The density will always be the same.
Using the velocity-time graph, the displacement can be calculated by the area under the velocity-time graph. At 3 seconds the total displacement is then equal to (4)(2) + (4 + 2)*1/2 = 11 m. Assuming that the starting point is at x = 0, then the particle at t=3s is at x=11 m.
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Answer:
d²x/dt² = - 4dx/dt - 4x is the required differential equation.
Explanation:
Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,
F = mg
kx = mg
k = mg/x
= 4 kg × 9.8 m/s²/2.45 m
= 39.2 kgm/s²/2.45 m
= 16 N/m
Now the drag force f = 16v where v is the velocity of the mass.
We now write an equation of motion for the forces on the mass. So,
F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass
-kx - 16v = 4a
-16x - 16v = 4a
16x + 16v = -4a
4x + 4v = -a where v = dx/dt and a = d²x/dt²
4x + 4dx/dt = -d²x/dt²
d²x/dt² = - 4dx/dt - 4x which is the required differential equation
This means it has TWO poles
The 2 poles are NORTH and SOUTH
Quantity of Charge , Q = ne
Where n = number of electrons
e = charge on one electron = -1.6 * 10 ^-19 C.
n = 50 * 10^31 electrons
Q = (50 * 10^31)*( -1.6 * 10 ^-19 ) = -8 * 10^13 C.
Note that the minus sign indicates that the charge is a negative charge.
