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Yuri [45]
1 year ago
3

Two runners are racing around a cicular track. The first runner is in the inside lane. The second runner is in the outside lane.

If the first runner wants to stay side-by-side with the second runner, what does her angular velocity need to be?.
Physics
1 answer:
kakasveta [241]1 year ago
3 0

In order to cowl an equal angular distance over time, their angular speed should be precisely same. A regular aspect-to-aspect dating suggests an same angular distance i.e., both have same velocity.

The definition of angular speed in physics is as follows: The vector illustration of rotation charge, or how speedy an object rotates or revolves in terms of any other factor, is referred to as angular speed. The time charge at which an item rotational or circles round an axis is called its angular speed. The charge at which an object or particle rotates approximately a middle or a selected factor in a finite quantity of time is called its angular speed. Additionally referred to as rotational speed. Radians in line with second (rad/s) is a unit of size for angular speed. The charge of alternate withinside the angular area of a rotating frame is called the angular speed. When an item movements alongside a directly route, its linear speed is the charge at which its displacement modifications in terms of time.

Learn extra approximately angular speed here: brainly.com/question/12446100

#SPJ4

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In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.43 km mark a
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Answer:

Acceleration, a = 0.101 m/s²

Explanation:

Average speed = total distance / total time.

At the 7.43km mark, total distance = 7.43km or 7430m

Total time = 25 * 60 s = 1500s

Average speed = 7430m/1500s = 4.95m/s

He then covers (10 - 7.43)km = 2.57 km = 2570 m

in t = 27m43.6s - 25min = 2m43.6s = 163.6 s

Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 - 60)s = 103.6 s.

From V = u + at; V = 4.95m/s + a *60s

Distance covered while accelerating is

s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²

Distance covered while at constant velocity, v after accelerating is

D = velocity * time

Where v = 4.95m/s + a*60s

D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²

Total distance covered after initial 7.43 km, S + D = 2570 m, so

2570 m = 297m + a*1800s² + 512.82m + a*6216s²

2570 = 809.82 + a*8016

a = 809.82m / 8016s² = 0.101 m/s²

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For convenience write RC=D and solve the ODE dV/dt=-V/D which gives 
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<span>When t=0, V=10 and this gives A=10, </span>
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