Answer:
By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.
here weight of the child =21kgx9.8m/s2 = 205.8N
the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.
torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.
net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.
b) Ta = 2.5x151 = 377.5N-m
Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.
c)Ta = 2x151 = 302N-m
Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.
Answer:
B) the average distance from the Earth to the Sun
Explanation:
The northward components of the resultant displacement is 40.96 m and the westward components of the resultant displacement of the bird from its nest is 28.68 m.
<h3>
Displacement of the bird</h3>
The displacement of the bird is the change in the position of the bird.
<h3>Vertical component of the bird's displacement </h3>
Vy₁ = -25 m x sin(55)
Vy₁ = -20.48 m
Vy₂ = 75 m x sin(55)
Vy₂ = 61.44 m
Total vertical displacement = 61.44 m - 20.48 m = 40.96 m
<h3>Horizontal component of the bird's displacement </h3>
Vx₁ = -25 m x cos(55)
Vx₁ = -14.34 m
Vx₂ = 75 m x cos(55)
Vx₂ = 43.02 m
Total horizontal displacement = 43.02 m - 14.34 m = 28.68 m
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Answer:
1.7323
Explanation:
To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.
From the data given we have to:
Where n means the index of refraction.
We need to calculate the index of refraction of the liquid, then applying Snell's law we have:
Replacing the values we have:
Therefore the refractive index for the liquid is 1.7323
