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andriy [413]
3 years ago
10

A small mailbag is released from a helicopter that is descending steadily at 1.20 m/s. (a) After 5.00 s, what is the speed of th

e mailbag?
Physics
1 answer:
gladu [14]3 years ago
3 0

Answer:

50.2m/s

Explanation:

Using first equation of motion we have

V= u+gt

V= final speed

U= initial speed

So v = 1.2m/s + 9.8(5s)

50.2m/s

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A skateboarder travels on a horizontal surface with an initial velocity of 4.0 m/s toward the south and a constant acceleration
Alenkinab [10]

Answer:

a) 0.32 m b) -2.4 m c) 1.08 m/s d) -4 m/s

Explanation:

a)

  • As the x and y axes (as chosen) are perpendicular each other, the movements along these axes are independent each other.
  • This means that we can use the kinematic equations for displacements along both axes.
  • In the x direction, as the only initial velocity is in the south direction (-y axis), the skateboarder is at rest, so we can write:

        x =\frac{1}{2}*a*t^{2} (1)

  • In the y-direction, as no acceleration is acting on the skateboarder, we can write  the following displacement equation:

        y = v_{0y} * t  (2)

  • For t = 0.6s, replacing by the givens, we get the position (displacement from the origin) on the x-axis, as follows:

       x =\frac{1}{2}*a*t^{2} =\frac{1}{2} * 1.8 m/s2*(0.6s)^{2}\\ x = 0.32 m

b)

  • From (2) we can get the position on the y-axis (displacement from the origin) as follows:

        y = v_{0y} * t  =  -4 m/s * 0.6 s = -2.4 m

c)

  • In the x- direction, we can find the component of the velocity along this direction, as follows:

        v_{fx} = a*t

  • Replacing by the values, we have:

        v_{fx} = a*t = 1.8 m/s2 * 0.6 s = 1.08 m/s

d)

  • As the skateboarder moves along the y-axis at a constant speed equal to her initial velocity, we  have:

        vfy = voy = -4 m/s

8 0
4 years ago
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