Answer:the resistance decrease
Explanation:
Answer:
B) waves speed up
C) waves bend away from the normal
Explanation:
The index of refraction of a material is the ratio between the speed of light in a vacuum and the speed of light in that medium:

where
c is the speed of light in a vacuum
v is the speed of light in the medium
We can re-arrange this equation as:

So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is
B) waves speed up
Moreover, when light enters a medium bends according to Snell's Law:

where
are the index of refraction of the 1st and 2nd medium
are the angles made by the incident ray and refracted ray with the normal to the interface
We can rewrite the equation as

So we see that if the index of refraction of the second medium is lower (
), then the ratio
is larger than 1, so the angle of refraction is larger than the angle of incidence:

This means that the wave will bend away from the normal. So the other correct option is
C) waves bend away from the normal
Answer:
Time period for Simple pendulum, 
Explanation:
The Simple Pendulum
Consider a small bob of mass
is tied to extensible string of length
that is fixed to rigid support. The bob is oscillating in the plane about verticle.
Let
is the angle made by string with vertical during oscillation.
Vertical component of the force on bob,
Negative sign shows that its opposing the motion of bob.
Taking
as very small angle then, 
Let
is the displacement made by bob from its mean position ,
then, 
so,
........(1)
Since, pendulum is in hormonic motion,
as we know, 
where
is the constant and 
.........(2)
From equation (1) and (2)


Since, 


Answer:
Explanation:
Given a square side loop of length 10cm
L=10cm=0.1m
Then, Area=L²
Area=0.1²
Area=0.01m²
Given that, frequency=60Hz
And magnetic field B=0.8T
a. Flux Φ
Flux is given as
Φ=BA Sin(wt)
w=2πf
Φ=BA Sin(2πft)
Φ=0.8×0.01 Sin(2×π×60t)
Φ=0.008Sin(120πt) Weber
b. EMF in loop
Emf is given as
EMF= -N dΦ/dt
Where N is number of turns
Φ=0.008Sin(120πt)
dΦ/dt= 0.008×120Cos(120πt)
dΦ/dt= 0.96Cos(120πt)
Emf=-NdΦ/dt
Emf=-0.96NCos(120πt). Volts
c. Current induced for a resistance of 1ohms
From ohms law, V=iR
Therefore, Emf=iR
i=EMF/R
i=-0.96NCos(120πt) / 1
i=-0.96NCos(120πt) Ampere
d. Power delivered to the loop
Power is given as
P=IV
P=-0.96NCos(120πt)•-0.96NCos(120πt)
P=0.92N²Cos²(120πt) Watt
e. Torque
Torque is given as
τ=iL²B
τ=-0.96NCos(120πt)•0.1²×0.8
τ=-0.00768NCos(120πt) Nm
Answer:
Force, F = 44 N
Explanation:
Given that,
Initial speed of the football, u = 0
Final speed, v = 15 m/s
The time of contact of the ball, t = 0.15 s
The mass of football, m = 0.44 kg
We need to find the average force exerted on the ball. It is given by the formula as :

So, the average force exerted on the ball is 44 N. Hence, this is the required solution.