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Advocard [28]
3 years ago
14

A student is bouncing on a trampoline. at her highest point, her feet are 65 cm above the trampoline. when she lands, the trampo

line sags 15 cm before propelling her back up. part a for how long is she in contact with the trampoline?
Physics
1 answer:
Ede4ka [16]3 years ago
3 0
<span>she is in contact with the trampoline for less than one second.</span>
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The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

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  • final velocity of the sprinter, v = 27 km/h
  • time of motion of the sprinter, t = 3.5 x 10⁻⁴ h

Convert the velocity of the sprinter to m/s;

initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\

The time of motion is seconds;

t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s

The sprinter’s average acceleration is calculated as follows;

a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2

Thus, the sprinter’s average acceleration is 1.98 m/s²

Learn more here:brainly.com/question/17280180

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