The position of a certain airplane during takeoff is given by x=1/2 *bt2, where b = 2.0 m/s2 and t = 0 corresponds to the instan
1 answer:
You might be interested in
Explanation:
Given that,
Initial speed of the airfield, u = 0
Final speed, v = 27.8 m/s
Acceleration of the airfield,
Length of the runway, d = 150 m
Let v' is the speed of the airplane to reach the required speed for takeoff. Finding v' using third equation of motion as :
This speed is not enough as the airfield must reach a speed before takeoff of at least 27.8 m/s. Now, the required length of the runways is :
So, the minimum length of the runways is 193.21 meters.
As v becomes zero at the highest point, i prefer considering different travelling directions so it will become less complicated.
dont forget to add the total time up .
also to master the skills, write down the "uvsat" may help (thats the way i found it easier to handle problems)
dont forget to add the total time up .
also to master the skills, write down the "uvsat" may help (thats the way i found it easier to handle problems)
The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.
<h3>What is Columb's law?</h3>The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.
They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.
The given data in the problem is;
q₁ is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C
F is the repulsive force = 36 mN =36 ×10⁶ N
d is the distance = 0.70 m
The Coulomb force is found as;
Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.
To learn more about Coulomb's law, refer to the link;
#SPJ2