Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 210kPa + 101.325kPa
P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)
T = 25°C = 298.15K
Final P and T values:
P = ?, T = 0°C = 273.15K
Set the initial and final P/T values equal to each other and solve for the final P:
311.325/298.15 = P/273.15
P = 285.220kPa
Subtract 101.325kPa to find the final gauge pressure:
285.220kPa - 101.325kPa = 183.895271kPa
The final gauge pressure is 184kPa or 26.7psi.
Answer:
The potential energy (P.E) at the top is 392 J
The kinetic energy (K.E) at the top is 0 J
The potential energy (P.E) at the halfway point is 196 J.
The kinetic energy (K.E) at the halfway point is 196 J.
Explanation:
Given;
mass of the rock, m = 2 kg
height of the cliff, h = 20 m
speed of the rock at the halfway point, v = 14 m/s
The potential energy (P.E) and kinetic energy (K.E) when its at the top;
P.E = mgh
P.E = (2)(9.8)(20)
P.E= 392 J
K.E = ¹/₂mv²
where;
v is velocity of the rock at the top of the cliff = 0
K.E = ¹/₂(2)(0)²
K.E = 0
The potential energy (P.E) and kinetic energy (K.E) at the halfway point;
P.E = mg(¹/₂h)
P.E = (2)(9.8)(¹/₂ x 20)
P.E = 196 J
K.E = ¹/₂mv²
where;
v is velocity of the rock at the halfway point = 14 m/s
K.E = ¹/₂(2)(14)²
K.E = 196 J.
It takes 392 joules of work to lift it.
It has 392 joules of gravitational potential energy up there.
Answer: 4.9 g/cm^3
Explanation.
!) Data
mass = 49 g
Volume = 10 cm^3
2) Formula
Density = mass / volume
Density = 49 g / 10cm^3 = 4.9 g/cm^3
This is a problem where you only need to know the formula and subtitute the numbers to calculate.
The formula of density is the very same definition: mass per unit of volume.
So, as you had the mass and the volume you just need to plug the values in the formula density = mass / volume.
Answer:
11.6 g/m³
Explanation:
The complete question is,
The device shows the relative humidity at 22 degrees celsius. What's the water vapor density if the maximum water vapor in air at this temperature is 20 gram/cubic meter? a device showing that at 22 degrees celsius the relative humidity is 58%.
Solution:
saturation of water in air is showed by the relative humidity. As it is 58% and not 100%. Lets scale density that we have, if humidity is 100%.
density is 20 gram/m³ when the humidity is 100%
When humidity is 58%
0.58 x 20 = 11.6 g/m³
Therefore, density is 11.6g/m³
