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nexus9112 [7]
3 years ago
14

An airplane maintains a speed of 585 km/h relative to the air it is flying through as it makes a trip to a city 815 km away to t

he north. (a) What time interval is required for the trip if the plane flies through a headwind blowing at 32.1 km/h toward the south? 1.47 Correct: Your answer is correct. h (b) What time interval is required if there is a tailwind with the same speed?
Physics
1 answer:
navik [9.2K]3 years ago
4 0

Answer:

a)   t = 1.47 h    b) t = 1.32 h

Explanation:

a)  In this problem the plane and the wind are in the same North-South direction, whereby the vector sum is reduced to the scalar sum (ordinary). Let's calculate the total speed

     v = v_{f}f - v_{w}

     v = 585 -32.1

     v = 552.9 km / h

We use the speed ratio in uniform motion

     v = x / t

     t = x / v

     t = 815 /552.9

     t = 1.47 h

b)  We repeat the calculation, but this time the wind is going in the direction of the plane

      v=  v_{f}f - v_{w}

      v 585 + 32.1

      v = 617.1 km / h

      t = 815 /617.1

      t = 1.32 h

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===>  Distance fallen from rest in free fall =

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                               (122.5 m) = (1/2) (9.8 m/s²) (time²)

Divide each side by (4.9 m/s²):   (122.5 m / 4.9 m/s²)  =  time²

                                                           (122.5/4.9) s²  =  time²

Take the square root of each side:    5.0 seconds


===> (Accelerating at 9.8 m/s², he will be dropping at
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===> With no air resistance, the horizontal component of velocity
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Horizontal distance = (10 m/s) x (5.0 s)  =  50 meters .

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                                 = √(10² + 49²)  =  50.01 m/s  arctan(10/49)

                                 =    50.01 m/s   at  11.5° from straight down,
                                                           
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7 0
3 years ago
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A gray kangaroo can bound across level ground with each jump carrying it 9.6 m from the takeoff point. Typically the kangaroo le
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Answer:

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(B) 1.3 m

Explanation:

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Angle of projection, theta = 28 degree

(A)

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u = 10.65 m/s

u = 11 m/s

(B)

Use the formula for maximum height

H = u^2 Sin ^2 theta / 2g

H =

10.65 × 10.65 × Sin^2 (28) / ( 2 × 9.8)

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H = 1 .3 m

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Answer:

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