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wlad13 [49]
3 years ago
12

Platinum has a density of 21.4 g/cm3. if thieves were to steal platinum from a bank using a small truck with a maximum payload o

f 900 lb, how many 1 l bars of the metal could they take?
Physics
1 answer:
slega [8]3 years ago
4 0
We will convert the 1dm3 in terms of cm3 as follows:

1dm^3 = (10 cm)^3
= 1000 cm^3

The mass of platinum is equal to 900 lb. 
Then we will convert the mass in terms of grams as follows:
1 lb = 453.6 g
900 = 900 x 453.6 g
= 408240 g

Then density of platinum is equal to 21.4 g/cm^3 
We will calculate the volume of platinum in mass 408240 g as follows:
Volume of platinum = mass of platinum / density of platinum
= 408240 g / 21.4 g/cm^3 
= 19076.6 cm^3  

The total volume of platinum is 19076.6 cm^3
The volume of platinum in 1 L bar is 1000cm^3
So, to calculate the number of bars we will use the formula as follows;
Number of bars = volume of platinum available / volume of platinum required in 1 L bar
= 19076.6 cm^3 / 1000 cm^3
= 19
So, the number of bars are 19.
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Answer:

Bulb 1 has more resistance.

Explanation:

Given that,

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For bulb 1,

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2 years ago
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
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Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

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final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

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where h is the difference in height between the two sides of the ramp

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         mg h = \frac{7}{5} (\frac{1}{2} m v²)

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This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

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   final point

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          mg y₁ = ½ m v² + m g y₂

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this is the speed of the box

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           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

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