The coefficient of friction between the soap and the floor is 0.081
If Juan steps on the soap with a force of 493 N, this is her weight, W. This weight also equals the normal reaction on the floor, N.
We know that frictional force F = μN where μ = coefficient of friction between soap and floor.
So, μ = F/N
Since F = 40 N and N = W = 493 N,
μ = F/N
μ = 40 N/493 N
μ = 0.081
So, the coefficient of friction between the soap and the floor is 0.081
Learn more about coefficient of friction here:
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Answer:
a)ΔV = 6.48 KV
b)ΔU =18.79 mJ
Explanation:
Given that
E= 1.8 KV/m
a)
We know that
Electric potential difference ΔV given as
ΔV = E .d
Here
E= 1.8 KV/m
d= 3.6 m
ΔV = E .d
ΔV = 1.8 x 3.6 KV
ΔV = 6.48 KV
b)
Given that
q=+2.90 µC
Change in electric potential energy ΔU given as
ΔU = q .ΔV
ΔU =18.79 mJ
Answer:
(a) 4.0334Ω
(b)parallel
Explanation:
for resistors connected in parallel;
Req =3.03Ω , R1 =12.18Ω
R2=1/0.2479
R2=4.0334Ω
(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.
Req = R1+R2
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Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
