Copper wire #1 has a length L and a radius b. Copper wire #2 has a length 2L and a radius 2b. Which statement about the resistance across the ends of the wires is true? The resistance of wire #1 is four times higher than that of wire #2. The resistance of wire #1 is equal to that of wire #2. The resistance of wire #1 is twice as high as that of wire #2. The resistance of wire #1 is half that of wire #2.

The resistance of wire #1 is twice as high as that of wire

Explanation:

For resistance of a wire the formula is as follows

R = ρ l / s where  ρ is a constant called specific resistance , l is length and s is cross sectional area . If r be the radius of cross section

R = ρ l / π r²

For first wire , putting the given values in the equation

R₁ = ρ L/ π b²

For second wire,

R₂ = ρ 2L/ π 4b²

= 2

The resistance of wire #1 is twice as high as that of wire

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student