Answer:
B
Explanation:
Because it has to increase
Answer:
The third charge placed is 0.80 m.
Explanation:
Given that,
Distance = 0.57 m
First charge = q
Third charge = 2q
We need to calculate the electrostatic force on charge q₁ due to q₂
Using formula of electrostatic force

When placed another charge q₃ at certain distance from origin, then the net force on charge q₁ due to both charges is

The net electrostatic force on the charge at the origin doubles.






Put the value into the formula



Hence, The third charge placed is 0.80 m from origin in x-axis.
Explanation :
We know that the slope of velocity -time graph gives the acceleration. Acceleration of an object is defined as the rate of change of velocity i.e.

Suppose a driver suddenly applies brakes. In this case the initial velocity of his or her vehicle is more and the final velocity is less.
So, the acceleration is negative in this case i.e. the object is decelerating.
A negative slope on the velocity versus time graph indicates that an object is not accelerating. This statement is false as the object is decelerating.
Answer:
v_{4}= 80.92[m/s] (Heading south)
Explanation:
In order to calculate this problem, we must use the linear moment conservation principle, which tells us that the linear moment is conserved before and after the collision. In this way, we can propose an equation for the solution of the unknown.
ΣPbefore = ΣPafter
where:
P = linear momentum [kg*m/s]
Let's take the southward movement as negative and the northward movement as positive.

where:
m₁ = mass of car 1 = 14650 [kg]
v₁ = velocity of car 1 = 18 [m/s]
m₂ = mass of car 2 = 3825 [kg]
v₂ = velocity of car 2 = 11 [m/s]
v₃ = velocity of car 1 after the collison = 6 [m/s]
v₄ = velocity of car 2 after the collision [m/s]
![-(14650*18)+(3825*11)=(14650*6)-(3825*v_{4})\\v_{4}=80.92[m/s]](https://tex.z-dn.net/?f=-%2814650%2A18%29%2B%283825%2A11%29%3D%2814650%2A6%29-%283825%2Av_%7B4%7D%29%5C%5Cv_%7B4%7D%3D80.92%5Bm%2Fs%5D)