What force must act on a 50.0-kg mass to give it an ancceleration of 0.30 m/s^2?
1 answer:
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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:
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and
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:
T=502.5N
Ax=171.8N
Explanation:
The computation of the tension T in the rope and the forces exerted by the pin at A is shown below:
vertical forces sum = Ay + Tsin20 + T - 245 - 883 = 0
Now
horizontal forces sum = Ax - Tcos70
Now Moment about B
-Ay × 4.8 + 245 × 2.4 + 883 × 1.8=0
Ay=453.6N
Now substitute in sum of vertical forces T=502.5N
Ax=171.8N
