What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed
1 answer:
Answer:
charge will be equal to
Explanation:
We have given mass of the particle m = 1.45 gram = 0.00145 kg
Acceleration due to gravity
Electric field E = 700 N/C
Electric force will be equal to , here q is charge and E is electric field
For particle to be stationary this force must be equal to force due to gravity , that is mg force
So qE = mg
So charge will be equal to
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Answer:
v = √[gR (sin θ - μcos θ)]
Explanation:
The free body diagram for the car is presented in the attached image to this answer.
The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.
Resolving the weight into the axis parallel and perpendicular to the inclined plane,
N = mg cos θ
And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ
Frictional force = Fr = μN = μmg cos θ
Centripetal force responsible for keeping the car in circular motion = (mv²/R)
So, a force balance in the plane parallel to the inclined plane shows that
Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)
(mv²/R) = (mg sin θ - μmg cos θ)
v² = R(g sin θ - μg cos θ)
v² = gR (sin θ - μcos θ)
v = √[gR (sin θ - μcos θ)]
Hope this Helps!!!
