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Bas_tet [7]
3 years ago
8

The nose of an ultralight plane is pointed south, and its airspeed indicator shows 28 m/s . the plane is in a 18 m/s wind blowin

g toward the southwest relative to the earth.
Physics
1 answer:
leva [86]3 years ago
6 0
<span>Here I think you have to find the velocity in x and y components where x is east and y is north
 So as air speed indicator shows the negative speed in y component and adding it in
  air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
 v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
 solving
  v= -12.7 m/s x-40.7 m/s y
 if magnitude of velocity or speed is required then
  speed= sqrt(12.7^2 + 40.7^2)
 speed= 42.63 m/s
 if angle is asked
  angle = arctan (40.7/12.7)
 angle = 72.67 degrees south of west</span>
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Svetlanka [38]

Answer:

a = 1.152s

b = 0.817 m

c = 7.29m/s

Explanation: let the following

From the first equation of linear motion

V = u+at..........1

parameters be represented as :

t = Time taken

v = Final velocity

a = Acceleration due to gravity = 9.8m/s²

u = Initial velocity = 4 m/s

s = Displacement

V = 0

Substitute the values into equation 1

0 = 4-9.8(t)

-4 = -9.8t

t = 4/9.8

t = 0.408s

From : s = ut+1/2at^2.........2

S = 4×0.408+0.5(-9.8)×0.408^2

S= 1.632-4.9(0.166)

S = 1.632-0.815

S = 0.817m

Her highest height above the board is 0.817 m

Total height she would fall is 0.817+1.90 = 2.717 m

From equation 2

s = ut+1/2at^2

2.717 m = 0t+0.5(9.8)t^2

2.717 m = 0+4.9t^2

2.717 m = 4.9t^2

2.717/4.9 = t^2

0.554 =t^2

t =√0.554

t = 0.744s

Hence, her feet were in the air for 0.744+0.408seconds

= 1.152s

Also recall from equation 1

V= u+at

V = 0+9.8(0.744)

V = 7.29m/s

Hence, the velocity when she hits the water is 7.29m/s

Finally,

a = 1.152s

b = 0.817 m

c = 7.29m/s

4 0
2 years ago
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On level ground a shell is fired with an initial velocity of 49.0 m/s at 70.0 โ above the horizontal and feels no appreciable ai
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Answer:

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Vertical component = 46.0 m/s

Explanation:

If we denote the initial velocity by <em>v</em> and the angle above the horizontal by <em>θ</em>,

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v_x = v\cos \theta

v_x = (49.0\text{ m/s})\cos\,(70.0^\circ) = 16.8\text{ m/s}

The vertical component is given by

v_y = v\sin\theta

v_x = (49.0\text{ m/s})\sin\,(70.0^\circ) = 46.0\text{ m/s}

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