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swat32
2 years ago
12

Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.8 km/h due east. Runner B is initia

lly 4.2 km east of the flagpole and is running with a constant velocity of 7.6 km/h due west. How far are the runners from the flagpole when their paths cross? Answer in units of km.
Physics
1 answer:
vampirchik [111]2 years ago
4 0

Answer:

0.39 km west

Explanation:

The position of Runner A is:

x = -5.7 + 8.8 t

The position of Runner B is:

x = 4.2 − 7.6 t

When the positions are equal:

-5.7 + 8.8 t = 4.2 − 7.6 t

16.4 t = 9.9

t = 0.604

Plug into either equation to find the position at this time:

x = -5.7 + 8.8 (0.604)

x = -0.39

The runners are 0.39 km west of the flagpole when they meet.

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Answer:

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Explanation:

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7 0
2 years ago
A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical a
VashaNatasha [74]

Answer:

-7.23 rad/s

Explanation:

Given that

Mass of the cockroach, m = 0.157 kg

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Speed of the cockroach, v = 2.92 m/s

Angular velocity of the rim, w = 3.89 rad/s

The initial angular momentum of rim is

Iw = 5.92*10^-3 * 3.89

Iw = 2.3*10^-2 kgm²/s

The initial angular momentum of cockroach about the axle of the disk is

L = -mvr

L = -0.157 * 2.92 * 0.149

L = -0.068 kgm²/s

This means that we can get the initial angular momentum of the system by summing both together

2.3*10^-2 + -0.068

L' = -0.045 kgm²/s

After the cockroach stops, the total inertia of the spinning disk is

I(f) = I + mr²

I(f) = 5.92*10^-3 + 0.157 * 0.149²

I(f) = 5.92*10^-3 + 3.49*10^-3

I(f) = 9.41*10^-3 kgm²

Final angular momentum of the disk is

L'' = I(f).w(f)

L''= 9.41*10^-3w(f)

Using the conservation of total angular momentum, we have

-0.068 = 9.41*10^-3w(f) + 0

w(f) = -0.068 / 9.41*10^-3

w(f) = -7.23 rad/s

Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed

b)

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Answer:

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