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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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There is no air in space astronauts in space cannot hear sounds from outside their spacesuits explain this

Semmy [17]

<span> In </span>space<span>, where there is no air, sound has no way to travel.</span>

7 0

3 years ago

) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi

natali 33 [55]

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

4 0

3 years ago

PLZ HELP!!!!

noname [10]

I believe the answer would be B) ...... sorry if i'm wrong

4 0

3 years ago

Read 2 more answers

A charge of +3.0 mC is distributed uniformly along the circumference of a circle with a radius of 20 cm. How much external energ

Marat540 [252]

Answer:

Work done = 4584.9 J

Explanation:

given: q1=3.0 mC = 3.0 × 10⁻³ C, r = 20 cm = 0.20 m, q1 = 34μC = 34 × 10⁻⁶ C

Solution:

Formula for the potential difference at the center of the circle

P.E = K × q1 q2 /r (Coulomb's constant k= 8.99 × 10⁹ N·m² / C²)

P.E = 8.99 × 10⁹ N·m² / C² × 3.0 × 10⁻³ C × 34 × 10⁻⁶ C / 0.20 m

P.E = 4584.9 J = Work done

3 0

3 years ago

satellite is placed in an elongated elliptical (not circular) orbit around the Earth. At the point in its orbitwhere it is close

GrogVix [38]

Answer:

a) v2=4147.72 m/s

b) stotal=5.53x10^6 m

Explanation:

a) the length from the center of the earth is equal to:

L1=1x10^6+((6.37/2)x10^6)=4.18x10^6 m

the velocity is 5.14 km/s=5.14x10^3 m/s

the farthest distance is equal to:

L2=2x10^6+((6.37/2)x10^6)=5.18x10^6 m

As the angular momentum is conserved, we have to:

I1=I2

m*L1*v1=m*L2*V2, where m is the mass of satelite

clearing v2:

v2=(L1*V1)/L2=(4.18x10^6*5.14x10^3)/5.18x10^6=4147.72 m/s

b) Using the Newton 3rd law:

vf^2=vi^2+2as

where:

a=g=9.8 m/s^2

vf=0

vi=5.14 km/s

s=?

Clearing s:

s=(vf^2-vi^2)/(2g)=((0-(5.14x10^3)^2)/(2*9.8)=1.35x10^6 m

the total distance is equal to:

stotal=s+L1=1.35x10^6+4.18x10^6=5.53x10^6 m

8 0

2 years ago