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3 years ago

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An astronaut stands by the rim of a crater on the moon, where the acceleration of gravity is 1.62 m/. To determine the depth of

the crater, she drops a rock and measures the time it takes for it to hit the bottom. If the time is 6.3 s, what is the depth of the crater?

1 answer:

leva [86]3 years ago

3 0

Answer;

=32.15 meters

Explanation;

Use the formula

s= ut+ 0.5 a (t)^2 to find out 's'

where s= distance traveled

u=initial velocity which is zero in this case

t= time taken to travel 's' distance

a=acceleration (due to gravity on moon i.e 1.62 m/s^2 )

Therefore;

S = 0.5 * 1.62 * 6.3 * 6.3

= 32.1489 meters

Thus; the crater is 32.15 meters deep.

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Answer:

internet of things.

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4 years ago

Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th

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Answer:

$T_f=5.0116^{\circ}C$

Explanation:

Given:

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<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

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Read 2 more answers

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

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at $t=0$

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$c=\hat{i}+\hat{j}$

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and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

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at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

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