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Eddi Din [679]
3 years ago
12

A cheetah is crouched 20 m to the east of an observer (Fig. 2.6a). At time the cheetah begins to run due east toward an antelope

that is 50 m to the east of the observer. During the first 2.0 s of the attack, the cheetah’s coordinate x varies with time according to the equation (a) Find the cheetah’s displacement between and (b) Find its average velocity during that interval. (c) Find its instantaneous velocity at by taking then then (d) ¢t = 0.1 s, 0.01 s, 0.001 s. Derive an t1 = 1.0 s t t 2 = 2.0 s. 1 = 1.0 s 15.0 m>s2 2t 2 x = 20 m + . t = 0 expression for the cheetah’s instantaneous velocity as a function of time, and use it to find at t=1.0s and t=2.0s
Physics
1 answer:
Korvikt [17]3 years ago
4 0

Answer:

i dont know but help me by marking me the brainliest

Explanation:

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Use this formula to solve this problem:
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An open organ pipe is 1.6m long. If the speed of sound is 343m/s, what are the pipes: a) fundamental , b) 1st overtone , &amp; c
Yakvenalex [24]

Answer:

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b) 214.375 Hz

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Explanation:

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