An electronic scoreboard of mass 500kg is suspended from a ceiling by four cables, each making an angle of 70° with the ceiling. The weight is evenly distributed. Determine the tensions in the cables, in newtons

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.

im not sure what this means: mc011-7.jpg

step-by-step explanation:

7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west

the answer is

d.

5.3066 m2

step-by-step explanation: