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makvit [3.9K]
3 years ago
13

Assume you have measured your little metal sphere's diameter to be 2.500 ± 0.005 cm and its mass to be 63.00 ± 0.05 g.

Physics
1 answer:
TiliK225 [7]3 years ago
5 0

Answer:

V = (8.2 ± 0.1) cm³ ,    % = 1.2%

Explanation:

The volume of the sphere is

             V = 4/3 π r³

     

The error in volume is

           ΔV = 4 / 3 π r² Δr

           DV = 4 π r² Δr

Let's calculate

         V = 4/3 π (2.500 / 2)³

         V = 8,181 cm³

         ΔV = 4 π 2,500²/4    0.005

         ΔV = 0.098 cm³

         V = (8.2 ± 0.1) cm³

The equation for the percentage error is

     % = ΔV / V 100

      % = 0.1 / 8.2 100

       % = 1.2%

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The reason I never forgot it is because it's SO useful SO often.  You really should memorize it.  And don't bury it too deep in your toolbox ... you'll be needing it again very soon. (In fact, if you had learned it the first time you saw it, you could have solved this problem on your own today.)

The problem doesn't tell us what planet this is happening on, so let's make it easy and just assume it's on Earth.  Then the 'acceleration' is Earth gravity, and that's 9.8 m/s² .

In 5 seconds:

D = 1/2 A T²

D = (1/2) (9.8 m/s²) (5 sec)²

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D = 122.5 meters


In 6 seconds:

D = 1/2 A T²

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"Two uniform identical solid spherical balls each of mass M and radius R" and moment of inertia about its center 2/5 MR2 are rel
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Answer:

he sphere that uses less time is sphere A

Explanation:

Let's start with ball A, for this let's use the kinematics relations

        v² = v₀² - 2g (y-y₀)

indicate that the sphere is released therefore its initial velocity is zero and when it reaches the floor its height is zero y = 0

         v² = 0 - 2 g (0- y₀)

         v = \sqrt{2g y_o}

         v = \sqrt{2 \ 9.8\ H}

         v = 4.427 √H

Now let's work the sphere B, in this case it rolls down a ramp, let's use the conservation of energy

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         Em₀ = U = m g y

final point. At the bottom of the ramp

         Em_f = K = ½ m v² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

          Em₀ = Em_f

          mg H = ½ m v² + ½ I w²

angular and linear velocity are related

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the momentorot of inertia indicates that it is worth

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           v = \sqrt{\frac{10}{7} \ g H}

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we can see that L> H

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