What happens when the temperature of a substance is slightly decreased?

Answer:

B Raise the object farther off the ground

Final answer:

To increase the gravitational potential energy of an object without altering its mass or gravity, you would need to raise the object to a greater height. The potential energy is determined by the object's mass, its height, and the gravitational force, as shown by the formula potential energy=mgh.

Explanation:

To increase the gravitational potential energy of an object without changing its mass and gravity, you could raise the object farther off the ground. This is because gravitational potential energy is a function of an object's mass, height, and acceleration due to gravity, as represented by the formula potential energy = mgh, m being mass, g being gravity, and h being height.

Moving the object to a greater height without acceleration or carrying the object with or without acceleration at the same height will not result in an increased potential energy. Only by raising the object to a higher position or altitude, you increase its potential energy.

In essence, the principle involves work done against the gravity. When an object is raised to a higher elevation, work is done against gravity. This work gets stored as potential energy in the object-Earth system. Thus, the higher the position of the object, the higher would be its gravitational potential energy.

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Final answer:

The electric fields at the points (5.0 cm, 0 cm), (-5.0 cm, 0 cm), and (0 cm, 5.0 cm) due to a -12 nC charge are -67.5x10^3 i^ N/C, 30x10^3 i^ N/C and -67.5x10^3 j^ N/C respectively.

Explanation:

The subject of this question is Electric Field, a part of physics. Specifically, we're looking at point charges and using the formula for the electric field due to a point charge, which is E = k|Q|/r², where k is the Coulomb's constant, Q is the charge and r is the distance to the point where the electric field is calculated.

1. At the position (5.0 cm, 0 cm), the distance from the charge is 4.0 cm. Thus, the electric field is E₁ = k|-12 x 10^(-9) C|/(0.04 m)² = -67.5 x 10^3 N/C. In terms of unit vectors, this is -67.5 x 10^3 i^ N/C, because the field points in the negative x direction.
2. At the position (-5.0 cm, 0 cm), the distance from the charge is 6.0 cm. Thus, the electric field is E₂ = k|-12 x 10^(-9) C|/(0.06 m)² = -30 x 10^3 N/C. In terms of unit vectors, this is 30 x 10^3 i^ N/C, because the field points in the positive x direction.
3. At the position (0 cm, 5.0 cm), the field is purely in the y-direction, and the point is 4.0 cm above the charge. Thus, the field is E₃ = k|-12 x 10^(-9) C|/(0.04 m)² = -67.5 x 10^3 N/C. In terms of unit vectors, this is -67.5 x 10^3 j^ N/C.

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By definition we have to:

Where,

Pabs: absolute pressure

Patm: atmospheric pressure

Pg: gage pressure

The atmospheric pressure is constant and its value is:

Then, by clearing gage pressure we have:

Substituting values we have:

Answer:

If the absolute pressure of a gas is 550.280 kPa, its gage pressure is:

D. 448.955 kPa

Answer:

Asymptotic-giant-branch stars have helium-burning shells inside the hydrogen-burning shells, whereas red-giant-branch stars have hydrogen-burning shells only. In either case, the accelerated fusion in the hydrogen-containing layer immediately over the core causes the star to expand.