That's for an accelerating mass, the mass being constant. Here, it's the water that is being moved, and the mass of that is related to the cross-sectional area (L2) and the velocity.welovephysics said:
since we do not know the mass of water ( we only know the density of water) , so we use (rho)(L^3 ) to find the mass of water ?haruspex said:
That would be true if we were considering a cube of water of side L, but L is the characteristic dimension of the model/prototype. How does the mass of water displaced in time t relate to L, t, ρ and v?welovephysics said:
ρ(L^3) [(L) / (T^-2) ] , which is ρ(L^3)v what are you trying to say ?haruspex said:
No, that's not it. The object moves at speed v. Think of it as a box of side L. What volume of water has to move aside to make way for it in time t?welovephysics said:
L^3 ?haruspex said:
How far does the box move in time t? What volume does the leading face of the box sweep through as the box advances that distance?welovephysics said:
the box will move by L in time , t , am i right ? the volume that the leading face of the box sweep through as the box advances that distance is (L^3) ?haruspex said:
It is moving at speed v, not L/t.welovephysics said:
so , what are you trying to say , i didnt get youharuspex said:
If a car length L moves at speed v for time t, how far does it go?welovephysics said:
car move by vtharuspex said:
is it correct ??haruspex said:
Yes, so try answering my post #8 again.welovephysics said:
distance moved = vt , volume of the leading face of the box sweep through as the box advances that distance is vt + Lharuspex said:
How do you get vt+L? Maybe it would help you to think about a more solid analogy. You are digging a tunnel. The excavator is a cube LxLxL. If it advances a distance vt, how much soil does it have to excavate? Draw a picture.welovephysics said:
sorry , the volume of the leading face of the box sweep through as the box advances that distance is (L^3) , what are you trying to say ?haruspex said:
No.welovephysics said:
noharuspex said:
So it can't be L3, right?welovephysics said:
what shape ? i am getting more confused nowharuspex said:
Lay a light box, LxLxL, on soft snow. Press down on it, pushing it down a distance vt. Remove the box. What shape hole have you made in the snow? What is its volume?welovephysics said:
volume = (L)(L)( L +vt) , shape = rectangular , is it correct ?haruspex said:
Nearly right. But we are only interested in the volume swept out by the leading face. That moves a distance vt, not L+vt.welovephysics said:
so the volume = (L)(L)(vt) ?haruspex said:
Yes!welovephysics said:
The formula for calculating resistance in a prototype boat is R = F/V, where R is the resistance, F is the force acting on the boat, and V is the velocity of the boat.
The force acting on the boat can be determined by considering all the external forces acting on the boat, such as the weight of the boat, the weight of any cargo, and the force of the water pushing against the boat. These forces can be calculated using basic physics principles.
The units used in the resistance formula should be consistent. For example, if the force is measured in newtons, then the velocity should be measured in meters per second. It is important to pay attention to units when working with formulas to ensure accurate results.
Yes, there are other factors that can affect the resistance in a prototype boat, such as the shape and size of the boat, the roughness of the surface of the boat, and the density of the water. These factors can be incorporated into the formula by using additional variables or coefficients.
The results of the resistance calculation can be used to determine which design elements are causing the most resistance and can be improved upon. For example, if the formula shows that the shape of the boat is creating a significant amount of resistance, the design can be modified to reduce this resistance and improve the boat's overall performance.
