If one bulb goes out then all the others won't light up because electricity will be cut off. It's a disadvantage because in a parallel circuit if one bulb burns out all the others will still be on because they won't be affected. I hope I've helped you ☺
Answer:
0.265
Explanation:
Draw a free body diagram. There are four forces:
Normal force Fn pushing up.
Weight force mg pulling down.
Tension force T at an angle θ.
Friction force Fn μ pushing left.
Sum the forces in the y direction:
∑F = ma
Fn + T sin θ − mg = 0
Fn = mg − T sin θ
Sum the forces in the x direction:
∑F = ma
T cos θ − Fn μ = 0
Fn μ = T cos θ
μ = T cos θ / Fn
μ = T cos θ / (mg − T sin θ)
Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:
μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)
μ = 0.265
Answer:
1340.2MW
Explanation:
Hi!
To solve this problem follow the steps below!
1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid
W=αhQ
α=specific weight for water =9.81KN/m^3
h=height=220m
Q=flow=690m^3/s
W=(690)(220)(9.81)=1489158Kw=1489.16MW
2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator
Wr=(0.9)(1489.16MW)=1340.2MW
the maximum possible electric power output is 1340.2MW
Answer: 71.93 *10^3 N/C
Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:
∫E*dr=Q inside/εo Q inside is given by: λ*L then,
E*2*π*r*L=λ*L/εo
E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C
If I remember correctly, it is the 3rd answer choice.
