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Helga [31]
3 years ago
11

If you double the dimensions of a cube what happens to the volume

Physics
1 answer:
Triss [41]3 years ago
8 0
By definition, a cube is a three-dimensional figure that have equal dimensions for all its sides. It comprises of two square bases, one on top and one on the bottom. The face sides are also squares. Therefore, the volume of a cube is equal to s³, where s is the measure of the side's length. To compare the change, let us assume values. First, suppose s=1. Then, we denote this volume as V₁.

V₁ = (1)³ = 1

Next, taking the double, s=2. The volume for this is denoted as V₂.

V₂ = (2)³ = 8

Taking the ratio of V₂ to V₁:

V₂/V₁ = 8

That means the scale factor is 8. When the side dimensions is doubled, the volume of the cube increases 8 times as great as the original volume.
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A bucket of water of mass 14.6 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.320
Svetach [21]

Answer:

a. T = 39.41 N

b. t = 1.76s

c. 150.78 N

Explanation:

Given:

Mass of bucket of water, Mb = 14.6 kg

Mass of cylinder, Mc = 11.1 kg

Diameter of cylinder, D =  0.320 m, or radius, r = D/2 = 0.16m

Displacement of the bucket from the top, that is the vertical displacement , y = 11.0 m

a. The tension in the rope while the bucket is falling is:

F = mg - T = ma

Where F= The force

m= mass

g= Acceleration due to gravity

T = tension in the rope

a = acceleration

T= m(g - a)

Then, calculating the angular acceleration of the pulley system in relation to its radial acceleration

T= 1/2Ma

Merging the two final equation so as to solve for a

M(g - a) = 1/2Ma

Make a the subject of the formula

Mg - Ma = 1/2Ma

1/2Ma + Ma = Mg

a (1/2 M + M) = Mg

Divide both side by (1/2 M + M)

a = Mg ÷ (1/2 M + M)

Inputing the given value in the formula above

g= 9.8m/s2

a = (14.6 kg) (9.8m/s2) ÷ 1/2 (11.1 kg) + 14.6 kg

a = 7.1007m/s2

Now it is easy to input the value into T= 1/2Ma

T = 1/2 (11.1 kg) (7.1007m/s2) = 39.41 N

B. Time of fall is:

Using one of the equation of motion

s = ut + 1/2 at^2

U = Initial velocity

t = time

a = acceleration

s= distance in this case displacement y

making t the subject of the formula

t = √(2s ÷ a)

u is 0 since the bucket starts from rest

so, t = √((2)(11.0 m) ÷ 7.1007m/s2)

t = 1.76s

c.  the force exerted on the cylinder by the axle = T + Mg

  = 42 N + (11.1 kg) (9.8m/s2)

= 150.78 N

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2 years ago
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Answer:

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Explanation:

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Answer:

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r =10 \sqrt{(Cos^{2} 6t)+(Sin^{2} 6t)

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