Answer:
v=18.97 m/s
Explanation:
the equation for the above question is v²=u²+2as
1.2 x (2.2 x 10⁵) = 264,000 Ω
0.8 x (2.2 x 10⁵) = 176,000 Ω
With a 'nominal' value of 220,000 Ω, it could actually be anywhere <em>between 176,000Ω and 264,000Ω</em> .
Incomplete question as the charge density is missing so I assume charge density of 3.90×10^−12 C/m².The complete one is here.
An electron is released from rest at a distance of 0 m from a large insulating sheet of charge that has uniform surface charge density 3.90×10^−12 C/m² . How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?
Answer:
Work=1.06×10⁻²¹J
Explanation:
Given Data
Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²
Charge density σ=3.90×10⁻¹² C/m²
The electron moves a distance d=3.00×10⁻²m
Electron charge e=-1.6×10⁻¹⁹C
To find
Work done
Solution
The electric field due is sheet is given as
E=σ/2ε₀

Now we need to find force on electron

Now for Work done on the electron
Answer: 1.135 L/s; 1.35 kg/s, 22.57 m/s
Explanation:
Given
Volume of bucket 
time to fill it 
Volume flow rate

The inner diameter of the hose 
diameter of the nozzle exit 
we can volume flow rate as

Mass flow rate
