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faust18 [17]
3 years ago
13

An airplane is heading due south at a speed of 690 km/h . A) If a wind begins blowing from the southwest at a speed of 90 km/h (

average), calculate the velocity (magnitude) of the plane relative to the ground.B) Calculate the velocity (direction) of the plane relative to the ground.C) Calculate how far from its intended position will it be after 11 min if the pilot takes no corrective action.

Physics
1 answer:
Afina-wow [57]3 years ago
4 0

Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h  east. c) 16,5 km NE of the desired position

Explanation:

Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where

 

see fig 1

Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then

90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,

this because we have an isosceles triangle, then the cathetus length is  

hypotenuse/\sqrt{2}

using Pythagoras, here the hypotenuse is 90, then the cathetus are of length

90/\sqrt{2} km/h= 63,6396 km/h.  

Now the total speed of the plane is

690km/h south + 63,6396 km/h north +63,6396 km/h east,

this is 626,3604 km/h south + 63,6396 km/h east,  here north is as if we had -south.

then using again Pythagoras we get the magnitude of the total speed it is

\sqrt{626,3604 ^2+63,6396^2} km/h=629,5851km/h,

the direction is calculated with respect to the south using trigonometry, we know the

sin x= cathetus opposed / hypotenuse,

then

x= sin^{-1}'frac{63,6396}{629,5851}=5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form

626,6396 km/h south + 63,6396km/h  east.

Finally since the detour is caused by the west speed component plus the slow down caused by the north component of the wind speed, we get

Xdetour{east}= 63,6396 km/h* (11 min* h)/(60 min)=11,6672 km=Xdetour{north} ,

since 11 min=11/60 hours=0.1833 hours.

Then the total detour from the expected position, the one it should have without the influence of the wind, we get  

Xdetour=[/tex]\sqrt{2*  11,6672x^{2} }[/tex]  = 16,5km at 45 degrees from east pointing north

The situation is sketched as follows  see fig 2

 

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A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m
lbvjy [14]

The equation 15v_{i} + 2*0 = (15 + 2)v_{f} (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.  

The horizontal momentum is given by:

p_{i} = p_{f}

m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}

Where:

  • m₁: is the mass of the lab cart = 15 kg
  • m₂: is the <em>mass </em>of the object dropped = 2 kg
  • v_{1}_{i}: is the initial velocity of the<em> lab cart </em>
  • v_{2}_{i}: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
  • v_{1}_{f}: is the final velocity of the<em> lab cart </em>
  • v_{2}_{f}: is the <em>final velocity</em> of the <em>object </em>

Then, the horizontal momentum is:

15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

15v_{1}_{i} + 2*0 = v_{f}(15 + 2)

Therefore, the equation 15v_{i} + 2*0 = (15 + 2)v_{f} represents the horizontal momentum (option 3).

Learn more about linear momentum here:

  • brainly.com/question/2141713?referrer=searchResults
  • brainly.com/question/2400186?referrer=searchResults

I hope it helps you!            

4 0
3 years ago
5. The Weeks family took a trip to the Dallas 200 for
zheka24 [161]

Answer: idk

Explanation: hahahaha

4 0
3 years ago
A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,
fomenos

Answer:

The momentum would be doubled

Explanation:

The magnitude of the momentum of the freight train is given by:

p=mv

where

m is the mass of the train

v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

m'=2m

therefore, the new momentum is

p'=m'v=(2m)v=2(mv)=2p

so, the momentum has also doubled.

7 0
3 years ago
Q3. You throw a ball into the air, it reaches a certain height and then comes back to you.
Grace [21]

The If a car is going round a curve , there is an acceleration because the direction of the velocity changes.

<h3>What is the direction of the velocity?</h3>

Now we know that if you throw the ball upwards, the motion is in opposite direction to gravity thus the ball is experiencing deceleration and the speed decreases. The velocity decreases and the acceleration is negative.

If the ball is coming down, then the ball is accelerated thus it speeds up and the direction of the acceleration is positive.

If a car is going round a curve, the vehicle is accelerating because the direction of the velocity changes even if its amount remains constant.

When a board is moving down a hill at 2 ms-1, it is experiencing an acceleration because the motion is in the same direction as gravity.

If a car is coming to a stop at a point, it experiences a deceleration and not an acceleration since the change of velocity with time is negative as the car comes to rest.

Learn more about acceleration:brainly.com/question/12550364

#SPJ1

5 0
1 year ago
A spring stretches 3.5 cm when a 9 g object is hung from it. The object is replaced with a block of mass 26 g which oscillates i
evablogger [386]

Answer:

The period of motion  of new mass T = 0.637 sec

Explanation:

Given data

Mass of object (m) = 9 gm = 0.009 kg

Δx = 3.5 cm = 0.035 m

We know that spring force is given by

F = m g

F = 0.009 × 9.81 = 0.08829 N

Spring constant

k = \frac{F}{x}

k = \frac{0.08829}{0.035}

k = 2.522 \frac{N}{m}

New mass(m_1) = 26 gm = 0.026 kg

Now the period of motion is given by

T = 2 \pi \sqrt{\frac{m}{k} }

T = 2 \pi \sqrt{\frac{0.026}{2.522} }

T = 0.637 sec

This is the period of motion  of new mass.

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2 years ago
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