Recent content by spidey64

really? no ideas at all? or am I going down the wrong path, or anything?

Ok, I need to calculate the increase in temperature of a nail hit by a hammer with force of 500N. The length of the nail is .06m, so the work done and, I'm assuming, the energy added to the nail (Q) is 30 Joules. The specific heat capacity of the nail is 450 J/kg *C(degrees Celsius). By using...

I've found that force = area X pressure, which of course makes sense...but given that my problem gives me the difference of pressures below and above as 4% atmospheric pressure, is the pressure value in that formula .04?

you could have a line graph with one axis representing mass and the other representing acceleration, then force would be shown as a perfectly diagonal line between the two

I know about Bernoulli's principle and that a difference of pressure below and above a wing create lift, but I cannot find anywhere a formula or any information on how to find the amount of lift given the surface area of the wing and the difference in pressure as a percentage of atmospheric...

alright, i think I'm getting it now, i got to go to bed, thanks for your help everyone!

action-reactions are in pairs of course, the pusher and the box is one, so then the friction is not part of that but it's own pair with box, is that the idea?

well since it doesn't mention friction, i'll just assume "no"

so is the frictional force unrelated to the push force in terms of action-reaction then?

the push force is the force of the interaction between the box and the pusher going both ways. the friction force is the force of the interaction between the box and the ground.

both the force of friction and the reaction force equal and oppose (essentially cancel out) the push force, how can there be two opposing forces (one acting on the box and one acting on the pusher)?

ok, so the box is in dynamic equilibrium...so what's the difference between the force of the box pushing back and the force of friction?

yeah, isn't what i just said basically about inertia? where the box has moving equilibrium once the pushing force equals the friction force. It's in equilibrium at that point because no net force is acting on the box because friction and push cancel each other.

so obviously the action-reaction is where two objects (me and the box) touch, with our forces pushing at each other. But my problem is with the sustained motion and the force of friction. The friction is needed in order for the action-reaction to take place. Maybe the friction merely reduces the...

i think this excerpt relates the best to the concept "the tires of a car push against the road while the road pushes back on the tires...the reaction force is what accounts for motion in this example. This force depends on friction; a person or car on ice, for example, may be unable to exert the...