Answer:
A real emf device has an internal resistance, but an ideal emf device does not.
Answer:
51 Ω.
Explanation:
We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:
Resistor 1 (R₁) = 40 Ω
Resistor 3 (R₃) = 70.8 Ω
Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?
Since the two resistors are in parallel connection, their equivalent can be obtained as follow:
R₁ₙ₃ = R₁ × R₃ / R₁ + R₃
R₁ₙ₃ = 40 × 70.8 / 40 + 70.8
R₁ₙ₃ = 2832 / 110.8
R₁ₙ₃ = 25.6 Ω
Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:
Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω
Resistor 2 (R₂) = 25.4 Ω
Equivalent Resistance (Rₑq) =?
Rₑq = R₁ₙ₃ + R₂ (series connection)
Rₑq = 25.6 + 25.4
Rₑq = 51 Ω
Therefore, the equivalent resistance of the group is 51 Ω.
I think is c because your seeing how your using you’re time.
Answer:
1.19 m/s²
Explanation:
The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so
f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)
Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²
Equating (1) and (2) we ave
2(√mg/μ)/f = T²g/4π²
Making g subject of the formula
g = 2π√(2√(m/μ)/f)/T
The period T = 316 s/100 = 3.16 s
Substituting the other values into , we have
g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16
g = 2π√(2 × 35.877/200 Hz)/3.16
g = 2π√(71.753/200 Hz)/3.16
g = 2π√(0.358)/3.16
g = 2π × 0.599/3.16
g = 1.19 m/s²