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ira [324]
3 years ago
9

A rock is thrown downward from the top of a 36.2-m-tall tower with an initial speed of 13 m/s. Assuming negligible air resistanc

e, what is the speed of the rock just before hitting the ground
Physics
1 answer:
vekshin13 years ago
4 0

Explanation:

u = 13m/s

s = 36.2m

a = 9.81m/s² (positive because it is acting on the same direction as the velocity)

Using Kinematics, we have v² = u² + 2as = (13)² + 2(9.81)(36.2) = 879.244m²/s². => v = 29.65m/s.

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lapo4ka [179]
Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).

Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr.  They'll only
change the units.

(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)

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Determine the magnitude of the current flowing through a 4.7 kilo ohms resistor if the voltage across it is (a) 1mV (b) 10 V (c)
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Answer:

213 nA

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We have a pretty straightforward question here.

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c) at V = 4e^-t

I = 4e^-t / 4.7 * 10^3

I = 0.000851e^-t A or 851e^-t μA

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