Answer: It doesn't because it's not that big.
Explanation:
The coldest part of the atmosphere is the mesosphere
We are given with conditions on temperature, pressure and volume of gas. In this problem, we apply ideal gas law PV = nRT to solve what are asked.
1. PV = nRT ; n = m/MM
16.5 x10^6 Pa * 0.003785 m3= n * 8.314 Pa m^3/ mol K * (23+273) K
n = 25.38 mol; mass = 812.08 grams.
2. V at STP is 22.4 L/mol
3. PV = nRT = 1.5 atm * 0.003785 *1000 L = <span>25.38 mol * 0.0821 L atm/mol K * T ; T = 2.72 K
4. P * 55 m3/1000 = 25.38 mol * </span><span>8.314 Pa m^3/ mol K</span><span> * (24+273) K
P = 1139. 45 kPa</span>
The buoyant force on the branch is given by:
F = pVg
F = buoyant force, p = water density, V = volume of branch submerged, g = gravitational acceleration
Given values:
p = 1000kg/m^3
V = 0.75x1.12m^3 (75% of total volume)
g = 9.81m/s^2
Plug in and solve for F:
F = 1000(0.75x1.12)(9.81)
F = 8240N
Answer:
Net force is 86.135 N , N 82.255° W
Explanation:
Resolving all forces vertically and horizontally.
First vertically.
20N = 20sin90 = 20
30N = 30sin(90-30)= 30sin60
=25.981
25N = 25sin0 = 0
35N =- 35sin80= -34.368
Total of vertical =
20+25.981+0-34.368
= 11.613N
For horizontal
20N =- 20cos 90 = 0
30N =- 30cos 30 = -25.981
25N = -25cos0 = -25
35N = -35cos 10 = -34.468
Total horizontal = 0-25.981-25-34.368= -85.349
Resultant force = √(11.613)²+(-85.349)²
= √(134.862) +(7284.452)
= √ 7419.314
= 86.135 N
Tan(tita°) = 11.613/-85.349
Tan(tita°) =- 0.136
tita° = Tan^-1(-0.136)
tita° = -7.745
North of West = 90-7.745
= 82.255°
