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:
The answer is "False"
Explanation:
The geologic time scale is the "schedule" for occasions in Earth history. It partitions time into named units of unique time called in descending order of duration "eons, eras, periods, epochs, and ages". The specification of those geologic time units depends on stratigraphy, which is the relationship and order of rock layers. The fossil structures that happen in the stones, nonetheless, give the central methods for setting up a geologic time scale, with the circumstance of the development and vanishing of far and wide species from the fossil record being used to outline the beginnings and endings of ages,, periods, and different stretches.
Geologic time is the broad time period involved by the geologic history of Earth. Formal geologic time starts toward the beginning of the Archean Eon (4.0 billion to 2.5 billion years back) and proceeds to the current day.
Answer:
Explanation:
Time dilation formula is
T = T₀ / √ 1-v²/c²
T₀ is time elapsed in moving reference , T time elapsed in stationary reference.
Here T₀ = 1 second
T = 1/√ 1-0.9² = 1/.4358 = 2.3 second
So 2.3 second will pass for each second on moving reference.
Answer:
Explanation:
This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".
In the y-stuff category:
v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)
Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)
a = -9.8 m/s/s
t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)
In the "x-stuff" category:
v₀ = 7.25 m/s (this is given)
Δx = ???
a = 0 (acceleration in this dimension is ALWAYS 0)
t = (we will solve for this in the y-dimension and plug it in here).
In the y dimension:
Δx = v₀t + 
and plugging in from the y-dimension info:
which simplifies to
so
which, to 2 significant digits is
t = 1.4 seconds
Now we will do the same in the x-dimension, using t = 1.4:
Δx = v₀t + and filling in the x-stuff:
Δx = 
Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:
Δx = 7.25(1.4) + 0 so
Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).
