If a problem says the acceleration is some positive value than solve using that value, a negative acceleration is said to be deceleration. E.g. a car decelerating at 10 m/sec can be said to be accelerating at -10 m/sec.
If a problem states decelerates at A, then use -A for acceleration in the classic equations which are for acceleration. If a problem says accelerates at a negative value like -A the use -A as the value for acceleration, it can also be said to be decelerating at A.
Mass of the dart = 0.01 kg
Speed at which the dart is thrown = 20 m/s
Kinetic Energy = (1/2) * mass * speed * speed
= (1/2) * (0.01) * (20) * (20) Joules
= (400 *0.01)/2 Joules
= 4/2 Joules
= 2 Joules
So the kinetic energy of the dart is 2 Joules. I hope this is the correct answer and it has helped you.
Answer:
(a) velocity at 9 rev/s is 34.3m/s and the velocity at 7rev/s is 40.4 m/s. These were values can be gotten by using the formula v = 2(pi)R/T. Where T is the period. T = 1/w(angular velocity).
(b) acceleration at 9.0rev/s is 1961 m/s²
(c) acceleration at 7.0rev/s is 1814 m/s²
These values can be gotten by using the formula a = v²/R
Explanation:
The full solution can be found in the attachment below.
Thank you for reading.
Answer: a. Stall AOA increases, CL max increases.
Explanation:
Deploying slats or slots on an airfoil affect CL max and Stall AOA by increasing Stall AOA and also increasing CL max.
It should be noted that the coefficient of lift CL would rise through the use of slots due to increase in boundary later energy. The slots also leads to delay of stall by through increase in AOA.
