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3 years ago

How many 3.2cm paperclips could fit on a football field

1 answer:

5 0

3429 paperclips because a football field is 120 yards and 120 yards converted to centimeters is 10972.8cm so 10972.8cm divided by 3.2 is 3429.

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A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.460 Hz. The pendulum ha

zlopas [31]

Answer:

The moment of inertia is $I =1.0697 \ kg m^2$

Explanation:

From the question we are told that

The frequency is $f = 0.460 \ Hz$

The mass of the pendulum is $m = 2.40 \ kg$

The location of the pivot from the center is $d = 0.380 \ m$

Generally the period of the simple harmonic motion is mathematically represented as

$T = 2 \pi * \sqrt{ \frac{I}{ m * g * d } }$

Where I is the moment of inertia about the pivot point , so making I the subject of the formula it

=> $I = [ \frac{T}{2 \pi } ]^2 * m* g * d$

But the period of this simple harmonic motion can also be represented mathematically as

$T = \frac{1}{f}$

substituting values

$T = \frac{1}{0.460}$

$T = 2.174 \ s$

$I = [ \frac{2.174}{2 * 3.142 } ]^2 * 2.40* 9.8 * 0.380$

$I =1.0697 \ kg m^2$

4 0

3 years ago

How do I do this physics problem about potential energy and kinetic energy?

larisa86 [58]

Ok i apologise for the messy working but I'll try and explain my attempt at logic

Also note i ignore any air resistance for this.

First i wrote the two equations I'd most likely need for this situation, the kinetic energy equation and the potential energy equation.

Because the energy right at the top of the swing motion is equal to the energy right in the "bottom" of the swing's motion (due to conservation of energy), i made the kinetic energy equal to the potential energy as indicated by Ek = Ep.

I also noted the "initial" and "final" height of the swing with hi and hf respectively.

So initially looking at this i thought, what the heck, there's no mass. Then i figured that using the conservation of energy law i could take the mass value from the Ek equation and use it in the Ep equation. So what i did was take the Ek equation and rearranged it for m as you can hopefully see. Then i substituted the rearranged Ek equation into the Ep equation.

So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).

Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.

The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).

I do hope I'm right and i suppose it's better than a blank piece of paper good luck my dude xx

4 0

3 years ago

If the moon were twice as far from years as it is now the following would be true

zubka84 [21]

The question is incomplete.

The distance between the Moon and Earth influences: 1) the attractive gravitational force between them, 2) the tides, 3) the eclipses, 4) the period of each full turn of the moon around the Earth.

Assuming the question refers to the gravitational attraction, we must use the fact that, as per, Newton's Universal Gravitaional Law, the attractive force between the two bodies is inversely related to the square distance that separates them.

Then, if the Moon were twice as far, the gravitational pull would be one fourth (1/4) of actual pull.

7 0

3 years ago

Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to

katen-ka-za [31]

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

$OB=0.40 km$