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SSSSS [86.1K]
3 years ago
10

James gently releases a ball at the top of a slope But does not push the ball. Space the ball rolls down the slope. Which force

causes the ball to move downhill
Physics
2 answers:
NARA [144]3 years ago
7 0
On an incline, the force causing the ball to move downwards would be gravity. Additionally, the component of gravity causing this ball to move downwards would be mgsintheta.

Hope this helps!
SIZIF [17.4K]3 years ago
5 0

Answer:

m g Sin theta

Explanation:

If a body s placed on an inclined plane, then there are two component of weight of the body . One component is along the plane and the other is perpendicular to the plane.

Let there is an inclined plane which makes an angle theta with the horizontal. A body of mass m is placed on the inclined plane.

The component of weight along the plane is m g Sin theta

The component of weight perpendicular to the plane is m g Cos theta.

So, the force acting on the ball to move downhill is the force acting along teh plane. That means m g Sin theta.

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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?
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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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