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.
Answer:
Explanation:
i = Imax sin2πft
given i = 180 , Imax = 200 , f = 50 , t = ?
Put the give values in the equation above
180 = 200 sin 2πft
sin 2πft = .9
sin2π x 50t = .9
sin 360 x 50 t = sin ( 360n + 64 )
360 x 50 t = 360n + 64
360 x 50 t = 64 , ( putting n = 0 for least value of t )
18000 t = 64
t = 3.55 ms .
When the body touches the ground two types of Forces will be generated. The Force product of the weight and the Normal Force. This is basically explained in Newton's third law in which we have that for every action there must also be a reaction. If the Force of the weight is pointing towards the earth, the reaction Force of the block will be opposite, that is, upwards and will be equivalent to its weight:
F = mg
Where,
m = mass
g = Gravitational acceleration
F = 5*9.8
F = 49N
Therefore the correct answer is E.
Answer:
a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).
b. third harmonic
c. to calculate frequency , we compare with general wave equation
y(x,t)=Acos(kx+ωt)
from ωt=742t
ω=742
ω=2*pi*f
742/2*pi
f=118.09Hz
Explanation:
A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 m and a mass of 3.38 grams. Assume that the ends of the rope are held fixed and that there is both this traveling wave and the reflected wave traveling in the opposite direction.
A) What is the wavefunction y(x,t) for the standing wave that is produced?
B) In which harmonic is the standing wave oscillating?
C) What is the frequency of the fundamental oscillation?
a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).
b. lambda=2L/n
when comparing the wave equation with the general wave equation , we get the wavelength to be
2*pi*x/lambda=6.98x
lambda=0.9m
we use the equation
lambda=2L/n
n=number of harmonics
L=length of string
0.9=2(1.35)/n
n=2.7/0.9
n=3
third harmonic
c. to calculate frequency , we compare with general wave equation
y(x,t)=Acos(kx+ωt)
from ωt=742t
ω=742
ω=2*pi*f
742/2*pi
f=118.09Hz
Answer:
Explanation:
position of centre of mass of door from surface of water
= 10 + 1.1 / 2
= 10.55 m
Pressure on centre of mass
atmospheric pressure + pressure due to water column
10 ⁵ + hdg
= 10⁵ + 10.55 x 1000 x 9.8
= 2.0339 x 10⁵ Pa
the net force acting on the door (normal to its surface)
= pressure at the centre x area of the door
= .9 x 1.1 x 2.0339 x 10⁵
= 2.01356 x 10⁵ N
pressure centre will be at 10.55 m below the surface.
When the car is filled with air or it is filled with water , in both the cases pressure centre will lie at the centre of the car .
