Answer:
(a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is 
Explanation:
Given that,
Charge q₁ = -4.00 μC
Inner radius = 3.13 m
Outer radius = 4.13 cm
Net charge q₂ = -6.43 μC
We need to calculate the charge on the outer surface
Using formula of charge
The charge on the inner surface is q.
We need to calculate the electric field outside the shell
Using formula of electric field
Put the value into the formula
Hence, (a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is
Scott needs to determine the density of a metallic rod. First, he should determine the mass of his sample on the laboratory balance. Second, he should measure the volume of his sample by water displacement. Finally, he can calculate the density by dividing mass/volume.
Hope this helped ;)
Answer:
C. 85%
Explanation:
A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat?
A. 15%
B. 30%
C. 85%
D. 100%
work done by the system will be
W=PdV
p=pressure
dV=change in volume
3tam will be changed to N/m^2
3*1.01*10^5
W=3.03*10^5*(1.5-1)
convert 0.5L to m^3
5*10^-4
W=3.03*10^5*5*10^-4
W=152J
therefore
to find the percentage used
152/1000*100
15%
100%-15%
85% uf the fuel's energy was lost to friction and heat
Answer:
A title
Explanation:
Because this is middle school.
Because gravity and it's force pushes an object down
