Answer:
The induced current is 0.084 A
Explanation:
the area given by the exercise is
A = 200 cm^2 = 200x10^-4 m^2
R = 5 Ω
N = 7 turns
The formula of the emf induced according to Faraday's law is equal to:
ε = (-N * dφ)/dt = (N*(b2-b1)*A)/dt
Replacing values:
ε = (7*(38 - 14) * (200x10^-4))/8x10^-3 = 0.42 V
the induced current is equal to:
I = ε /R = 0.42/5 = 0.084 A
Combine all of the x's on one side of the equation and then finish the problem!
Answer:
Yes, ultraviolet light can turn a rubber into solid due to prolong exposure.
Explanation:
A rubber is a material with an elastic property, causing it to be deform by an external force but takes its shape when the force is removed. Light is an electromagnetic wave which causes the sensation of vision. It transfers energy to a medium during propagation through the medium.
Generally, most light do not cause hardness of a rubber. But an ultraviolet light can cause rubber to become solid over a period of time. This is possible if there is a prolong exposure of the rubber, and because of the evaporation of volatiles in the polymer material. Ultraviolet light are known to cause a rubber to become solid.
Answer:
The torque on the wrench is 4.188 Nm
Explanation:
Let r = xi + yj where is the distance of the applied force to the origin.
Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,
r = 0.18i + 0.055j
The applied force f = 88i - 23j
The torque τ = r × F
So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j
= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j
= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0 since i × i = 0, j × j = 0, i × j = k and j × i = -k
= 0 - 4.14k + 0.0484(-k) + 0
= -4.14k - 0.0484k
= -4.1884k Nm
≅ -4.188k Nm
So, the torque on the wrench is 4.188 Nm
Answer:
Explanation:
Two frequencies with magnitude 150 Hz and 750 Hz are given
For Pipe open at both sides
fundamental frequency is 150 Hz as it is smaller
frequency of pipe is given by
where L=length of Pipe
v=velocity of sound
for n=1
and f=750 is for n=5
thus there are three resonance frequencies for n=2,3 and 4
For Pipe closed at one end
frequency is given by
for n=0
for n=2
Thus there is one additional resonance corresponding to n=1 , between 
and 
