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Alik [6]
3 years ago
11

A plane has a takeoff speed of 88.3 m/s and can accelerate at a rate of 3m/s to reach that speed. How long does the plane take t

o take-off.
Physics
1 answer:
Mrrafil [7]3 years ago
3 0

v = \frac{d}{t}

and

a = \frac{v}{t}

We have acceleration and velocity so:

3 = \frac{v}{t}

88.3 = \frac{d}{t}

In the acceleration equation we can isolate for v and then plug it back into the other equation to solve...

So...

3t = v

88.3 = 3t

Divide by three and

t = 29.4 s

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A 3" diameter germanium wafer that is 0.020" thick at 300K has 1.015 x 10^17 As atoms added to it. What is the resistivity of th
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Answer:

0.546 ohm / μm

Explanation:

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N = 1.015 * 10^17

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Resistivity = 1/qNu

Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)

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jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh
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Answer:

C. 590 mph

\vert v_{cj}\vert=589.49\ mph

Explanation:

Given:

  • velocity of jet, v_j=500\ mph
  • direction of velocity of jet, east relative to the ground
  • velocity of Cessna, v_c=150\ mph
  • direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

\vec v_j=500\ \hat i\ mph

&

\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)

\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph

Now the vector of relative velocity of Cessna with respect to jet:

\vec v_{cj}=\vec v_j-\vec v_c

\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )

\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph

Now the magnitude of this velocity:

\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}

\vert v_{cj}\vert=589.49\ mph is the relative velocity of Cessna with respect to the jet.

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