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Yuki888 [10]
3 years ago
7

A model airplane is flying horizontally due north at 44 ​mi/hr when it encounters a horizontal crosswind blowing east at 44 ​mi/

hr and a downdraft blowing vertically downward at 22 ​mi/hr. a. Find the position vector that represents the velocity of the plane relative to the ground. b. Find the speed of the plane relative to the ground.
Physics
1 answer:
Alik [6]3 years ago
7 0

Explanation:

Let i, j and k represents east, north and upward direction respectively.

Velocity due north, v_a=44j\ mi/hr

Velocity of the crosswind, v_w=44i\ mi/hr

Velocity of downdraft, v_d=-22k\ mi/hr (downward direction)

(a) Let v is the position vector that represents the velocity of the plane relative to the ground. It is given by :

v=44i+44j-22k

(b) The speed of the plane relative to the ground can be calculated as :

v=\sqrt{44^2+44^2+22^2}

v = 66 m/s

Hence, this is the required solution.

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NeTakaya

Answer:

Value of magnitude of acceleration will be 9.8m/sec^2

Explanation:

It is given that when a stone is dropped it takes 2.2 sec to reach the ground

(a) As the stone is dropped from the top of building

So its initial velocity u_i will be 0 m /sec

(b) As the stone is free falling and there is no external force applied on it so its acceleration will be equal to acceleration due to gravity

So value of magnitude of acceleration will be equal to 9.8m/sec^2

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A radar station sends out a 250000 Hz sound wave at a speed of 340 m/s. The sound wave bounces off a weather ballon and returns
Eddi Din [679]

Answers:

a)The balloon is 68 m away of the radar station

b) The direction of the balloon is towards the radar station

Explanation:

We can solve this problem with the Doppler shift equation:

f'=\frac{V+V_{o}}{V-V_{s}} f  (1)

Where:

f=250,000 Hz is the actual frequency of the sound wave

f'=240,000 Hz is the "observed" frequency

V=340 m/s is the velocity of sound

V_{o}=0 m/s is the velocity of the observer, which is stationary

V_{s} is the velocity of the source, which is the balloon

Isolating V_{s}:

V_{s}=\frac{V(f'-f)}{f'}  (2)

V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}  (3)

V_{s}=-14.16 m/s (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.

Now that we have the velocity of the balloon (hence its speed, the positive value) and the time (t=4.8 s) given as data, we can find the distance:

d=V_{s}t (5)

d=(14.16 m/s)(4.8 s) (6)

Finally:

d=68 m (8) This is the distance of the balloon from the radar station

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A block is at rest on a incline plane as shown in the diagram.As angle is increased, the componet of the blocks weight parallel
Elodia [21]

Answer:

The component of block weight parallel to the plane, Wₓ = W cosθ

Explanation:

Let the weight of the block due to gravitation is W

The direction of the weight is vertically down

Let θ be the angle formed with the vertical weight of the block and the incline.

Taking two components of weight one along the vertical weight and another component perpendicular to it.

Then the component `of weight long the parallel of the plane is

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6 0
3 years ago
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