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Leona [35]
3 years ago
10

Gulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells.

If a seagull drops a shell from rest at a height of 16 m, how fast is the shell moving when it hits the rocks
Physics
1 answer:
kumpel [21]3 years ago
8 0

Answer:

v = 17.71 m / s

Explanation:

We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity

                v² = v₀² - 2 g (y -y₀)

                v² = 0 - 2g (y -y₀)

when it hits the stone the height is zero and part of the height of the seagull I

              v² = 2g y₀

              v = Ra (2g i)

let's calculate

              v =√ (2 9.8 16)

              v = 17.71 m / s

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A car initially traveling at 17.1 mph comes to rest in 9.7s what was its acceleration in this time?
ra1l [238]

Answer:

a=-.78m/s^2

Explanation:

Δv=at

  • Δv is the difference in velocity before and after a given time.
  • a is the acceleration of the object during this time.
  • t is time

(v_f-v_i)=at is another way to write this equation.

  • The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δv=(v_f-v_i)

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

  • I rearranged this equation to solve for a, but this is a step that you don't need to take, it's just good to get in the habit of doing this.
  • Plug in the given values. Note that our final velocity is 0, because the car travels until at <em>rest</em>.

a=\frac{v_f-v_i}{t}\\a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}

  • Our initial velocity is in mph, something not in standard units, so if not changed, you will get an incorrect answer. What you need to do is cancel out the units your prior value had using division and multiplication, and at the same time multiply and divide the correct numbers and units into your equation. Or look up a converter.

a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}\\a=\frac{0m/s-7.6m/s}{9.7s} \\a=\frac{-7.6m/s}{9.7s}

  • if you converted correctly, your answer for v_f will be ≅ 7.6m/s.
  • Now divide. Notice that the units for acceleration are m/s^2 or <em>meters per second, per second</em>.

a=\frac{-7.6m/s}{9.7s}\\a=-.78m/s^2

  • Our final answer is <em>negative </em>because the car is <em>slowing down</em>. Do not square this answer as the square symbol only applies to the units, not the magnitude.
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