The magnetic force experienced by the proton is given by
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and
the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so
and we can ignore it in the formula.
For Netwon's second law, the force is also equal to the proton mass times its acceleration:
So we have
from which we can find the magnitude of the field:
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Answer:
The ladder is 3.014 m tall.
Explanation:
To solve this problem, we must use the following formula:
v = x/t
where v represents the woman’s velocity, x represents the distance she climbed (the height of the ladder), and t represents the time it took her to move this distance
If we plug in the values we are given for the problem, we get:
v = x/t
2.20 = x/1.37
To solve this equation for x (the height of the ladder), we must multiply both sides by 1.37. If we do this, we get:
x = (2.20 * 1.37)
x = 3.014 m
Therefore, the ladder is 3.014 m tall.
Hope this helps!
Answer:
0.217 m/s
Explanation:
The protons in the beam passes undeflected when the electric force is equal to the magnetic force:
qE = qvB
where
q is the proton's charge
E is the magnitude of the electric field
v is the speed of the protons
B is the magnitude of the magnetic field
Re-arranging the equation,
And by substituting
E = 0.5 N/C
B = 2.3 T
We find
