To answer this item, it is assumed that the gas in the cylinder is ideal such that it follows the equation,
PV = nRT
when V is to be calculated,
V = nRT/P
V = (4)(0.0821 L.atm/molK)(300 K) / (400 kPa/101.325 kPa/atm)
V = 24.95 L
Thus, the volume of gas in the cylinder is 24.95 L.
Answer:
A. The energy is not truly lost: it transforms into forms that are not easily put to use.
Explanation:
This is due to the fact that when we use energy, it seems to have disappear, but it really hasn't. In fact, when we use energy, it is changing into another form that has to be converted in order for energy to be present once again.
Answer:
The pressure on the ground is about 9779.5 Pascal.
The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side. See more details further below.
Explanation:
Start with the formula for pressure
(pressure P) = (Force F) / (Area A)
In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

The density of water is 997 kg/m^3, so the mass of the barrel is:

and so the weight is

and so the pressure is

This answers the first part of the question.
The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.
Answer:
1.25 kgm²/sec
Explanation:
Disk inertia, Jd =
Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²
Disk angular speed =
ωd = 0.1047 * 30 = 3.1416 rad/sec
Hollow cylinder inertia =
Jc = 3.7 * 0.40² = 0.592 kgm²
Initial Kinetic Energy of the disk
Ekd = 1/2 * Jd * ωd²
Ekd = 0.148 * 9.87
Ekd = 1.4607 joule
Ekd = (Jc + 1/2*Jd) * ω²
Final angular speed =
ω² = Ekd/(Jc+1/2*Jd)
ω² = 1.4607/(0.592+0.148)
ω² = 1.4607/0.74
ω² = 1.974
ω = √1.974
ω = 1.405 rad/sec
Final angular momentum =
L = (Jd+Jc) * ω
L = 0.888 * 1.405
L = 1.25 kgm²/sec
Answer:
1. Count the atoms of each element in the reactants and the products.
2. Use coefficients; place them in front of the compounds as needed.
Explanation: