Answer:
Explanation:
According to question,
Charge 1 and charge 2 are
The distance between charges is 2 m
We need to find the force with which two positive charges repel. It is called electrostatic force of repulsion. It can be given by :
So, the electric force of repulsion is .
Answer:
a) 20 seconds
b) No.
Explanation:
t = Time taken for jet to stop
u = Initial velocity = 100 m/s (given in the question)
v = Final velocity = 0 (because the jet will stop at the end)
s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)
a = Acceleration = -5.00 m/s² (slowing down, so it is negative)
a) Equation of motion
The time required for the plane to slow down from the moment it touches the ground is 20 seconds.
The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.
The gas is in a rigid container: this means that its volume remains constant. Therefore, we can use Gay-Lussac law, which states that for a gas at constant volume, the pressure is directly proportional to the temperature. The law can be written as follows:
Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:
So, the new pressure is 2 atm.