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
Answer:
Explanation:
I dont know if this will help but A two force member is a body that has forces (and only forces, no moments) acting on it in only two locations. In order to have a two force member in static equilibrium, the net force at each location must be equal, opposite, and collinear.
Answer:
Charge, 
Explanation:
It is given that,
Electric field strength, E = 180000 N/C
Distance from a small object, r = 2.8 cm = 0.028 m
Electric field at a point is given by :

Q is the charge on an object



So, the charge on the object is
. Hence, this is the required solution.
Answer:
291.598 N-m
291.6 N-m
Explanation:
Let's first take a look at the free bodily diagrammatic representation.
The first diagram will aid us in answering question (a), so as the second diagram will facilitate effective understanding when solving for question (b).
Let's first determine our angle θ from the diagram
To find angle θ ; we have :
tan θ = 
tan θ = 
tan θ = 1.333
θ = tan⁻¹ (1.333)
θ = 53.13°
Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.
We have:

where Force(F) = Force in the cord AC = 1350 N and θ = 53.13° ; we have:




Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m
b) From the second diagram, taking the moment at point B
,
we have:



where Force(F) = 1350 N and θ = 53.13° ; we have:



Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m
Using the given formula with v0=56 ft/s and h=40 ft
h = -16t2 + v0t
40 = -16t2 + 56t
16t2 - 56t + 40 = 0
Solving the quadratic equation:
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32
We have two possible solutions
t1 = (56+24)/32 = 2.5
t2 = (56-24)/32 = 1
So initially the ball reach a height of 40 ft in 1 second.