I'm stuck on the same question, as well :(
Supposing there's no air
resistance, horizontal velocity is constant, which makes it very easy to solve
for the amount of time that the rock was in the air.
Initial horizontal
velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s
15.5m / 10.3923m/s = 1.49s
So the rock was in the air for 1.49 seconds. </span>
<span>
Now that we know that, we can use the following kinematics
equation:
d = v i * t + 1/2 * a * t^2
Where d is the difference in y position, t is the time that
the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>
<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s
So:
d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2
d = 8.94 + -10.89</span>
d = -1.95<span>
<span>This means that the initial y position is 1.95 m higher than
where the rock lands. </span></span>
The electrostatic force is directly proportional to the product of the charges, by Coulomb's law.
F α Qq
If the charges are now half the initial charges:
<span>F α (1/2)Q *(1/2)q
</span>
F α (1/4)Q<span>q
The new force when the charges are each halved is (1/4) the first initial force experienced at full charge.</span>
Explanation:
C.
Object A will require more force to be set in motion but will travel faster than object B.
2. true
