Answer:
Explanation:
P(v) = 16 / v + 10⁻³ v³
differentiating on both sides
dP / dt = - 16 / v² + 3 x 10⁻³ v²
For maxima and minima , the condition is
dP / dt = - 16 / v² + 3 x 10⁻³ v² = 0
v² = 160 / 3 x 10²
v² = 73 m/s
v = 8.54 m /s
To know the condition of minima
again differentiating
d²P / dt² = - 16 x -2 / v² + 6 x 10⁻³ x v
= 32 / v³ + 6 x 10⁻³ x v
= + ve quantity
So at v_p = 8.54 m /s , power consumption will be minimum .
(b) the exergy destroyed during this process.
A) The exergy of the refrigerant at the initial and final states are :
B) The exergy destroyed during this process is : - 1048.4397 kJ
Given data :
Mass ( M ) = 5 kg
P1 = 0.7 Mpa = P2
T1 = 60°C = 333 k
To = 24°C = 297 k
P2 = 100 kPa
A) Determine the exergy at initial and final states
At initial state :
U = 274.01 kJ/Kg , V = 0.034875 m³/kg , S = 1.0256 KJ/kg.k
exergy ( Ф ) at initial state = M ( U + P₂V - T₀S )
= 5 ( 274.01 + 100* 10³ * 0.034875 - 297 * 1.0256)
≈ - 135.5285 kJ
At final state :
U = 84.44 kJ / kg , V = 0.0008261 m³/kg, S = 0.31958 kJ/kg.k
exergy ( ( Ф ) at final state = M ( U + P₂V - T₀S )
= -51.96 kJ
B) Determine the exergy destroyed
exergy destroyed = To * M ( S2 - S1 )
= 297 * 5 ( 0.31958 - 1.0256 )
= - 1048.4397 KJ
Hence we can conclude that A) The exergy of the refrigerant at the initial and final states are : Initial state = - 135.5285 kJ, Final state = -51.96 kJ and The exergy destroyed during this process is : - 1048.4397 kJ
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Exergy of refrigerant-134a at initial and final states is obtained from property tables and by multiplying the mass of the refrigerant with its specific exergy at each state. The difference in exergy between the two states represents the exergy destroyed.
To solve the given question, we need the property values of
refrigerant-134a
at the initial and the final states.
At an initial state of 0.7 MPa and 60°C, the specific exergy for refrigerant-134a can be obtained from property tables which are standard in thermodynamics textbooks. Same for the final state at 0.7 MPa and 24°C, the specific exergy can be obtained from the same property tables.
The exergy of the refrigerant at the initial and the final states can be calculated by multiplying the mass of the refrigerant with its specific exergy at each state.
Exergy destruction during this process can be calculated using the relation between exergy change and exergy destruction. The exergy change of a system between initial and final states is equal to the difference of the exergy of the system at final and initial states.
Based on the second law of thermodynamics, the difference in exergy should be equal to the exergy destroyed during the process.
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The position vector of the bullet has components
The bullet hits the ground when , which corresponds to time :
The bullet travels 168 m horizontally, which would require a muzzle velocity such that
In the given physics problem, the bullet travels horizontally 168 meters before hitting the ground from a height of 1.4 meters. By calculating the time it takes for the bullet to fall to the ground due to gravity and then applying that time to the horizontal distance traveled, we find that the speed of the bullet when it exited the rifle was approximately 313.43 m/s.
The scenario defined is a classic Physics problem where an object is fired horizontally and falls to the ground due to gravity. We can calculate the horizontal speed of the bullet using the equations of motion associated with the vertical, free-fall motion of the bullet.
Gravity causes the bullet to fall to the ground. As we know that the height from the ground is 1.4 meters, we can calculate the time taken for the bullet to hit the ground using the equation: time = sqrt(2 * height / g), where g is the gravitational constant (approx. 9.8 m/s^2).
Substituting the given value, we get time = sqrt(2 * 1.4 / 9.8), which is around 0.536 seconds. The bullet travels 168 meters in this time horizontally, therefore its horizontal speed will be distance / time, which is 168 meters / 0.536 seconds = 313.43 m/s. So, Madelin's bullet had a speed of around 313.43 m/s when it exited the rifle.
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Answer:
Explanation:
Given
Wavelength of incoming light
We know
Energy associated with this frequency
where h=Planck's constant
Energy of one mole of Photon
To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.
To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).
Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.
Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).
Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).
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While a power supply tester can be a useful tool for quickly checking voltage output, it might not reveal all the potential issues a faulty power supply can cause.
Even if a power supply tester shows that the voltage output of a power supply is within acceptable limits, it's still possible that the power supply may be faulty. Here's why:
1. Voltage Under Load: A power supply tester might only measure the voltage output under no load or very light load conditions.
A faulty power supply might provide the correct voltage at low loads but fail to deliver stable voltage under high loads, which could lead to system instability or crashes.
2. Voltage Ripple and Noise: Power supplies are expected to provide a stable and clean output voltage.
3. Short Circuits or Overloads: A power supply tester typically doesn't simulate the behavior of a real system.
4. Intermittent Issues: Faulty power supplies can exhibit intermittent issues. The power supply might work fine during the testing but fail when subjected to extended periods of operation or specific conditions.
5. Quality of Components: A power supply tester might not assess the quality of individual components within the power supply.
6. Compatibility Issues: Some power supplies might not be fully compatible with certain computer hardware. Even if the voltage seems fine, compatibility issues can still cause problems.
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Answer:
Explanation:
object distance u = 38.5 cm ( negative )
focal length f = 17.5 cm ( negative )
mirror formula
1 / v + 1 / u = 1 / f
1 / v - 1 / 38.5 = - 1 / 17.5
1 / v = - 1 / 17.5 + 1 / 38.5
= - 0 .03116
v = - 1 / .03116 = - 32 cm
Image will be formed in front of the mirror at 32 cm distance .