a system absorb 500 J of heat and the same time 400J of work is done one the system find change in internal enery ?
1 answer:
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Answer:
E. Some charges in the region are positive, and some are negative.
Explanation:
Electric potential is given as;
where;
W is the work done in moving a charge between two points which have a difference in potential
Q is quantity of charge in the given region
If the electric potential at a given point in the region is zero, then sum of the charges in the given region must be equal to zero. For the charges to sum to zero, some will be positive while some will be negative,.
Therefore, the correct statement in the given options is "E"
E. Some charges in the region are positive, and some are negative.
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows: