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3 years ago

What is the surface area to volume ratio of this cube

1 answer:

3 0

I can't see that cube from here.

But if the length of the side of the cube is ' K ' units,
then the surface area of the cube is 6K² units², and
the volume of the cube is K³ units³.

The ratio of the surface area to the volume is

(6K² units²) / (K³ units³) = (6) / (K units) .

So for example, if the side of the cube is 2 inches, then
the ratio of surface area to volume is "3 per inch".

That's the answer. I did the whole thing in order to earn
the points, but I don't expect you to understand much of it,
because I see from your username that you suck at math.
I'm sorry you decided that. Now that you've put up the
brick wall, it'll be even harder for any math to find its way
in there, and you'll miss out on a lot of the fun.

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A turntable with a moment of inertia of 7.2 × − ⋅ rotates freely with an angular speed of 6.5 ⁄ . Riding on the rim of the turnt

irina1246 [14]

Answer:

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Explanation:

6 0

2 years ago

Is the maximum displacement of a point on a vibrating body or wave.

Georgia [21]

The correct answer is amplitude

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3 years ago

A mass of 8 kg moves at a rate of 14.3 m/sec. what is the KE developed by the mass?

Lynna [10]

Kinetic energy is the energy associated with the motion of an object. It's a scalar quantity, there is no direction associated with KE and it has no components. $KE = \frac{mv ^{2} }{2} = \frac{8 *14.3 ^{2} }{2} = 4 *204.49 = 817.96J$ . Therefore Kinetic energy is 817.96J.

6 0

3 years ago

The Doppler effect occurs when a source of sound moves

Dmitrij [34]

The Doppler effect occurs when a source of sound or light
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4 0

3 years ago

Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const

bagirrra123 [75]

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

$S = ut + \frac{1}{2}at^{2}$

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

Learn more about distance, velocity and acceleration at: brainly.com/question/14344386

#SPJ1

7 0

2 years ago