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Express the dimensional analysis of power

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Answers

2 parts
a

125 m125 \, \text{m}

b

50 m/s50 \, \text{m/s}

Step-by-step solution

6 steps

Here's the solution for question 7:

Given:
Initial velocity, u=0 m/su = 0 \, \text{m/s} (since the stone is dropped)
Time, t=5 st = 5 \, \text{s}
Acceleration due to gravity, g=10 m/s2g = 10 \, \text{m/s}^2

Part a· 3 steps

How deep is the well?

  1. Use the kinematic equation for displacement

    s=ut+12gt2s = ut + \frac{1}{2}gt^2

  2. Substitute the given values into the equation

    s=(0 m/s)(5 s)+12(10 m/s2)(5 s)2s = (0 \, \text{m/s})(5 \, \text{s}) + \frac{1}{2}(10 \, \text{m/s}^2)(5 \, \text{s})^2

  3. Calculate the depth

    s=0+12(10 m/s2)(25 s2)s = 0 + \frac{1}{2}(10 \, \text{m/s}^2)(25 \, \text{s}^2) s=5 m/s2×25 s2s = 5 \, \text{m/s}^2 \times 25 \, \text{s}^2 s=125 ms = 125 \, \text{m}

    The depth of the well is 125 m\boxed{125 \, \text{m}}.

Part b

50 m/s50 \, \text{m/s}

What is the velocity when it hits the water surface?

  1. Use the kinematic equation for final velocity

    v=u+gtv = u + gt

  2. Substitute the given values into the equation

    v=0 m/s+(10 m/s2)(5 s)v = 0 \, \text{m/s} + (10 \, \text{m/s}^2)(5 \, \text{s})

  3. Calculate the final velocity

    v=50 m/sv = 50 \, \text{m/s}

    The velocity when it hits the water surface is 50 m/s\boxed{50 \, \text{m/s}}.

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