To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.
The stagnation temperature can be defined as

Where
T = Static temperature
V = Velocity of Fluid
Specific Heat
Re-arrange to find the static temperature we have that



Now the pressure of helium by using the Adiabatic pressure temperature is

Where,
= Stagnation pressure of the fluid
k = Specific heat ratio
Replacing we have that


Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa
<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>
Answer:
They both work together in order to create a good experience for the passenger, but I think that the answer is changing acceleration because when the rollercoaster changes acceleration it gives it speed.
Explanation:
I really hope this helps.
Answer: quatrain
Explanation:
Reading the Shakespeare's "Sonnet 100", we can infer that the underlined section is referred to as a quatrain.
The quatrain simply refers to a type of stanza that is made up of four lines. For example, based on the information given, we can deduce that the rhyme scheme for the second quatrain is given as cdcd.
Answer:
Force per unit plate area is 0.1344 
Solution:
As per the question:
The spacing between each wall and the plate, d = 10 mm = 0.01 m
Absolute viscosity of the liquid, 
Speed, v = 35 mm/s = 0.035 m/s
Now,
Suppose the drag force that exist between each wall and plate is F and F' respectively:
Net Drag Force = F' + F''

where
= shear stress
A = Cross - sectional Area
Therefore,
Net Drag Force, F = 

Also
F = 
where
= dynamic coefficient of viscosity
Pressure, P = 
Therefore,


Answer:
C = 59.17 nF
Q = 2.6
Explanation:
given data
frequencies = 40k Hz
frequencies = 90k Hz
solution
we take here R, L C take in series
so cut off frequency is express as
Wc1 =
= 40000
wc2 =
= 90000
so here
wc2 - wc1 will be
wc2 - wc1 = 90000 - 40000 = 50000
so
= 50000
we consider here R is 500
so L =
L = 10 m H
and here total cut off frequency is
total cut off frequency = 40000 + 90000 = 130000
so capacitance will be
capacitance C is =
so C = 59.17 nF
quality factor Q will be
Q =
Q = 2.6