If we assume that Earth is a solid sphere, then I = (2/5)mr² = (2/5) * 5.98e24kg * (6.371e6m)² = 9.71e37 kg·m²
torque τ = Iα, but also τ = F*r = 3.98e7N * 6.371e6m = 2.54e14 N·m Since τ = τ, 2.54e14 N·m = 9.17e37kg·m² * α α = 3.61e-23 rad/s²
ωo = 2πrads / (24h*3600s/h) = 7.27e-5 rad/s ω1 = 2πrads / (28h*3600s/h) = 6.23e-5 rad/s
Plug in numbers t = 2/5 * (5.98 * 10^24 kg) * (6.37*10^6 m) * (7.27 * 10^-5 rad/s - 6.23 * 10^-5 rad/s) / (3.98*10^7 N) = 3.98*10^22 s = 1.26*10^17 yrs
<span>The platform scale consists of a combination of third and first class levers so that the load on one lever becomes the effort that moves the next lever.Through this arrangement, a small weight can balance a massive object. If x=450 mm,determine the required mass of the counterweight S required to balance a 90-kg load, l.</span>
Answer:

Explanation:

Cross multiply


cross multiply

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14m/s
Explanation:
Given parameters:
Height of the ball = 10m
Unknown:
Velocity of fall or final velocity = ?
Solution:
We are going to use the appropriate equation of motion to solve this problem.
The object is falling with respect to gravity.
V² = U² + 2gH
where V is the final velocity
U is the initial velocity
g is the acceleration due to gravity 9.8m/s²
H is the height of fall
The initial velocity here is zero and
V² = 2 x 9.8 x 10 = 196
V = 14m/s
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Motion problems brainly.com/question/5248528
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The correct answer is false cause how can u fit your finger in a wall something it's to small