Answer:
The electric field will be zero at x = ± ∞.
Explanation:
Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.
We know that,
The electric field is

The electric field vector due to charge one

The electric field vector due to charge second

We need to calculate the electric field
Using formula of net electric field


Put the value into the formula




Put the value into the formula


If x = ∞, then the equation is be satisfied.
Hence, The electric field will be zero at x = ± ∞.
At sea level, the size amid the 2 alkanes lets for pentane to simmer at a lower temperature than hexane. Phenol has a higher boiling point due to hydrogen bonding High altitude would have the same order while low pressure only cuts the temperature at which a solvent boils. Boiling has to do with molecular size, the occurrence/nonappearance of hydrogen bonds, and other steric issues.
So the answer would be pentane high altitude, hexane high altitude, hexane sea level, hexanol sea level. In order of boil first to boil last. This is clarified because altitude has a better effect on vapor pressure (and hence boiling points) than inter-molecular forces.
C) Radiation that comes from Earth...... Hope it helps, Have a nice day :)
Answer:
0.79 s
Explanation:
We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that 
Now, we find the acceleration:

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

Answer:
Loss, 
Explanation:
Given that,
Mass of particle 1, 
Mass of particle 2, 
Speed of particle 1, 
Speed of particle 2, 
To find,
The magnitude of the loss in kinetic energy after the collision.
Solve,
Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.
Applying the conservation of momentum to find the speed of two particles after the collision.



V = 6.71 m/s
Initial kinetic energy before the collision,



Final kinetic energy after the collision,



Lost in kinetic energy,



Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.