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hoa [83]
3 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]3 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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Calculate the equivalent of 20 degrees Celsius in degrees Fahrenheit and Kelvin.
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Answer:

68 °F, 293.15 K

Explanation:

Fahrenheit, Kelvin and Celsius are the different scales of temperature in which temperature is measured.

Given : T = 20°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

<u>T = (20 + 273.15) K = 293.15 K </u>

The conversion of T( °C) to T(F) is shown below:

T (°F) = (T (°C) × 9/5) + 32

So,

<u>T (°F) = (20 × 9/5) + 32 = 68 °F</u>

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3 years ago
Objects have a tendency to resist changing their motion. This property is called: *
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Answer:

The answer your looking for is option 2 - Inertia

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4 years ago
A 0.095 kg tennis ball is traveling to the right at 40 m/s when it bounces of a wall and travels in the opposite direction it ca
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given that

mass of ball = 0.095 kg

initial velocity of ball towards the wall = 40 m/s

final velocity of the ball after it rebound = 30 m/s

now change in momentum is given as

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\Delta P = 0.095(30 - (-40))

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Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016 Na+ ions arrive at the negative electrode and 3.
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Answer:

10.6 mA

Explanation:

t = time interval = 1.00 s

q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C

n₁ = number of Na⁺ ions = 2.68 x 10¹⁶

q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C

n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶

q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C

i₁ = Current due to Na⁺ ions = \frac{q_{1}}{t} = \frac{0.004288}{1} = 0.004288 A

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Current passing between the electrodes is given as

i = i₁ + i₂

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i = 0.01056 A

i = 10.6 x 10⁻³ A

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