Answer:
Potential Energy
Explanation:
Potential energy is the energy stored in an object due to it's position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height.
Answer:
E = 2.7 x 10¹⁶ J
Explanation:
The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:
where,
E = Energy Released = ?
m = mass of material reduced = 0.3 kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
<u>E = 2.7 x 10¹⁶ J</u>
The density of the nickel was greater than that of the quarter and penny, thus, the results supports the hypothesis.
<h3>What is density of substance?</h3>
The density of a substance is a measure of how tightly-packed the particles of the substance are.
Density is calculated as the ratio of the mass of the substance and the volume of the substance.
The hypothesis of the lab to compare the densities of a penny, a nickel, and a quarter is:
- If the nickel has a greater density than the quarter and penny, then it will have a greater mass to volume ratio. If the nickel has a lower density than the quarter and penny, then it will have a lower mass-to-volume ratio.
The average mass and the average volume of a penny, a nickel, and a quarter are then used to determine the density of each coin.
Based on obtained results, it would be found that the density of the nickel was greater than that of the quarter and penny. Therefore, the results supports the hypothesis.
In conclusion, the density of a substance depends on the mass and the volume.
Learn more about density at: brainly.com/question/1354972
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Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
