Answer:
V1 = 1.721 * V2
Explanation:
To start with, we assume that both lift forces are equal, such that
L2 = L1
1 is that of the level at 10000 m, and 2 is that of the level at sea level.
Next, we try and substitute the general formula for both forces such that
C(l).ρ1/2.V1².A = C(l).ρ2/2.V2².A
On further simplification, we have
ρ1.V1² = ρ2.V2², making V1 subject of formula, we have
V1 = √(ρ2/ρ1). V2²
Using the values of density for air at 10000 m and at sea level(source is US standard atmosphere), we have
V1 = √(1.225/0.4135) * V2
V1 = √2.9625 * V2
V1 = 1.721 * V2
Answer:
Amount of fuel used per year is supposed to be 34150 KJ/kg
3-SAT ≤p TSP
If P ¹ NP, then no NP-complete problem can be solved in polynomial time.
both the statements are true.
<u>Explanation:</u>
- 3-SAT ≤p TSP due to any complete problem of NP to other problem by exits of reductions.
- If P ¹ NP, then 3-SAT ≤p 2-SAT are the polynomial time algorithm are not for 3-SAT. In P, 2-SAT is found, 3- SAT polynomial time algorithm implies the exit of reductions. 3 SAT does not have polynomial time algorithm when P≠NP.
- If P ¹ NP, then no NP-complete problem can be solved in polynomial time. because for the NP complete problem individually gets the polynomial time algorithm for the others. It may be in P for all the problems, the implication of latter is P≠NP.
Answer:
See explaination
Explanation:
int RED=10; int BLUE=11; int GREEN=12; int BUTTON1=8; int BUTTON2=9; void setup() { pinMode(RED, OUTPUT); pinMode(BLUE, OUTPUT); pinMode(GREEN, OUTPUT); pinMode(BUTTON1, INPUT); pinMode(BUTTON2, OUTPUT); } void loop() { int BTN1_STATE=digitalRead(BUTTON1); int BTN2_STATE=digitalRead(BUTTON2); if(BTN1_STATE==HIGH) { digitalWrite(BLUE, HIGH); delay(1000); // Wait for 1 second digitalWrite(BLUE, LOW); } if(BTN2_STATE==HIGH) { digitalWrite(RED, HIGH); delay(4000); // Wait for 4 seconds digitalWrite(RED, LOW); } if(BTN1_STATE==HIGH && BTN2_STATE==HIGH) { digitalWrite(GREEN, HIGH); delay(2000); // Wait for 2 second digitalWrite(GREEN, LOW); } }