MCQOPTIONS
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MCQOPTIONS
This section includes 11 Mcqs, each offering curated multiple-choice questions to sharpen your Aircraft Performance knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
What is the energy change in the aircraft when the thrust is 244N and the distance travelled is 1230km? |
| A. | 198000 N-Km |
| B. | 484000 N-Km |
| C. | 300120 N-Km |
| D. | 234553 N-Km |
| Answer» D. 234553 N-Km | |
| 2. |
The energy change is given by __________ |
| A. | ΔE=\(\frac{T^2}{S^2}\) |
| B. | ΔE=\(\frac{T^2}{S}\) |
| C. | ΔE=\(\frac{T}{S}\) |
| D. | ΔE=T x S |
| Answer» D. ΔE=T x S | |
| 3. |
The difference between the lift-off speed and take-off safety speed must be small. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 4. |
In the calculations the horizontal airborne distance is assumed to be much greater than 35 feet. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 5. |
The airborne distance is given by __________ |
| A. | SA=\(\frac{W}{(F_N-D)_{av}}\Big\{\frac{V_{2}^2+V_{LOF}^2}{2g}+35\Big\}\) |
| B. | SA=\(\frac{W}{(F_N+D)_{av}}\Big\{\frac{V_{2}^2+V_{LOF}^2}{2g}+35\Big\}\) |
| C. | SA=\(\frac{W}{(F_N+D)_{av}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\) |
| D. | SA=\(\frac{W}{(F_N-D)_{av}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\) |
| Answer» E. | |
| 6. |
The formula for ground run is given by the formula __________ |
| A. | SG=\(\frac{V_{LOF}^2}{2g(A+BV^2)_{V_{LOF}}}\) |
| B. | SG=\(\frac{V_{LOF}^2}{2g(A+BV^2)_{0.7V_{LOF}}}\) |
| C. | SG=\(\frac{V_{LOF}^2}{2g(A-BV^2)_{0.7V_{LOF}}}\) |
| D. | SG=\(\frac{V_{LOF}^2}{2g(A-BV^2)_{V_{LOF}}}\) |
| Answer» D. SG=\(\frac{V_{LOF}^2}{2g(A-BV^2)_{V_{LOF}}}\) | |
| 7. |
The round bracket in the accelerating force represents the drag- lift ratio at the ground angle of attack. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 8. |
The curly bracket in the accelerating force represents the net propulsive thrust- weight ratio. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 9. |
Which of the following is the correct formula for accelerating force? |
| A. | \(\frac{F}{W}=\Big\{\frac{F_N}{W}+\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\) |
| B. | \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R+sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\) |
| C. | \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\) |
| D. | \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}+\mu_R\Big)\) |
| Answer» D. \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}+\mu_R\Big)\) | |
| 10. |
Which of the following are the correct equations for take-off run? |
| A. | FN+D-WsinγR-μrR=mV̇ |
| B. | R-L=WcosγR |
| C. | R-L=WsinγR |
| D. | FN-D-WsinγR-μrR=mV̇ |
| Answer» E. | |
| 11. |
What are the additional forces acting on the aircraft? |
| A. | Runway friction |
| B. | Lift |
| C. | Thrust |
| D. | Power |
| Answer» B. Lift | |