Downward force of magnitude 5 N is exerted on the book by the force of <em>gravity</em>. We call that the "weight" of the book.
At the same time, upward force of 5 N is exerted on the book by the table. This one is called the "normal force".
Since the vertical forces on the book are 'balanced' (add up to zero), the book just lays there on the table, and does not accelerate.
volume of balloon
= 4/3 T R3
= 4/3 x 3.14 x 6.953
= 1405.47 m3
uplift force
= volume of balloon x density of air x 9.8
= = 1405.47 x 1.29 x 9.8
= 1813.05 x 9.8 N
weight of helium gas
= volume of balloon x density of helium x
9.8
= 1405.47 x .179 x 9.8
= 251.58 x 9.8 N
Weight of other mass = 930 x 9.8 N Total weight acting downwards
= 251.58 x 9.8 +930 x 9.8
= 1181.58 x 9.8 N
If W be extra weight the uplift can balance
1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8
1181.58+W=1813.05
W= 631.47 kg
Answer:
a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0
Explanation:
a)
- The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
- For a point charge, it can be expressed as follows:
- As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
- This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
- In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at any other corner, as follows:
- The potential at point C is 8.99*10³ V
b)
- The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:
- The work needed is 0.045 J.
c)
- If we replace one of the charges creating the potential at the point C, by one of the same magnitude, but opposite sign, we will have the following equation:
- This means that the potential due to both charges is 0, at point C.
d)
- If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.