Answer:
8, 8 W
Explanation:
The useful power of 1 Light Emitting Diode is
Total power required is 1.6 W
Number of Light Emitting Diodes would be
The number of Light Emitting Diodes is 8.
Power would be
The power that is required to run the Light Emitting Diodes is 8 W
<span>The speed of longitudinal waves, S, in a thin rod = âšYoung modulus / density , where Y is in N/m^2.
So, S = âšYoung modulus/ density. Squaring both sides, we have, S^2 = Young Modulus/ density.
So, Young Modulus = S^2 * density; where S is the speed of the longitudinal wave.
Then Substiting into the eqn we have (5.1 *10^3)^2 * 2.7 * 10^3 = 26.01 * 10^6 * 2.7 *10^6 = 26.01 * 2.7 * 10^ (6+3) = 70.227 * 10 ^9</span>
Answer:
false.
Explanation:
Ok, we define average velocity as the sum of the initial and final velocity divided by two.
Remember that the velocity is a vector, so it has a direction.
Then when she goes from the 1st end to the other, the velocity is positive
When she goes back, the velocity is negative
if both cases the magnitude of the velocity, the speed, is the same, then the average velocity is:
AV = (V + (-V))/2 = 0
While the average speed is the quotient between the total distance traveled (twice the length of the pool) and the time it took to travel it.
So we already can see that the average velocity will not be equal to half of the average speed.
The statement is false
Answer:
Da=(1/4)Db
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
When s = Da, t = t
When s = Db, t = 2t
Dividing the two equations
Hence, Da=(1/4)Db
Answer:
the average velocity of car A between t1 and t2greater is greater than the average velocity of B berween t1 and t2
Explanation:
Velocity is displacement over time,
Displacement is the distance covered relative to the initial starting position
For A:
at time ti, A moved from Xo to 2Xo, displacement is 2Xo.
at time t2 a moves with speed 3V, hence, his new position will be 3Xo from 2Xo which will be at 5Xo. A's displacement is 5Xo from starting point.
For B:
at time ti, B moved from Xo to 2Xo, displacement is 2Xo.
at time t2 a moves with speed V in the opposite position so he'll be back to his starting point, hence, his new position will be at Xo. A's displacement is 0 from his starting point.
