Suppose you fill two rubber balloons with air, suspend both of them from the same point, and let them hang down on strings of eq
1 answer:
The charge on each the balloon is 100nC or 1.2 × 10^-7 C
Consider two balloons of diameter 0.200m each with a mass of 1.00g hanging apart with 0.0500m separation on the ends of string making angles of 10.0° with the vertical.
The charge on each balloon can be found from
or 100nC
An electric charge is the property of matter where it has more or fewer electrons than protons in its atoms. Electrons carry a negative charge and protons carry a positive charge. The matter is positively charged if it contains more protons than electrons and negatively charged if it contains more electrons than protons.
Learn more about the charge here:
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To solve for force, you need to get the product of mass and acceleration.
F = ma
Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms
As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.
a = change in velocity/time
To get the change in velocity, you get the difference between the initial velocity and final velocity:
The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.
So we can put it into our formula now:
WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.
x 
= 
So your new time is 0.02s. Now we put this time into the formula:
As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.
So now we have our acceleration. Now using this, we can get our force.
Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is
but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first.
Our new mass is 0.150kg.
To make things clearer, let us write down all our new values:
m = 0.150kg
a = -2,500
Now that all our units match, we can put that into our formula:
The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.
F = ma
Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms
As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.
a = change in velocity/time
To get the change in velocity, you get the difference between the initial velocity and final velocity:
The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.
So we can put it into our formula now:
WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.
So your new time is 0.02s. Now we put this time into the formula:
As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.
So now we have our acceleration. Now using this, we can get our force.
Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is
Our new mass is 0.150kg.
To make things clearer, let us write down all our new values:
m = 0.150kg
a = -2,500
Now that all our units match, we can put that into our formula:
The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.