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ICE Princess25 [194]
3 years ago
11

What is the distance covered by a Freely falling object 5 seconds after being dropped ? After 6 seconds?

Physics
1 answer:
mario62 [17]3 years ago
5 0

This year is 60 years since I learned this stuff, and one of the things I always remembered is the formula for the distance a dropped object falls:

D = 1/2 A T²

Distance = (1/2) (acceleration) (time²)

The reason I never forgot it is because it's SO useful SO often.  You really should memorize it.  And don't bury it too deep in your toolbox ... you'll be needing it again very soon. (In fact, if you had learned it the first time you saw it, you could have solved this problem on your own today.)

The problem doesn't tell us what planet this is happening on, so let's make it easy and just assume it's on Earth.  Then the 'acceleration' is Earth gravity, and that's 9.8 m/s² .

In 5 seconds:

D = 1/2 A T²

D = (1/2) (9.8 m/s²) (5 sec)²

D = (4.9 m/s²) (25 sec²)

D = 122.5 meters


In 6 seconds:

D = 1/2 A T²

D = (1/2) (9.8 m/s²) (6 sec)²

D = (4.9 m/s²) (36 sec²)

D = 176 meters


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a) For y = 102 mA, R = 98.039 ohms

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b) Check explanatios for b

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According to ohm's law, V = IR

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y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]

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A linear equation is of the form y = Gx

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x = \left[\begin{array}{ccc}100\end{array}\right]

\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right] =  \left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] =  \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5}  \end{array}\right]

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