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seraphim [82]
3 years ago
5

How much force is needed to accelerate a 100 kg mass at a rate of 2.5 m/s^2

Physics
1 answer:
krok68 [10]3 years ago
4 0
F= x
M=100
A=2.5

X= (100) (2.5)

x= 250 N!
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An airplane changes its velocity uniformly from 150 m/s to 60 m/s in 15 s. Calculate:
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Does the earths magnetic field change with time?
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A stone is thrown horizontally at 60.0 m/sm/s from the top of a very tall cliff. Calculate its horizontal position and vertical
svp [43]

Answer:

X-Positions:                                         Y-Positions

x(0) = 0                                                   y(0) = 0

x(2) = 120 m                                           y(2) = 19.6 m

x(4) = 240 m                                          y(4) = 78.4 m

x(6) = 360 m                                          y(6) = 176.4 m

x(8) = 480 m                                          y(8) = 313 m

x(10) = 600m                                         y (10) = 490 m

Explanation:

X-Positions

  • First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
  • After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
  • So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:

       x = v_{ox} * t = 60.0 m/s * t(s)

  • It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.

Y-Positions

  • We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
  • As both axes are  perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
  • At any moment, it is subject to the acceleration of gravity, g.
  • As the acceleration is constant, we can find the vertical displacement (taking the  height of the cliff as the initial reference level), using the following kinematic equation:

       y = \frac{1}{2} * g* t^{2} = \frac{1}{2} * 9.8 m/s2 * t(s)^{2}

  • Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
  • y(2) = 2* 9.8 m/s2 = 19.6 m
  • y(4) = 8* 9.8 m/s2 = 78.4 m
  • y(6) = 18*9.8 m/s2 = 176.4 m
  • y(8) = 32*9.8 m/s2 = 313.6 m
  • y(10)= 50 * 9.8 m/s2 = 490.0 m
5 0
3 years ago
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* taking the distance traveled and divide it by the time spent

* taking the velocities in each time interval and then finding the weighted average by the time fraction

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Let's apply this last equation

               

Total time is

               t = t₁ + t₂

               t = 10 + 10 = 20   min

              v_average = 10/20 10 + 10/20 20

              v_average = 10/2 + 20/2

              v_average = 15 m / s

7 0
3 years ago
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