Answer:
a. Wavelength = λ = 20 cm
b. Next distance of maximum intensity will be 40 cm
Explanation:
a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.
Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.
This we get the equation:
(nλ + λ/2) - nλ = 30-20
λ/2 = 10
λ = 20 cm
b. at what distance, sound intensity will be maximum again.
For next point calculation for maximum sound intensity, the path difference must be increased (n+1) λ. The distance must increase by λ/2 from the point of zero intensity.
= 30 + λ/2
= 30 + 20/2
=30+10
=40 cm
Power is the energy in a system per time. It will have units of Watts which is equal to joules per second. It can be expressed as:
P = E / t
where E = Force x distance
P = Fd / t
t = Fd / P
t = 8 (9.72) / 3.0
t = 25.92 s
Answer:

Explanation:
= normal force acting on the coin
Normal force in the upward direction balances the weight of the coin, hence

= frequency of rotation
Angular velocity of turntable is hence given as

= distance from the axis of rotation
= minimum coefficient of static friction
static frictional force is given as

The static frictional force provides the necessary centripetal force , hence
Centripetal force = Static frictional force

Answer:
hello your question is incomplete attached below is the complete question
answer :
a) I1 = I2
b) J1 > J2
c) E 1 > E2
d) ( vd1 ) > ( vd2 )
Explanation:
a) The currents in the two segments are the same i.e. I1 = I2 and this is because the segments are connected in series
b) Comparing the current densities J1 and J2 in the two segments
note : current density ∝ 1 / area
The area of the second segment is > the area of first segment therefore
J1 > J2
J1 ( current density of first segment )
J2 ( current density of second segment )
c) Comparing the electric field strengths E1 and E2
note : electric field strength ∝ current density
since current density of first segment is > current density of second segment and conductivity of the materials are the same hence
E 1 > E2
d) Comparing the drift speeds Vd1 and Vd2
( vd1 ) > ( vd2 )
this because ; vd ∝ current density
Answer:
<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>
Explanation:
<em>Frequency</em> is defined as number of oscillation per second ie back and forth swings done in single second.
Here it is given that the object oscillates 20 times in 10 seconds.
So f =
= 2Hz
The <em>time period</em> is defined as time taken by the object to complete one full oscillation.
T = 
T=
=0.5 s
<em>Thus the object has frequency of 2 Hz and time period of 0.5 s.</em>