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3 years ago

A joger travels 8.0 kilometers in 1.25 hours what is the average speed?

Physics

2 answers:

gogolik [260]3 years ago

8 0

8/1.25 = 6.4 kmph
The answer is 6.4 kilometers per hour

Svetach [21]3 years ago

3 0

ΔS = 8.0 km

ΔT = 1.25 h

Vm = ΔS / ΔT

Vm = 8.0 / 1.25

Vm = 6.4 Km/h

hope this helps!

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When a river flows into an ocean, it slows down and deposits materials in its alluvial fan/delta

nikklg [1K]

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3 years ago

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Allushta [10]

Answer:

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Mariulka [41]

Answer:

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3 years ago

A day on a distant planet observed orbiting a nearby star is 21.5 hr. Also, a year on the planet lasts 69.3 Earth days. In other

serg [7]

Answer:

Part A

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Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, $\omega_{rotation}$ = Angle turned/Time

∴ $\omega_{rotation}$ = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, $\omega_{rotation}$ = 8.11781 × 10⁻⁵ rad/s

The angular speed of rotation of the plane $\omega_{rotation}$ = 8.11781 × 10⁻⁵ rad/s

Part B

The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days

Therefore, the average angular speed of the planet around its neighboring star, $\omega _{Star}$ , is given as follows;

$\omega _{Orbit}$ = 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s

The average angular speed of orbit, $\omega _{Orbit}$ = 1.04938 × 10⁻⁶ rad/s

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Gnoma [55]

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4 years ago