An activity that is relatively short in time <10 seconds and has few repetitions predominantly uses the ATP/PC energy system. The cellular respiration procedure that changes food energy into ATP which is a form of energy is largely reliant on oxygen obtainability. During exercise the source and request of oxygen obtainable to muscle is unnatural by period and strength and by the individual’s cardiorespiratory suitability level.
Steps of the ATP-PC system:
1. Primarily, ATP kept in the myosin cross-bridges which is microscopic contractile parts of muscle is broken down to issue energy for muscle shrinkage. This action consents the by-products of ATP breakdown which are the adenosine diphosphate and one single phosphate all on its own.
2. Phosphocreatine is then broken down by the enzyme creatine kinase into creatine and phosphate.
3. The energy free in the breakdown of PC permits ADP and Pi to rejoin creating more ATP. This newly made ATP can now be broken down to issue energy to fuel activity.
The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm
<u>Explanation:</u>
H₂ + Br₂ ⇒ 2HBr
PH₂ = 0.782atm
PBr₂ = 0.493atm
Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹
At equilibrium:
Let 2x = pressure of HBr
PH₂ = 0.782 -x
PBr₂ = 0.493 - x
Kp = (2x)^2 / (0.782-x)(0.493-x)
Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.
Then,
Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)
x = 1.2X10⁻¹¹
PHBr = 2x = 2.4 X 10⁻¹¹ atm
Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm
Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C <em>in kelvin</em> t1=75+273
t1=348K
T2=130°C <em>in kelvin</em> t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
<em>putting values:</em>
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
<em>by simplifying:</em>
Tfinal=363K
As an object accelerates i.e., change it's velocity(either direction or speed), the position of the object depends on two factor; If the acceleration was direction based then it might have a zero displacement for eg: if it travels in circle. or it might have a net displacement if it travels in a straight line, quantitatively
where,
s = displacement
u = initial velocity
v = final velocity
a = acceleration
t = time
Now, for the hypothesis;
There is no direct relationship between fan speed and acceleration but anyways generally speaking if we do have a relationship that with more fan speed we have a larger displacement of air i.e., a more force i.e., greater acceleration
Thus, it can be said, well not exactly scientific, that with a greater fan speed there will be greater acceleration. if fan speed is increased then acceleration will be more.
:)
Answer:
a) (95.4 i^ + 282.6 j^) N
, b) 298.27 N 71.3º and c) F' = 298.27 N θ = 251.4º
Explanation:
a) Let's use trigonometry to break down Jennifer's strength
sin θ = Fjy / Fj
cos θ = Fjx / Fj
Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be
T = 90 -32 = 58º
Fjy = Fj sin 58
Fjx = FJ cos 58
Fjx = 180 cos 58 = 95.4 N
Fjy = 180 sin 58 = 152.6 N
Andrea's force is
Fa = 130.0 j ^
We perform the summary of force on each axis
X axis
Fx = Fjx
Fx = 95.4 N
Axis y
Fy = Fjy + Fa
Fy = 152.6 + 130
Fy = 282.6 N
F = (95.4 i ^ + 282.6 j ^) N
b) Let's use the Pythagorean theorem and trigonometry
F² = Fx² + Fy²
F = √ (95.4² + 282.6²)
F = √ (88963)
F = 298.27 N
tan θ = Fy / Fx
θ = tan-1 (282.6 / 95.4)
θ = tan-1 (2,962)
θ = 71.3º
c) To avoid the movement they must apply a force of equal magnitude, but opposite direction
F' = 298.27 N
θ' = 180 + 71.3
θ = 251.4º
