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pochemuha
3 years ago
5

An astronaut on a distant planet wants to determine its acceleration due to gravity. the astronaut throws a rock straight up wit

h a velocity of +16 m/s and measures a time of 21.2 s before the rock returns to his hand. what is the acceleration (magnitude and direction) due to gravity on this planet? (indicate direction by the sign of the acceleration.)
Physics
1 answer:
Tamiku [17]3 years ago
7 0
Define
u = 16 m/s, the vertical launch velocity
g = acceleration due to gravity, measured positive downward
s = vertical distance traveled
t  = 21.2 s, total time of travel.

The vertical motion obeys the equation
s = ut - (1/2)gt²

When the rock is at ground level, s = 0.
Therefore
(16 m/s)(21.2 s) - 0.5*(g m/s²)*(21.2 s)² = 0
339.2 - 224.72g = 0
g = 1.5094 m/s²

Answer:
The acceleration due to gravity is 1.509 m/s² measured positive downward.


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What happens when tectonic plates move away from each other? Molten matter cools and sinks towards the core. Lava travels away f
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A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter
kifflom [539]

Answer:

a). 87.5 mA or 87.5 x10^{-3}A

b). 1.78 \frac{m}{s}

Explanation:

d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}

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a).

I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\  I=87.5 x10^{-3}A

b).

I=n*abs (q)*V_{d}*A

A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2}  \\A=5.309x10^{-6}

V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}

8 0
3 years ago
calculate the work done in kilo joules in lifting a mass of 20kg at steady velocity through a vertical height of 20m
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Answer:

\huge\boxed{\sf Work\ done = 4 kJ}

Explanation:

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We know that,

<h3>P.E. = mgh </h3>

Where,

m = mass = 20 kg

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h = vertical height = 20 m

So,

<h3>Work done = mgh</h3>

Work done = (20)(10)(20)

Work done = 4000 joules

Work done = 4 kJ

\rule[225]{225}{2}

5 0
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