<span>Every 10s 5 waves; t1 = 2s for each wave
When v = 1.5m/s, 3 waves in 10s t2 = 10 / 3s
Calculating the frequency in first case f1 = 5 / 10 = 0.5
Calculating the frequency in second case f2 = 3 / 10 = 0.3
Using the Doppler formula f = (1-v/c) f0
For the formula f = f2, v = velocity of boat= 1.5 m/s, f0 = f1, c is velocity of wave
0.3 = 0.5 x (1 - 1.5/c) => 1.5/c = 1 - 0.6 => 1.5/c = 0.4 => c = 1.5/0.4
Velocity of the wave = 3.75 m/s</span>
Answer:
B
i am not fully sure but i believe its the most accurate its either A or B
The answer to the given question above is option C. In a classical conditioning experiment, you ring a bell but no longer provide dinner to your subject, a beagle and at last, the CS no longer evokes the CR so therefore, <span> you've accomplished behavioral EXTINCTION. Hope this helps.</span>
Answer:0.506 N
Explanation:
Given
Charge on first balloon 
Charge on second balloon is 
Distance between them 
Electrostatic Repulsive force is given by

Where K is constant



Answer
given,
constant speed of cart on right side = 2 m/s
diameter of nozzle = 50 mm = 0.05 m
discharge flow through nozzle = 0.04 m³
One-fourth of the discharge flows down the incline
three-fourths flows up the incline
Power = ?
Normal force i.e. Fn acting on the cart

v is the velocity of jet
Q = A V


v = 20.37 m/s
u be the speed of cart assuming it to be u = 2 m/s
angle angle of inclination be 60°
now,

F n = 2295 N
now force along x direction



Power of the cart
P = F x v
P = 1987.52 x 20.37
P = 40485 watt
P = 40.5 kW