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lions [1.4K]
3 years ago
13

An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed o

f the plane (its speed in still air) is 320.0 km/h (about 200 mi/h), in which direction should the pilot head?

Physics
1 answer:
SashulF [63]3 years ago
8 0

Answer:

14.48° North of West.

Explanation:

The conditions mentioned in the question has been drawn in the figure below where,

  • The vector CA represents the velocity of airplane,
  • θ is the angle from west pointing the direction of airplane in which it is required to travel,
  • Vector AB is the direction of wind, and,
  • Vector CB is the resultant direction.

As ΔABC is a right angle triangle, for calculating the direction of airplane so that it can travel to west, we need to calculate the direction of vector CA which can be calculated with the help of sine theta.

So,

Sinθ= \frac{80}{320} = \frac{1}{4}

Thus the value of θ = 14.48°

So, the pilot should head towards 14.48° North of West.

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A 103 kg horizontal platform is a uniform disk of radius 1.71 m and can rotate about the vertical axis through its center. A 68.
Andreyy89

Answer:

I_{total}=220.64 kg*m^{2}

Explanation:

The moment of inertia of the system is equal to the each population and the platform inertia so

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I_{disk}=\frac{1}{2}*m_{disk}*(r_{p})^{2}

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I_{p}=\frac{1}{2}*m_{p}*(r_{p})^{2}

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I_{d}=\frac{1}{2}*m_{d}*(r_{d})^{2}

The Inertia of the system is the sum of each mass taking into account that all exert the force of inertia:

I_{total}=I_{disk}+I_{p}+I_{d}

I_{total}=\frac{1}{2}*103kg*(1.71)^{2}+\frac{1}{2}*68.9kg*(1.09)^{2}+\frac{1}{2}*27.7kg*(1.45)^{2}

I_{total}=220.64 kg*m^{2}

5 0
3 years ago
3.7. A dog searching for a bone walks 3.50 m south, then 8.20 m at an angle 23.1 degrees north of east, and finally 15.0 m west.
andriy [413]

Answer:

Explanation:

We shall represent displacement of dog in vector form , in terms of i , j , i representing east  and  j representing north .

Dog travels 3.5 m south .

Displacement D₁ = - 3.5 j

then dog travels 8.2 m , 23.1 degree north of east

Displacement D₂ = 8.2 cos23.1 i + 8.2 sin23 j

D₂ = 8.2 cos23.1 i + 8.2 sin23.1  j

= 7.54 i + 3.22 j  

Third displacement

D₃ = - 15i

Total displacement = D₁ + D₂ + D₃

= - 3.5 j + 7.54 i + 3.22 j  -15i

= - 7.46 i - 0.28 j

Magnitude of displacement = √ ( 7.46² + .28²)

= √(55.65 + .08 )

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b ) Direction of displacement

If Ф be angle , displacement makes with west direction

TanФ =  .08 / 55.65 = .00143

Ф = .082 degree south of west or almost west .

From east , this angle = 180 + .082 = 180.082 , counterclockwise .

5 0
2 years ago
Describe how pieces of rope move as waves pass
MaRussiya [10]
Well, as the waves move it moves the rope as if its trying to take shape of it. Since the rope it light it will move along the ocean and the ocean will keep pushing up on the rope. (even without the waves the water is pushing the rope up so it can take its shape)

Maybe that'll help
6 0
3 years ago
Mechanical waves can travel through _______.
eimsori [14]

mediums only, mechanical waves always propagate through medium

5 0
2 years ago
Julie blows a bubble. At first, the pressure of the gas in the bubble is 4kPa. The bubble floats into the air and expands. When
Andrew [12]

Answer:

V₁ = 1.75 m³

Explanation:

Assuming the gas to be an ideal gas. At constant temperature, the relationship between the volume and temperature of an ideal gas is given by Boyle's Law as follows:

P_{1}V_{1} = P_{2}V_{2}

where,

P₁ = Initial Pressure of the Gas = 4 KPa

V₁ = Initial Volume of the Gas = ?

P₂ = Final Pressure of the Gas = 2 KPa

V₂ = Final Volume of the Gas = 3.5 m³

Therefore,

(4\ KPa)V_{1} = (2\ KPa)(3.5\ m^{3})\\\\V_{1}=\frac{2\ KPa}{4\ KPa}(3.5\ m^{3})\\\\

<u>V₁ = 1.75 m³</u>

4 0
3 years ago
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