Answer:
Explanation:
From the question we are told that:
Number of battery 
Voltage source
Lamp Power
Generally the equation for Resistance is mathematically given by
Therefore
Generally the equation for Current is mathematically given by
The twice as heavy weight will hit the ground with more force, or impact.
Impulse, denoted as J, is defined by the change in momentum. Since we have our initial and our final, we can solve for the change in momentum.
0.495 m/s
Explanation
the formula for the terminal velocity is given by:
![\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\ \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2mg%7D%7B%5Csigma%20AC%7D%7D%20%5C%5C%20%5Ctext%7Bwhere%7D%20%5C%5C%20%20%5Cend%7Bgathered%7D)
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient
then
Step 1
Let's find the mass
now, replace
![\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2mg%7D%7B%5Csigma%20AC%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2%280.002kg%29%289.81%5Ctext%7B%20%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%7D%7B%282%5Ccdot10%5E3%5Cfrac%7B%5Coperatorname%7Bkg%7D%7D%7Bm%5E3%7D%29%280.0001m%5E2%290.8%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.03924%5Cfrac%7B%5Coperatorname%7Bkg%7Dm%7D%7Bs%5E2%7D%7D%7B0.16%5Cfrac%7B%5Coperatorname%7Bkg%7D%7D%7Bm%5E%7B%7D%7D%7D%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B0.2452%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%7D%20%5C%5C%20v%3D0.495%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
hence, the answer is 0.495 m/s
Is this like a zoom or something like that
