Answer:
(a) Magnitude: 14.4 N
(b) Away from the +6 µC charge
Explanation:
As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:
K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.
Let's say that a force that goes toward the +6 µC charge is positive, then:
The magnitude will be:
, away from the +6 µC charge
You can find
1) time to hit the ground
2) initial velocity
3) speed when it hits the ground
Equations
Vx = Vxo
x = Vx * t
Vy = Vyo + gt
Vyo = 0
Vy = gt
y = yo - Vyo - gt^2 / 2
=> yo - y = gt^2 / 2
1) time to hit the ground
=> 8.0 = g t^2 / 2 => t^2 = 8.0m * 2 / 9.81 m/s^2 = 1.631 s^2
=> t = √1.631 s^2 = 1.28 s
2) initial velocity
Vxo = x / t = 6.5m / 1.28s = 5.08 m/s
3) speed when it hits the ground
Vy = g*t = 9.81 m/s * 1.28s = 12.56 m/s
V^2 = Vy^2 + Vx^2 = (12.56 m/s)^2 + (5.08 m/s)^2 = 183.56 m^2 / s^2
=> V = √ (183.56 m^2 / s^2) = 13.55 m/s
Answer:0.5
Explanation:
Given
Piece of taffy slams in to sticks to another identical piece and stuck into it.
Let m be the mass of taffy and u be the initial velocity of taffy and v be the final velocity of system.
Conserving momentum
Initial kinetic energy
Final Kinetic Energy
change in kinetic energy 
change in kinetic energy will contribute in heat energy
thus fraction of kinetic energy converted in to heat 
If he's traveling at 5.2 m/s, then he covers
5.2 meters in the 1st second
5.2 meters in the 2nd second
5.2 meters in the 3rd second
5.2 meters in the 4th second
.
.
.
.
5.2 meters in the 14th second, and
5.2 meters in the 15th second.
He covers 5.2 meters 15 times in 15 seconds.
To find the total distance, you could write down 5.2 meters
15 times and add them all up. Or you could use the newly-
discovered, labor-saving process called 'multiplication' to do
the same thing. With multiplication, it can be done in one line:
(5.2 m/s) x (15 sec) = 78 meters .
