Respuesta :

#a

  • -2<0

[tex]\\ \sf\longmapsto f(x)=x^2[/tex]

[tex]\\ \sf\longmapsto f(-2)[/tex]

[tex]\\ \sf\longmapsto (-2)^2[/tex]

[tex]\\ \sf\longmapsto 4[/tex]

#b

  • 0=0

[tex]\\ \sf\longmapsto f(x)=2[/tex]

[tex]\\ \sf\longmapsto f(0)=2[/tex]

#c

  • 2>0

[tex]\\ \sf\longmapsto f(x)=2x+1[/tex]

[tex]\\ \sf\longmapsto f(2)[/tex]

[tex]\\ \sf\longmapsto 2(2)+1[/tex]

[tex]\\ \sf\longmapsto 4+1[/tex]

[tex]\\ \sf\longmapsto 5[/tex]

rityahir1591

Answer:

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°

Step-by-step 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

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