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jekas [21]
3 years ago
7

Two planes take off from the same airport at the same time using different runways. One plane travels on a bearing Upper S 14 de

grees Upper W at 440 miles per hour. The other plane travels on a bearing Upper N 75 degrees Upper W at 420 miles per hour. How far are the planes from each other 3 hours after​ takeoff?
Physics
1 answer:
Mashcka [7]3 years ago
6 0

Answer:

 Let's imagine this is the diagram for the question,

                 •    

3*440=       /  |

               /     |

1320mi  ↑       |

           ↑         ↓

          /           ↓

        /             |

      • ) 89⁰     |

       \              |   "x" miles

         \            ↑

           \          ↑

            ↓         |

              ↓       |      

3*420=      \     |

1260 mi      \   |

                        •

From the Cosine Rule:  x² = y² + z² - 2 y z cos X⁰

x² = 1320² + 1260² - 2 * 1320 * 1260 * cos 89⁰ =  

1742400 + 1587600 - 3326400 * cos 89⁰ = 3271946315  

x ≈ 1809 mi

Hence, the planes are 1809 miles from each other 3 hours after​ takeoff?

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Answer:

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Due to increase in size of the ions, the lattice energy decreases while the lattice energy increases as the charge of the ions increases.

When the size increase, the distance between the nuclei also increase leading a decrease the force of attraction between the nuclei

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In a series lrc circuit, the frequency at which the circuit is at resonance is f0. If you double the resistance, the inductance,
taurus [48]

When you double capacitance and inductance, the new resonance frequency becomes f/2.

  • Resonance frequency:

The resonance frequency of RLC series circuit, is the frequency at which the capacity reactance is equal to inductive reactance.

It can also be defined as the natural frequency of an object where it tends to vibrate at a higher amplitude.

Xc = Xl

which gives the value for resonance frequency:

f = \frac{1}{2\pi \sqrt{LC} }

where;

f is the resonance frequency

L is the inductance

C is the capacitance

When you double capacitance and inductance, the new resonance frequency becomes;

f' = \frac{1}{2\pi \sqrt{2L2C} }

f' = \frac{1}{2\pi \sqrt{4LC} }

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f' = \frac{1}{2} f

Thus from above,

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One scientist is using an electromagnetic wave to map the dust between stars. The electromagnetic wave has shorter wavelength th
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3 years ago
1. A piece of metal weighs 50.0 N in air, 36.0 N in water, and 41.0 N in an unknown
denis23 [38]

Answer:

a) 3.37 x 10^{3} kg/m^3

b) 6.42kg/m^{3}

Explanation:

a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .

Weight of metal in air = 50N = mg implies the mass of metal is 5kg.

Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x 10^{-4}  kg/m^{3}. So density of metal = mass of metal / volume of metal = 5 / 14 x 10^{-4}  kg/m^{3} = 3.37 x 10^{3} kg/m^3

b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/m^{3}

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