Answer:
1. 6.672 kPa
2. 49.05 mm of mercury
Explanation:
h = 6400 m
Absolute pressure, p = 46 kPa = 46000 Pa
density of air, d = 0.823 kg/m^3
density of mercury, D = 13600 kg/m^3
(a) Absolute pressure = Atmospheric pressure + pressure due to height
46000 = Atmospheric pressure + h x d x g
Atmospheric pressure = 46000 - 6400 x 0.823 x 10 = 6672 Pa = 6.672 kPa
(b) To convert the pressure into mercury pressure
Atmospheric pressure = H x D x g
Where, H is the height of mercury, D be the density of mercury, g be the acceleration due to gravity
6672 = H x 13600 x 10
H = 0.04905 m
H = 49.05 mm of mercury
Answer:
The Answer is A Hope I helped you :D Have a Great Day!
Explanation:
Answer:
The length of the stick is 0.28 m.
The time the stick take to move is 0.97 ns.
Explanation:
Given that,
Relative speed of stick v= 0.96 c
Speed of light 
Proper length of stick = 1 m
We need to calculate the length of the stick
Using formula of length

Put the value into the formula



We need to calculate the time the stick take to move
Using formula of time

Put the value into the formula



Hence, The length of the stick is 0.28 m.
The time the stick take to move is 0.97 ns.
Answer:
the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers
Explanation:
The equilibrium wage rate and the equilibrium quantity of labor are found as the point where the equation of demand intercepts the equation of supply, so the equilibrium quantity of labor is:

15 - (1/200) L = 5 + (1/200) L
15 - 5 = (1/200) L + (1/200) L
10 = (2/200) L
(10*200)/2 = L
1000 = L
Then, the equilibrium wage rate is calculated using either the equation of demand for labor or the equation of supply of labor. If we use the equation of demand for labor, we get:
W = 15 - (1/200) L
W = 15 - (1/200) 1000
W = 10
Finally, the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers