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Serhud [2]
3 years ago
14

A solid sphere and a solid cube are made from the same material. The sphere would just fit within the cube, if it could. Both be

gin at the same temperature, and both are heated to the same temperature. Which object, if either, has the greatest change in volume?a. Both have the same change in volume.b.The cube.c. Insufficient information is given for an answer.d. The sphere.
Physics
1 answer:
algol [13]3 years ago
3 0

Answer:

Explanation:

A solid is Sphere is somehow place inside a solid cube and the system is heated .

Material of cube and Sphere is same therefore coefficient of volume expansion is same for both shapes.

change in volume is given by

\Delta V=\gamma V_0\Delta T

where \Delta V=change\ in\ volume

\gamma =coefficient\ of\ volume\ expansion

\Delta T=Change\ in\ Temperature

V_0=Initial\ Volume

As initial volume of Cube is more than the sphere therefore Cube will expand more keeping other two factors same.

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A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co
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(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
  • The wheels under the wind do not pass through the center line.
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Position of the center mass of the plane is calculated as follows;

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<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

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