- #1
Skotster
- 8
- 0
Second one:
Protons are projected at an angle of 300 with an initial speed, given by V0=8.2*105 m/s, in a region of an electric field E=-390j N/C, is present. The proton is to hit a point 1.27 mm in the pos-x direction from the launch point. Find (a) the two porjections angles (theta) that will result in a hit, and (b) the total time of flight for each of these two trajectories.
Here is what I can get for this one.
(theta)=cos-1(V0x/V0)
(theta)=sin-1(V0y/V0)
F=q*E=-1.15*10-16 N
a=F/m=-6.89*1010 m/s2
First one: Solved
on an analogue clock the minute hand aligns perfectly with the hour hand once and hour. 12:00 is the first time, when are the exact other times this happens. (Hint: in only happens 11 times, not 12 and the hour and minute hand constantly move)
time=3927.272727s*n
where n is the number of times the min hand has ligned up with the hour hand
ex:
1st time...1:05:27.2727
Protons are projected at an angle of 300 with an initial speed, given by V0=8.2*105 m/s, in a region of an electric field E=-390j N/C, is present. The proton is to hit a point 1.27 mm in the pos-x direction from the launch point. Find (a) the two porjections angles (theta) that will result in a hit, and (b) the total time of flight for each of these two trajectories.
Here is what I can get for this one.
(theta)=cos-1(V0x/V0)
(theta)=sin-1(V0y/V0)
F=q*E=-1.15*10-16 N
a=F/m=-6.89*1010 m/s2
First one: Solved
on an analogue clock the minute hand aligns perfectly with the hour hand once and hour. 12:00 is the first time, when are the exact other times this happens. (Hint: in only happens 11 times, not 12 and the hour and minute hand constantly move)
time=3927.272727s*n
where n is the number of times the min hand has ligned up with the hour hand
ex:
1st time...1:05:27.2727
Last edited: