<span>The repelling of the support magnet decreases friction. is the answer you're looking for . :)
hope i helped - beanz</span>
Incomplete question.The Complete question is here
A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.
a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.
b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.
Answer:
(a)ω = 1 rad/s
(b)t = 2.41 s
Explanation:
(a) initial angular momentum = final angular momentum
0 = L for disk + L............... for runner
0 = Iω² - mv²r ...................they're opposite in direction
0 = (MR²/2)(ω²) - mv²r
................where is ω is angular speed which is required in part (a) of question
0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)
0=200ω²-200
200=200ω²
ω = 1 rad/s
b.)
lets assume the "starting point" is a point marked on the disk.
The person's angular speed is
v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s
As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.
(angle) + (angle disk turns) = 2π
(1.6 rad/s)(t) + ωt = 2π
t[1.6 rad/s + 1 rad/s] = 2π
t = 2.41 s
Answer:
metals,nonmetals, and inert gases
Explanation:
The reading will be 0.00. that is the least count<span> of </span>stopwatch<span>.</span>
Answer:
Explanation:
a ) Height to be cleared = 5 - 1.6 = 3.4 m
Horizontal distance to be cleared = 5 m .
angle of throw = 56°
here y = 3.4 , x = 5 , θ = 56
equation of trajectory
y = x tanθ - 1/2 g ( x/ucosθ)²
3.4 = 5 tan56 - 1/2 g ( 5/ucos56)²
3.4 = 7.4 - 122.5 / .3125u²
122.5 / .3125u² = 4
u² = 98
u = 9.9 m /s
Range = u² sin 2 x 56 / g
= 9.9 x 9.9 x .927 / 9.8
= 9.27 m
horizontal distance be-yond the fence will the rock land on the ground
= 9.27 - 5
= 4.27 m
