Why is current a scalar quantity?
1 answer:
Despite current has a magnitude and a direction, like vectors, it is a scalar because it doesn't obey laws of vector addition. For instance, if we consider a junction of 
in a circuit, and two currents entering this junction, we know that the resultant current is just the algebraic sum of the two currents, not the vector sum, so it is not a vector quantity.
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Answer:
1.28 m
Explanation:
As shown in the diagram attached,
According to the principle of moment,
For a body at equilibrium,
Sum of clockwise moment = sum of anticlockwise moment.
Taking moment about the pivot,
W₁(1.6)+W(0.133) = W₂(x)............... Equation 1
Where W₁ = Weight of the first child, Wₓ = Weight of the seesaw, W₂ = weight of the second child, x = distance of the second child from the pivot.
But,
W = mg
Where g = 9.8 m/s², m = mass of the body
Therefore,
W₁ = 26×9.8 = 254.8 N,
Wₓ = 18×9.8 = 176.4 N
W₂ = 34.4×9.8 = 337.12 N
Substitute these values into equation 1
(254.8×1.6)+(176.4×0.133) = 337.12(x)
407.68+23.4612 = 337.12x
337.12x = 431.1412
x = 431.1412/337.12
x = 1.2789
x ≈ 1.28 m
1) The total mechanical energy of the rock is:
where U is the gravitational potential energy and K the kinetic energy.
Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
where
is the mass, 
is the gravitational acceleration and 
is the height.
Putting the numbers in, we find the potential energy
2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
where U is the gravitational potential energy and K the kinetic energy.
Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
where
Putting the numbers in, we find the potential energy
2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is: