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love history [14]
3 years ago
12

Not only did Skid get shot out of a cannon when he was a clown in the circus, but he was also launched into the air by a vertica

lly compressed spring. Skid of 55 kg is standing on a massless platform attached to a spring of constant 8700 N/m that is compressed 125 cm. Skid is launched from rest. He eventually leaves the platform and reaches a peak high above the platform. (a) What is the velocity of Skid when he leaves the platform
Physics
1 answer:
tiny-mole [99]3 years ago
8 0

Answer:

v = 16.11 m / s

Explanation:

For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation

starting point. When the spring is compressed

        Em₀ = K_e + U = ½ k x² + m g x ’

final point. The point where it leaves the platform

        Em_f = K = ½ m v²

energy is conserved

         Em₀ = Em_f

         ½ k x² + m g x ’= ½ m v²

         v² = \frac{k}{m}  x² + g x

let's calculate

         v² = \frac{8700}{55}  1.25² + 9.8 1.25

         v² = 247.159 + 12.25 = 259.409

         v = 16.11 m / s

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Suppose that a freely falling object were somehow equipped with an odometer. Would the readings of distance fallen each second i
maks197457 [2]

Answer:

1  greater distances fallen in successive seconds

Explanation:

When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement

If we use the kinematic formula we can find the position of the body

       Y = Vo t + ½ to t2

Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity

Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2

Let's look for the position for successive times

t (s)      Y (m)

  1          -4.9

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The sign indicates that the positive sense is up

It can be clearly seen that the distance is greatly increased every second that passes

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3 years ago
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3 years ago
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sergiy2304 [10]

Answer:

true

Explanation:

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3 years ago
The energy in the typical thunderstorm is about 1.4×1011j. a typical lightning flash transfers 30 c across a potential differenc
iragen [17]
780 seconds, or 13 minutes.

In the future, please use proper capitalization. There's a significant difference in the meaning between mV and MV. One of them indicated millivolts while the other indicates megavolts. For this problem, I'll make the following assumptions about the values presented. They are:
Total energy = 1.4x10^11 Joules (J)
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First, let's determine the power discharged by each bolt. That would be the current multiplied by the voltage, so
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Now that we know how many joules are dissipated per flash, let's determine how flashes are needed.
1.4x10^11 / 9x10^8 = 1.56E+02 = 156

Since each flash takes 5 seconds, that means that it will take about 5 * 156 = 780 seconds which is about 780/60 = 13 minutes.
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3 years ago
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