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3 years ago

One cubic meter (1.0 m3

1 answer:

8 0

Because the masses that you give are for blocks that are 1 cubic meter in volume, they also serve as the densities for the two metals that you are comparing.

mass = density*volume 

volume = (4/3)*pi*r^3 

volume of iron sphere = (4/3)*3.14*0.0201^3 = 3.40*10^-5 m^3 

mass of iron sphere = 7860* 3.40*10^-5 m^3 = 0.27 kg = mass of Aluminum Sphere 

Volume of Al Sphere = 0.27/2700 = 9.90*10^-5 m^3 

Radius = cube root (volume / (4/3) / pi) = 2.87 cm. 

I did this using the MS calculator, and I'm not 100% sure on the numerical answer, but the process is what you need to do to solve the problem. You should double check my answer.

hope this helped :)

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A mover brings a box up the stairs in 10 seconds. If he applied a force of 20 N over a distance 10 m on the box, calculate the p

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Answer:

20 Watts

Explanation:

Work = force × distance

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3 years ago

Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?

shtirl [24]

Answer:

$L=6.21m$

Explanation:

For the simple pendulum problem we need to remember that:

$\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0$ ,

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$\omega^{2}=\frac{g}{L}$ ,

where $\omega$ is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

$\omega=\frac{2\pi}{T}$ ,

where T is the oscillation period. Now, we can easily solve for L:

$(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m$

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3 years ago

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Hunter-Best [27]

Answer:

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3 years ago

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JulijaS [17]

The answer is definitely C.) their molecules move at the same average speed

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You move a 2.5 kg book from a shelf that is 1.2 m above the ground to a shelf that is 2.6 m above the ground. What is the change

Sophie [7]

The change in potential energy of an object is given by
$U=mg \Delta h$
where
m is the mass of the object
g is the gravitational acceleration
$\delta h$ is the increase in altitude of the object

In our problem, $m=2.5 kg$ is the mass of the book, $g=9.81 m/s^2$ and
$\Delta h=2.6 m -1.2 m=1.4 m$ is the increase in altitude of the book, so its variation of potential energy is
$U=mg\Delta h=(2.5 kg)(9.81 m/s^2)(1.4 m)=34.3 J$

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3 years ago