The voltage across the secondary winding is 10,000 V, and the voltage across the primary winding is 25,000 V. If the primary win
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Answer:
20.4
Explanation:
To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.
First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.
Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.
Applied force = 1055 N (490 + 565)
Friction force= Unknown
To find the friction force, use the kinetic friction formula, Friction = μkN
μk is the coefficient, which the problem includes- it is 0.613.
N is the normal force, which we have to find.
*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.
Weight = mg = (40)(9.8) = 392 N
Normal force = 392 N
Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N
Friction = 240.296 N
Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N
Horizontal Net Force: 814.704 N
Now that we know the net force, plug in the numbers for the formula
∑F= ma.
814.704 = (40.0)(a)
*Divide on both sides)
a = 20.3676 m/s^2
Round it to 3 significant figures, to get:
20.4
let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.
the mathematical formula for potential is
for positive charges the potential is positive and is negative for negative charges.
the formula for electric field is given as-
for positive charges,the line filed is away from it and for negative charges the filed is towards it.
we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.
here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.
now let the two charges of same nature.let these are positively charged.
here we can not find a point between two charges and on the line joining two charges where the potential is zero.
but at the mid point of the line joining two charges the filed is zero.