Answer:
1st digestion
2nd Most chemical changes in digestion occur in the small intestine
Explanation:
Answer:
<u>Amplitude - remains the same</u>
<u>Frequency - increases</u>
<u>Period - decreases</u>
<u>Velocity - remains the same.</u>
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Explanation:
The amplitude of the wave remains the same since you are not changing the distance your hand moves and the amplitude of the wave depends on how much distance your hand covers while moving.
The frequency of your wave increases since now you are moving your hand more number of times in the same period i.e. your hand is moving faster in one second. So, the frequency of your wave increases.
The period is the time taken by the wave to travel a certain distance. Since your hand is now moving faster, the wave will travel faster and will take less time to cover the same distance hence, we can say that its period will decrease.
The velocity of a wave depends on the medium in which it is travelling. Your wave was previously travelling in air and the new wave is also travelling in the same medium so the velocity of the wave remains unchanged.
Both speed and velocity measure how fast something is moving. Hence, since speed is not a vector it does not require direction.
Speed only depends on the distance covered by the object and how much time it took to cover that distance. It does not care about the direction, it just depend on the length of the entire path covered. Whereas, velocity depends on magnitude as well as the direction of the object's motion for example: When an airplane is traveling, It is better to deal the motion with velocity and when we are moving in a car inside the city, we should deal the motion with speed.
Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
