= 4pi/3 R3
= 4pi/3 × (0.125 m)^3
= 8.2 × 10^-3
m3
Fnet = [(10^3 kg/m3
) × (8.2 × 10^-3
m3
) - 0.54 kg] × 9.81 m/s2
= 75.1 N
No fixed answer, depends on the surface.
The force required to overcome friction = the coefficient of friction x the reactive force (as in, the force the object exerts on the surface you want to move it over). So, you have to know the coefficient of friction, and simply times it by 600 in this case to know the force required. Oddly, once friction has been overcome, the required force drops slightly to keep it moving.
No. Frictional force is independant of surface area.
Force divided by surface area is called pressure, if that's what you mean.
smooth surface.
Force it below the surface of water and measure the force required.
no, it's not the same, there is more force below the surface, because it will have the force of the whole bridge and the force of the things on the bridge.
That's a trick question, right? Jupiter has no surface.
yes
No fixed answer, depends on the surface.
The force required to overcome friction = the coefficient of friction x the reactive force (as in, the force the object exerts on the surface you want to move it over). So, you have to know the coefficient of friction, and simply times it by 600 in this case to know the force required. Oddly, once friction has been overcome, the required force drops slightly to keep it moving.
gravity
You would weigh the most on the surface, where the most gravitational energy is below you.As you enter the Earth, gravitational force is also exerted from the mass above, meaning that when you reach the center, the gravitational force will be equal from all directions, leaving you weightless.As you depart the surface of the Earth, you will lose weight as the gravitational force weakens with distance.
The best way is to find the centre of surface of planar area. Then the force due to hydrostatic pressure will be:F = d h0 g S,where:F is force,d is density of fluid,h0 is depth at the centre of surface,S is surface of the area.It works because when we consider the centre of surface, there will exactly as much surface with lesser pressure effecting on it as there is surface below the centre point where the pressure is higher.The net force vector will be perpendicular to the area at the centre of surface point.
As long as any part of the object ... doesn't matter how much ... is below the surface of the water ... doesn't matter how far ... there is buoyant force on it.
As long as any part of the object ... doesn't matter how much ... is below the surface of the water ... doesn't matter how far ... there is buoyant force on it.
As long as any part of the object ... doesn't matter how much ... is below the surface of the water ... doesn't matter how far ... there is buoyant force on it.