The attractive electrostatic force between the point charges +8.44 ✕ 10-6 and q has a magnitude of 0.961 n when the separation b
1 answer:
The electrostatic force between two charges Q1 and q is given by
where
ke is the Coulomb's constant
Q1 is the first charge
q is the second charge
r is the distance between the two charges
Re-arranging the formula, we have
and since we know the value of the force F, of the charge Q1 and the distance r between the two charges, we can calculate the value of q:
And since the force is attractive, the two charges must have opposite sign, so the charge q must have negative sign.
where
ke is the Coulomb's constant
Q1 is the first charge
q is the second charge
r is the distance between the two charges
Re-arranging the formula, we have
and since we know the value of the force F, of the charge Q1 and the distance r between the two charges, we can calculate the value of q:
And since the force is attractive, the two charges must have opposite sign, so the charge q must have negative sign.
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To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity:
where
is the final velocity
is the initial velocity
is the acceleration due to gravity 
is the time
2. Kinematic equation for distance:
where
is the distance
is the initial velocity
is the final velocity
is the acceleration due to gravity
is the time
First, we are going to convert 21.82 mi/h to ft/s:
Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore:
. We also know that it initial speed is 31.21 ft/s, so 
. Lets replace those values in our formula to find :
seconds
Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
ft
Now we can add the height of the building and the maximum height of the object:
ft
Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
or 
Since time cannot be negative,
is the time it takes the object to fall to the ground.
Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
ft/s
We can conclude that the speed of the object when it hits the ground is 110.25 ft/s
1. Kinematic equation for final velocity:
where
2. Kinematic equation for distance:
where
First, we are going to convert 21.82 mi/h to ft/s:
Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore:
Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
Now we can add the height of the building and the maximum height of the object:
Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
Since time cannot be negative,
Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
We can conclude that the speed of the object when it hits the ground is 110.25 ft/s