One way you could think about it is that you used science of electronics to develop the technology of a computer
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The bigger the starting force, the further the distance moved by the car?
Answer:
Explanation:
As an example, ice on steel has a low coefficient of friction – the two materials slide past each other easily – while rubber on pavement has a high coefficient of friction – the materials do not slide past each other easily. The coefficients of friction ranges from near 0 to greater than 1.
Answer:
4.36 rad/s
Explanation:
Radius of platform r = 2.97 m
rotational inertia I = 358 kg·m^2
Initial angular speed w = 1.96 rad/s
Mass of student m = 69.5 kg
Rotational inertia of student at the rim = mr^2 = 69.5 x 2.97^2 = 613.05 kg.m^2
Therefore initial rotational momentum of system = w( Ip + Is)
= 1.96 x (358 + 613.05)
= 1903.258 kg.rad.m^2/s
When she walks to a radius of 1.06 m
I = mr^2 = 69.5 x 1.06^2 = 78.09 kg·m^2
Rotational momentuem of system = w(358 + 78.09) = 436.09w
Due to conservation of momentum, we equate both momenta
436.09w = 1903.258
w = 4.36 rad/s
Explanation:
We'll need two equations.
v² = v₀² + 2a(x - x₀)
where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.
x = x₀ + ½ (v + v₀)t
where t is time.
Given:
v = 47.5 m/s
v₀ = 34.3 m/s
x - x₀ = 40100 m
Find: a and t
(47.5)² = (34.3)² + 2a(40100)
a = 0.0135 m/s²
40100 = ½ (47.5 + 34.3)t
t = 980 s
