Basking in the sun, a 1.10 kg lizard lies on a flat rock tilted at an angle of 15.0° with respect to the horizontal. What is the magnitude of the normal force exerted by the rock on the lizard? a. 10.5 N c. 10.4 N b. 2.8 N d. -10.5 N

Respuesta :

10.4 N

Given

m = 1.10 kg

θ = 15.0°

g = 9.81 m/s2

Solution

Fnet, y = ΣF y = Fn − Fg, y = 0

Fn = Fg, y = Fg

cosθ = mgcosθ

Fn = (1.10 kg)(9.81 m/s2

)(cos15.0°) = 10.4 N

The normal force exerted by the rock will be "10.42 N".

Given values:

  • Mass of lizard, m = 1.10 kg
  • Angle, [tex]\Theta[/tex] = 15°

As we know the formula,

→ [tex]N' = mg Cos \Theta[/tex]

By substituting the values, we get

       [tex]= 1.10\times 9.81\times Cos15^{\circ}[/tex]

       [tex]= 10.42 \ N[/tex]

Thus the above approach i.e., "option c" is right.  

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