Given:
Object in circular motion 25 m/s
1 second to go quarter circle
Required:
Centripetal acceleration:
Solution:
Acceleration = v2/r
Where v is the velocity and r is
the radian
Substituting the values into the
equation,
Acceleration = v2/r = (25
m/s)2/(4*pi/180) = 8952.47 m2/s2
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Answer:
b. The side the boy is sitting on will tilt downward
Explanation:
Initially, the seesaw is balanced because the torque exerted by the boy is equal to the torque exerted by the girl:

where
Wb is the weight of the boy
db is the distance of the boy from the pivot
Wg is the weight of the girl
dg is the distance of the girl from the pivot
When the boy moves backward, the distance of the boy from the pivot (
increases, therefore the torques are no longer balanced: the torque exerted by the boy will be larger, and therefore the side of the boy will tilt downward.
The electric field strength is inversely related to the square of the distance.so the strength of the electric field is given by

Here,
is constant depend upon medium and its value is
and q is charge and r is the distance.
Given
and we know the charge of proton,
.
Therefore,
