A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri
1 answer:
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Okay, haven't done physics in years, let's see if I remember this.
So Coulomb's Law states that
so if we double the charge on 
and double the distance to 
we plug these into the equation to find
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So we see the new force is exactly 1/2 of the old force so your answer should be
if I can remember my physics correctly.
So Coulomb's Law states that
<span>
So we see the new force is exactly 1/2 of the old force so your answer should be
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
<em>The object could fall from six times the original height and still be safe</em>
Explanation:
<u>Free Falling</u>
When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.
If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of
When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy
Where v is the speed just before hitting the ground
If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from
Solving for h
If the drop had occurred in the Moon, then
Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g
This means the object could fall from six times the original height and still be safe