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REY [17]
4 years ago
6

A space probe approaches a planet and goes into a low orbit. If the orbiting probe's velocity is

Physics
1 answer:
Nezavi [6.7K]4 years ago
4 0

here given that the velocity of the probe is

v_x = - v sin\omega t

v_y = v cos\omega t

now at initial position where t = 0

v_{xi} = - v sin0 = 0

v_{yi} = v cos 0 = v

Now after t = 24 minutes we need to find final components of velocity

v_{xf} = - v sin(1.20* 10^{-3} * 24*60) = -v sin(1.728)

v_{yf} =  v cos(1.20* 10^{-3} * 24*60) = v cos(1.728)

now as we know that acceleration is given as

a = \frac{v_f - v_i}{t}

Now for x direction of motion

a_x = \frac{v_{xf} - v_{xi}}{t}

a_x = \frac{- v sin(1.728) - 0}{24*60}

a_x = \frac{-7.77*10^3 * 0.99}{24*60}

a_x = -5.33 m/s^2

Now for y direction of motion

a_y = \frac{v_{yf} - v_{yi}}{t}

a_y = \frac{ v cos(1.728) - v}{24*60}

a_y = \frac{7.77*10^3 * (-0.16) - 7.77 * 10^3}{24*60}

a_y = -6.24 m/s^2

now in order to find the magnitude of acceleration we can say

a = \sqrt{a_x^2 + a_y^2}

a = \sqrt{5.33^2 + 6.24^2} = 8.2 m/s^2

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A canoe, which has a speed of 2.40 m/s in still water, is headed North as it crosses a river flowing East at 1.60m/s. What is th
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The resultant velocity of the canoe is 2.88 m/s.

The Resultant velocity of the river can be calculated using Pythagoras Theorem

<h3> </h3><h3>Pythagoras Theorem: </h3>

In a right angle triangle, The square of the hypotenuse is equal to the sum of the square of the two other sides.

Note: In a cardinal point, The north and the east are at right angle.

<h3>Formula:</h3>
  • a² = b²+c²............... Euqation 1
<h3 /><h3>Where:</h3>
  • a = resultant velocity of the canoe
  • b = speed of the canoe
  • c = velocity of the flowing river.

From the question,

<h3>Given:</h3>
  • b = 2.4 m/s
  • c = 1.6 m/s

Substitute these values into equation 1

  • a² = 2.4²+1.6²
  • a² = 5.76+2.56
  • a² = 8.32
  • a = √8.32
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Hence, the resultant velocity of the canoe is 2.88 m/s

Learn more about resultant velocity here: brainly.com/question/24767211

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