Answer:
Approximately 
, if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.
Explanation:
Let 
and 
denote the pressure of this gas before and after the changes.
Let 
and 
denote the volume of this gas before and after the changes.
Let 
and 
denote the temperature (in degrees Kelvins) of this gas before and after the changes.
Let 
and 
denote the quantity (number of moles of gas particles) in this gas before and after the changes.
Assume that this gas is an ideal gas. By the ideal gas law, the ratios 
and 
should both be equal to the ideal gas constant, 
.
In other words:
.
.
Combine the two equations (equate the right-hand side) to obtain:
.
Rearrange this equation for an expression for , the temperature of this gas after the changes:
.
Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: 
, 
.
![\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20T_2%20%26%3D%20%5Cfrac%7BP_2%7D%7BP_1%7D%20%5Ccdot%20%5Cfrac%7BV_2%7D%7BV_1%7D%20%5Ccdot%20%5Cfrac%7Bn_1%7D%7Bn_2%7D%5Ccdot%20T_1%20%5C%5C%5B0.5em%5D%20%26%3D%20%5Cfrac%7B1.67%5C%3B%20%5Crm%20atm%7D%7B1.12%5C%3B%20%5Crm%20atm%7D%20%5Ctimes%20%5Cfrac%7B11.2%5C%3B%20%5Crm%20m%5E%7B3%7D%7D%7B15.7%5C%3B%20%5Crm%20m%5E%7B3%7D%7D%20%5Ctimes%201%20%5Ctimes%20245%5C%3B%20%5Crm%20K%20%5C%5C%5B0.5em%5D%20%26%5Capprox%20261%5C%3B%20%5Crm%20K%5Cend%7Baligned%7D)
.
Answer:
Applied pressure is 1.08 10⁵ Pa
Explanation:
This exercise is a direct application of Boyle's law, which is the application of the state equation for the case of constant temperature.
PV = nR T
If T is constant, we write the expression for any two points
Po Vo = p1V1
From the statement the initial pressure is the atmospheric pressure 1.01 10⁵ Pa, so we clear and calculate
1 Pa = 1 N / m2
P1 = Po Vo / V1
P1 = 1.01 10⁵ 20/18.7
P1 = 1.08 10⁵ Pa
The distance the bird flies is (17,000 + 600 + 14,400) = 32,000 km.
His average speed is (32,000 km)/(122 days) = 262.3 km/day.
The bird ends up (17,000 - 600 + 14,400) = 30,800 km north
from where he started.
His average velocity is (30,800 km)/(122 days) = 252.5 km/day north .
The gravitational potential energy has the formula:
PE = mgh
where PE is in joules
m is the mass of the object in kg
g is the acceleration due to gravity
h is the height in m
since you did not give the value of h, i cannot fully answer your quesstion.
PE = (58/1000 kg) (9.81m/s2) h
PE = 0.57h
just substitute the value of h in the equation
............................................................................... Hello wonderful person <3
