Question

Can somebody help me out with this question?

Answer 1

If the four riders have a total mass of 230 kg, what is the tension in the left cable just before release?

Express your answer with the appropriate units.

15° 26°

Explanation:

T1 cos 15 = T2 cos 26

T1 = T2 cos 26 / cos 15

T1 = 0.930 T2 ------------(1)

T1 sin 15 + T2 sin 26 = m g

substitute eqn(1)

(0.930 T2 ) sin 15 + T2 sin 26 = 230 * 9.8

0.241 T2 + 0.438 T2 = 2254

Tension in right cable T2 = 3319 N

substitute in eqn (1)

T1 = 0.930 * 3319

Tension in left cable T1 = 3087 N

Answer 2

#a

• -2<0

`\\ \sf\longmapsto f(x)=x^2`

`\\ \sf\longmapsto f(-2)`

`\\ \sf\longmapsto (-2)^2`

`\\ \sf\longmapsto 4`

#b

• 0=0

`\\ \sf\longmapsto f(x)=2`

`\\ \sf\longmapsto f(0)=2`

#c

• 2>0

`\\ \sf\longmapsto f(x)=2x+1`

`\\ \sf\longmapsto f(2)`

`\\ \sf\longmapsto 2(2)+1`

`\\ \sf\longmapsto 4+1`

`\\ \sf\longmapsto 5`