,every electromagnetic wave moves with a velocity equal to the velocity of light in vacuum and that is equal to 3×10^8 m/s.
From the equation of the wave,we know that velocity of wave is the product of frequency and wavelength.
Mathematically wave velocity
where f is the frequency of the wave and
is the wavelength.
As per the question 

Here 
Hence frequency of the wave 


Here Hertz [Hz] is the unit of frequency.
Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA =
/ ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.