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3 years ago

6

Austin left the park traveling 4 mph. Then, 3 hours later, Wyatt left traveling the same direction at 10 mph. How long until Wya

tt catches up with Austin?

1 answer:

bixtya [17]3 years ago

8 0

ANSWER: 2 hours

EXPLANATION: After three hours Austin has traveled 12 miles. Wyatt will start up and in an hour will be at 10 miles, but by that time Austin will be at 16 miles. One more mile and Wyatt will be at 20 miles and so will Austin. So that would make the answer 2 hours.

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Arrange the step in order to describe what happens to a liquid when it is heated

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Answer:

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Answer:

$a_c=1.44\ m/s^2$

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A plane flying horizontally at an altitude of 1 mi and a speed of 560 mi/h passes directly over a radar station. Find the rate a

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Given:

altitude, x = 1 mile

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To find the rate,

$\frac{dx}{dt} = 0$

Now, from the right angle triangle in fig 1.

Applying pythagoras theorem:

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differentiating the above eqn w.r.t 't' :

$2h\frac{dh}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ (1)

Now, putting values in eqn (1):

$2h\frac{dh}{dt} = 2\times 1\times 0 + 2y\frac{dy}{dt}$

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A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

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Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

$\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ$

which is approximately 41º west of north.

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