Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Get immediate and reliable solutions to your questions from a community of experienced experts on our Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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.
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
