Answer:
Detailed step wise solution is attached below
Explanation:
(a) wavelength of the initial note 2.34 meters
(b) wavelength of the final note 0.389 meters
(d) pressure amplitude of the final note 0.09 Pa
(e) displacement amplitude of the initial note 4.78*10^(-7) meters
(f) displacement amplitude of the final note 3.95*10^(-8) meters
Answer:
10.2 Watt
Explanation:
= number of turns in flat coil = 160
= area = 0.20 m²
B₀= initial magnetic field = 0.40 T
= final magnetic field = - 0.40 T
Change in magnetic field is given as
ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T
= time taken for the magnetic field to change = 2.0 s
Induced emf is given as


= 12.8 volts
= Resistance of the coil = 16 Ω
Power is given as


= 10.2 Watt
Answer:
Ae/A* = 1.115
Explanation:
Let the reservoir pressure be 
Let the exit pressure be 
Ratio of reservoir pressure and exit pressure

= 3.182
For the above value of pressure ratio
Obtain the area ratio from the isentropic flow table
Ae/A* = 1.115
The value of pressure ratio is Ae/A* = 1.115
Answer:
Tension, T = 0.0115 N
Explanation:
Given that,
Mass of the plastic ball, m = 1.1 g
Length of the string, l = 56 cm
A charged rod brought near the ball exerts a horizontal electrical force F on it, causing the ball to swing out to a 21.0 degree angle and remain there. According to attached figure :

T is tension in the string

So, the tension in the string is 0.0115 N.
Answer:
tsunami
Explanation:
A tsunami is a series of waves that come ashore at interval 10-45 mintues