Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp
ress your answer with the appropriate units. Enter positive value if the force is in the positive xx-direction and negative value if the force is in the negative xx-direction.
1 answer:
Answer:
The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.
Explanation:
The application of coulonb's law is used to approach the question as shown in the attached file.
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Answer:
The maximum height reached is 4.63 m.
Explanation:
Given:
Initial speed of the man (u) = 31.0 m/s
Speed at the top of trajectory () = 29.5 m/s
Acceleration due to gravity (g) = 9.8 m/s²
When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.
Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.
So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.
Now, initial velocity can be rewritten in terms of its components as:
Where, are the initial horizontal and vertical velocities of the man.
Now, plug in the given values and simplify. This gives,
Now, we know that, for a projectile motion, the maximum height is given as:
Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,
Therefore, the maximum height reached is 4.63 m.
Answer:
4800N
Explanation:
Lets assume,
Mass of first object = m₁
Mass of second object = m₂
Distance between the two objects = r
Thus the force between the two objects will be
where, G = Universal gravitational constant
Given, F = 2400N
New mass of second object = 2m₂
Now, the force will be
Thus, F₂ = 4800N