With total velocity
, the horizontal component of velocity is

Then the horizontal displacement over the first minute is

Answer:
0.20 A
Explanation:
The current in the wire is given by Ohm's law:

where
I is the current
V is the voltage
R is the resistance
In this problem,
V = 3.0 V is the voltage
is the resistance
Substituting into the formula,

Answer:
The object's maximum speed remains unchanged.
Explanation:
The speed of a particle in SHM is given by :

Maximum speed is, 
If A' = 2A and T' = 2T



So, the maximum speed of the object remains the same i.e. it remains unchanged. Hence, this is the required solution.
<span>C) Just before he starts to fall to the ground
Remember that "gravitational potential energy" is the energy the object potentially has if it were to drop from the current height. So let's look at the options and see what's correct.
A) When he initially lands
* He's currently near the lowest height. So his gravitational potential energy is quite low. So this is the wrong choice.
B) As he runs forward to plant the pole
* He's on the ground. So has almost no height. So almost no gravitational potential energy. So this too is wrong.
C) Just before he starts to fall to the ground
* Ah. He's at his maximum height. So his potential gravitational potential energy is maximized. This is the correct choice.
D) After take-off as the pole expands and lifts the vaulter toward the bar
* He's at a low height. So once again, this is a bad choice.</span>