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IgorC [24]
3 years ago
7

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

at is the Cessna's speed relative to the 747? A 634 mph B 578 mph C 590 mph D 444 mph E 610 mph

Physics
1 answer:
Greeley [361]3 years ago
8 0

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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A rocket is launched from the surface of Earth with a speed v0 that will allow the rocket to escape the gravitational field of E
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option ( a ) is correct .

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8 0
3 years ago
Sara walks part way around a swimming pool. She walks 50 yards north, then
Mekhanik [1.2K]

Answer:

20 Yards

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|---20----|

|            |

| 50       |50

|---D--->|

Start      End

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7 0
3 years ago
Donna drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 8 hours. When Donna drove
Dennis_Churaev [7]

Answer:

d=360 miles

Donna lives 360 miles from the mountains.

Explanation:

Conceptual analysis

We apply the formula to calculate uniform moving distance[

d=v*t   Formula (1)

d: distance in miles

t: time in hours

v: speed in miles/hour

Development of problem

The distance Donna traveled to the mountains is equal to the distance back home, equal to d,then,we pose the kinematic equations for d, applying formula 1:

travel data to the mountains: t₁= 8 hours ,  v=v₁

d= v₁*t₁=8*v₁ Equation (1)

data back home : t₂=4hours ,  v=v₂=v₁+45

d=v₂*t₂=(v₁+45)*4=4v₁+180 Equation (2)

Equation (1)=Equation (2)

8*v₁=4v₁+180

8*v₁-4v₁=180

4v₁=180

v₁=180÷4=45 miles/hour

we replace v₁=45 miles/hour in equation (1)

d=8hour*45miles/hour

d=360 miles

8 0
2 years ago
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