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Sagot :
Certainly! Let's examine the validity of the given formulae for power and pressure by analyzing their units step-by-step.
### Analyzing the Formula for Power [tex]\( P = mv^2 \)[/tex]
1. Power (P):
- Power is measured in Watts (W), which is equivalent to Joules per second (J/s).
- The unit of a Joule (J) is [tex]\( \text{kg} \cdot \frac{\text{m}^2}{\text{s}^2} \)[/tex].
2. Mass (m):
- Mass is measured in kilograms (kg).
3. Velocity (v):
- Velocity is measured in meters per second (m/s).
4. Units of [tex]\( P = mv^2 \)[/tex]:
- Let’s substitute the units into the formula:
[tex]\[ P = m \cdot v^2 = \text{kg} \cdot \left(\frac{\text{m}}{\text{s}}\right)^2 \][/tex]
- This gives us:
[tex]\[ P = \text{kg} \cdot \frac{\text{m}^2}{\text{s}^2} \][/tex]
- The unit [tex]\(\text{kg} \cdot \frac{\text{m}^2}{\text{s}^2}\)[/tex] corresponds to a Joule (J), not Watts (W).
Therefore, the unit analysis shows that [tex]\( P = mv^2 \)[/tex] results in Joules, not Watts, making this formula invalid for power.
### Analyzing the Formula for Pressure [tex]\( P = \frac{mv}{A} \)[/tex]
1. Pressure (P):
- Pressure is measured in Pascals (Pa), which is equivalent to Newtons per square meter (N/m[tex]\(^2\)[/tex]).
- The unit of a Newton (N) is [tex]\(\text{kg} \cdot \frac{\text{m}}{\text{s}^2}\)[/tex].
2. Mass (m):
- Mass is measured in kilograms (kg).
3. Velocity (v):
- Velocity is measured in meters per second (m/s).
4. Area (A):
- Area is measured in square meters (m[tex]\(^2\)[/tex]).
5. Units of [tex]\( P = \frac{mv}{A} \)[/tex]:
- Let’s substitute the units into the formula:
[tex]\[ P = \frac{m \cdot v}{A} = \frac{\text{kg} \cdot \frac{\text{m}}{\text{s}}}{\text{m}^2} \][/tex]
- This simplifies to:
[tex]\[ P = \frac{\text{kg} \cdot \text{m}}{\text{s} \cdot \text{m}^2} = \frac{\text{kg}}{\text{s} \cdot \text{m}} \][/tex]
- The unit [tex]\(\frac{\text{kg}}{\text{s} \cdot \text{m}}\)[/tex] does not correspond to Pascals (Pa), which should be [tex]\(\text{kg} \cdot \frac{\text{m}}{\text{s}^2 \cdot \text{m}^2} = \frac{\text{kg}}{\text{m} \cdot \text{s}^2}\)[/tex].
Therefore, the unit analysis shows that [tex]\( P = \frac{mv}{A} \)[/tex] does not result in Pascals, making this formula invalid for pressure.
In conclusion, both given formulae are invalid according to the unit analysis.
### Analyzing the Formula for Power [tex]\( P = mv^2 \)[/tex]
1. Power (P):
- Power is measured in Watts (W), which is equivalent to Joules per second (J/s).
- The unit of a Joule (J) is [tex]\( \text{kg} \cdot \frac{\text{m}^2}{\text{s}^2} \)[/tex].
2. Mass (m):
- Mass is measured in kilograms (kg).
3. Velocity (v):
- Velocity is measured in meters per second (m/s).
4. Units of [tex]\( P = mv^2 \)[/tex]:
- Let’s substitute the units into the formula:
[tex]\[ P = m \cdot v^2 = \text{kg} \cdot \left(\frac{\text{m}}{\text{s}}\right)^2 \][/tex]
- This gives us:
[tex]\[ P = \text{kg} \cdot \frac{\text{m}^2}{\text{s}^2} \][/tex]
- The unit [tex]\(\text{kg} \cdot \frac{\text{m}^2}{\text{s}^2}\)[/tex] corresponds to a Joule (J), not Watts (W).
Therefore, the unit analysis shows that [tex]\( P = mv^2 \)[/tex] results in Joules, not Watts, making this formula invalid for power.
### Analyzing the Formula for Pressure [tex]\( P = \frac{mv}{A} \)[/tex]
1. Pressure (P):
- Pressure is measured in Pascals (Pa), which is equivalent to Newtons per square meter (N/m[tex]\(^2\)[/tex]).
- The unit of a Newton (N) is [tex]\(\text{kg} \cdot \frac{\text{m}}{\text{s}^2}\)[/tex].
2. Mass (m):
- Mass is measured in kilograms (kg).
3. Velocity (v):
- Velocity is measured in meters per second (m/s).
4. Area (A):
- Area is measured in square meters (m[tex]\(^2\)[/tex]).
5. Units of [tex]\( P = \frac{mv}{A} \)[/tex]:
- Let’s substitute the units into the formula:
[tex]\[ P = \frac{m \cdot v}{A} = \frac{\text{kg} \cdot \frac{\text{m}}{\text{s}}}{\text{m}^2} \][/tex]
- This simplifies to:
[tex]\[ P = \frac{\text{kg} \cdot \text{m}}{\text{s} \cdot \text{m}^2} = \frac{\text{kg}}{\text{s} \cdot \text{m}} \][/tex]
- The unit [tex]\(\frac{\text{kg}}{\text{s} \cdot \text{m}}\)[/tex] does not correspond to Pascals (Pa), which should be [tex]\(\text{kg} \cdot \frac{\text{m}}{\text{s}^2 \cdot \text{m}^2} = \frac{\text{kg}}{\text{m} \cdot \text{s}^2}\)[/tex].
Therefore, the unit analysis shows that [tex]\( P = \frac{mv}{A} \)[/tex] does not result in Pascals, making this formula invalid for pressure.
In conclusion, both given formulae are invalid according to the unit analysis.
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