155Ω
Explanation:
R = R ref ( 1 + ∝ ( T - Tref)
where R = conduction resistance at temperature T
R ref = conductor resistance at reference temperature
∝ = temperature coefficient of resistance for conductor
T = conduction temperature in degrees Celsius
T ref = reference temperature that ∝ is specified at for the conductor material
T = 600 k - 273 k = 327 °C
Tref = 300 - 273 K = 27 °C
R = 50 Ω ( 1 + 0.007 ( 327 - 27) )
R = 155Ω
The answer is C, individuals copy works to view at a later time.
Answer:
T
Explanation:
= magnitude of current in each wire = 2.0 A
= length of the side of the square = 4 cm = 0.04 m
= length of the diagonal of the square = 
a = (0.04) = 0.057 m
= magnitude of magnetic field by wires at A and C
T
= magnitude of magnetic field by wire at B
T
Net magnitude of the magnetic field at D is given as
T
Answer: rp/re= me/mp= 544 * 10^-6.
Explanation: To calculate this problem we have to consider the circular movement by the electron and proton inside a magnetic field.
Then the dynamic equation for the circular movement is given by:
Fcentripetal= m*ω^2.r
q*v*B=m*ω^2.r
we write this for each particle then we have the following:
q*v*B=me* ω^2*re
q*v*B=mp* ω^2*rp
rp/re=me/mp=9.1*10^-31/1.67*10^-27=544*10^-6
