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Nimfa-mama [501]
3 years ago
12

What is the kinetic energy of a 1700 kg car traveling at a speed of 30 m/s (â65 mph)? does your answer to part b depend on the c

ar's mass?
Physics
1 answer:
MArishka [77]3 years ago
6 0
The kinetic energy of an object is given by
K= \frac{1}{2} mv^2
where m is the mass of the object and v its velocity.

For the car in the problem, m=1700 kg and v=30 m/s, so the kinetic energy of the car is
K= \frac{1}{2}mv^2= \frac{1}{2}(1700 kg)(30 m/s)^2=7.65\cdot 10^5 J

and as we can see, yes, the answer depends on the car's mass.
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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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