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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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3 years ago
A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v
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Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, d_{\frac{1}{2}}.

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

                                          d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, d_{\text{remain}} , in which the driver reduces the speed to 40km/hr is

                                             d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}.

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by  t_{\text{remian}}.

                                              t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}.

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

                                                     \text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}

Therefore, the average speed of the car is 50 km/hr.

8 0
2 years ago
A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik
shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

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x =12\pm 0.27

y=1.05

equation of trajectory of ball is given by

y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }

for x=12.27

1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }

u=10.69

for x=11.73

1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }

u=11.436 m/s

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Answer:

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Explanation:

Solution

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