Answer:
37.42 m/s
Explanation:
We know that apparent frequency, 
is given by
where f is the given frequency in this case 392, V is the speed of sound in air which is given as 343 and 
is the speed of car which is unknown, \bar f is given as 440 Hz
Answer:
(A) Q = 321.1C (B) I = 42.8A
Explanation:
(a)Given I = 55A−(0.65A/s2)t²
I = dQ/dt
dQ = I×dt
To get an expression for Q we integrate with respect to t.
So Q = ∫I×dt =∫[55−(0.65)t²]dt
Q = [55t – 0.65/3×t³]
Q between t=0 and t= 7.5s
Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]
Q = 321.1C
(b) For a constant current I in the same time interval
I = Q/t = 321.1/7.5 = 42.8A.
Answer:
The answer is given in the attachment
Explanation:
Answer:
When you are performing spike it's most effective to strike the ball from the right or left side at a sharp downward angle. Whether you are spiking the ball from the right or left front position, position yourself behind the 10-foot line (attack line), which is the line that is about four steps away from the net.
Answer:
V = 9.682 × 10^(-6) V
Explanation:
Given data
thick = 190 µm
wide = 4.20 mm
magnitude B = 0.78 T
current i = 32 A
to find out
Calculate V
solution
we know v formula that is
V = magnitude× current / (no of charge carriers ×thickness × e
here we know that number of charge carriers/unit volume for copper = 8.47 x 10^28 electrons/m³
so put all value we get
V = magnitude× current / (no of charge carriers ×thickness × e
V = 0.78 × 32 / (8.47 x 10^28 × 190 × 1.602 x 10^(-19)
V = 9.682 × 10^(-6) V
