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Rufina [12.5K]
3 years ago
10

A parachutist jumps out of an airplane and accelerates with gravity to a maximum velocity of 58.8 m/s in 6.00 seconds. She then

pulls the parachute cord and after a 4.00-second constant deceleration, descends at 10.0 m/s for 60.0 seconds, reaching the ground. From what height did the parachutist jump?
Physics
1 answer:
MrRissso [65]3 years ago
7 0

Answer:

914m

Explanation:

We must divide the parachutist motion into three parts:

First we have free fall motion:

y_1=\frac{v_f^2-v_0^2}{2g}\\y_2=\frac{(58.8\frac{m}{s})^2-0^2}{2(9.8\frac{m}{s^2})}=176.4m

Then, we have a uniformly accelerated motion. The initial speed is this part will be the same final speed as the previous part.

a=\frac{v_f-v_0}{t}\\a=\frac{10.0\frac{m}{s}-58.8\frac{m}{s}}{4s}=-12.2\frac{m}{s^2}\\y_2=\frac{v_f^2-v_0^2}{2a}\\y_2=\frac{(10.0\frac{m}{s})^2-(58.8\frac{m}{s})^2}{2(-12.2\frac{m}{s^2})}=137.6m

Finally, we have a uniform linear motion:

y_3=v*t\\y_3=10\frac{m}{s}*60s=600m

The total heigh will be the sum of all heights:

y=y_1+y_2+y_3\\y=176.4m+137.6m+600m=914m

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Can you help me with my homework please ? Science
geniusboy [140]
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1. S=72m/37s

Divide

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Kira's average speed is 1.94m/s.

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natita [175]

Answer:

The phase difference is 0.659 rad.

Explanation:

Given that,

Distance between two identical loudspeakers d= 1.00  m

Distance between speakers and listener r= 4.00 m

Frequency = 300 Hz

Suppose we need to find the phase difference in radian between the waves from the speakers when they reach the observer

We need to calculate the r'

Using Pythagorean theorem

r'=\sqrt{d^2+r^2}

Where, d = distance between two identical loudspeakers

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Put the value into the formula

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We need to calculate the wavelength

Using formula of wavelength

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Using formula of phase difference

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\phi=\dfrac{2\pi\times0.12}{1.143}

\phi=0.659\ rad

Hence, The phase difference is 0.659 rad.

7 0
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