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hoa [83]
2 years ago
11

An airplane flies between two points on the ground that are 500 km apart. the destination is directly north of the origination o

f the flight. the plane flies with an air speed of 120 m/s. if a constant wind blows at 10.0 m/s due west during the flight, what direction must the plane fly relative to the air to arrive at the destination?
Physics
1 answer:
Ilya [14]2 years ago
5 0

<u>Answer:</u>

 Plane must fly 4.78⁰ due north to east at a speed of 120 m/s to reach destination.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of plane = 120 m/s

  Wind speed  = 10.0 m/s due west  = -10 m/s

  So plane must have an horizontal speed of 10 m/s to cancel this wind speed

  Horizontal speed of plane = u cos θ = 120*cos θ = 10

  Angle θ =85.22⁰ from positive horizontal axis.

  Angle from North axis towards East required = 90 - 85.22 = 4.78⁰

 So plane must fly 4.78⁰ due north to east at a speed of 120 m/s to reach destination.

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Human muscles have an efficiency of about __________. A. 9 to 17% B. 18 to 26% C. 27 to 35% D. 36 to 44%
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The correct answer would be B. 18 to 26%.

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Levi observed properties of four different waves and recorded observations about each one in his chart. A 2-column table with 4
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Wave W is a sound wave, Waves X and Y are light waves, and it is impossible to tell what kind of wave Wave Z is.

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X       travels fastest through air,

Y       travels more slowly through water than air    

Z       travels more slowly at cool temperatures

W appears to be sound wave  as sound travels fastest through metal .

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Explanation:

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The temperature of the water before drinking is T_w  =  3.8 ^oC

The temperature of the body is T_b  =  36.6^oC

Generally the amount of heat required to move the water from its former temperature to the body temperature is

H=  m_w  *  c_w * \Delta T

Here c_w is the specific heat of water with value c_w = 4.18 J/g^oC

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Here Z_w is the molar mass of sweat with value

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=> n = \frac{5.7 *10^2}{18.015}

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Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

H_v  =  n  *  L_v

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H_t =  H +  H_v

=> H_t =  7.8 *10^{4} +  1.29 *10^{6}

=> H_t =  1.37 *10^{6} \ J

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