Answer:
The question does not state how the answer is to be entered. I would use 10% because that is most common.
Explanation:
You are using 300 J of energy to get 30 J of light energy. The fact that you can account for the left over energy is not relevant.
Efficiency = Work Out / Work In
Efficiency = 30 J / 300 J = 0.1
If you need this as a %, you can multiply by 100
Efficiency = 0.1 * 100 = 10%
Answer: 10.36m/s
How? Just divide 200m by 19.3 and you will get how fast he ran per m/s
Answer:
Option (d) is correct.
Explanation:
Work done is given by :
W = Fd, where F is force and d is displacement
Unit of work done :
The SI unit of force is Newton (N) and that of displacement is meter (m). So, the unit of work done is N-m. It is call Joule. It means that the unit of work done is Joule.
Power is given by rate at which the work is done. It is given by :
P = W/t, W is work done and t is time
Unit of power:
Unit of work is Joule (J) and that of time is second (s). It means that the unit of power is Watt and it is equal to Joule/second
Hence, the correct option is (d) "The unit for work is a joule. The unit for power is a watt, which is a joule per second".
Answer:
The x-coordinate of the particle is 24 m.
Explanation:
In order to obtain the x-coordinate of the particle, you have to apply the equations for Two Dimension Motion
Xf=Xo+Voxt+0.5axt²(I)
Yf=Yo+Voyt+0.5ayt² (II)
Where Xo, Yo are the initial positions, Xf and Yf are the final positions, Vox and Voy are the initial velocities, ax and ay are the accerelations in x and y directions, t is the time.
The particle starts from rest from the origin, therefore:
Vox=Voy=0
Xo=Yo=0
Replacing Yf=12, Yo=0 and Voy=0 in (I) and solving for t:
12=0+(0)t+ 0.5(1.0)t²
12=0.5t²
Dividing by 0.5 and extracting thr squareroot both sides:
t=√12/0.5
t=√24 = 2√6
Replacing t=2√6, ax=2.0,Xo=0 and Vox=0 in (I) to obain the x-coordinate:
Xf=0+0t+0.5(2.0)(2√6)²
Xf= 24 m
Answer:
Los movimientos verticales o circulación convectiva consisten en un intercambio de lugar entre el aire caliente de las capas bajas y el aire frío de las superiores. Este movimiento se debe a la diferencia de densidades de las distintas masas de aire. Los movimientos horizontales se basan en las diferencias de presión.
