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Ludmilka [50]
2 years ago
13

A jogger runs 5 km in 0.6 h, then 9 km in 0.5 hr. What is the average speed of the jogger

Physics
1 answer:
vesna_86 [32]2 years ago
7 0

Note: The values given in the answer are rounded off to the nearest hundredth.

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A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr
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1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration a=g=9.8 m/s^2 directed downward, initial vertical position d=750 m, and initial vertical velocity v_0 = 0. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

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v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

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In order to do that, we use again the same SUVAT equation substituting v_f with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m

Which means that the heigth of the packet was

h=750 m-187.4 m=562.6 m

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How are frequency and wave period related?
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For any wave ...

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<span>virtual, upright, and magnified</span>
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