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.
<span>Using conservation of energy and momentum you can solve this question. M_l = mass of linebacker
M_ h = mass of halfback
V_l = velocity of linebacker
V_h = velocity of halfback
So for conservation of momentum,
rho = mv
M_l x V_li + M_h x V_hi = M_l x V_lf + M_h x V_hf
For conservation of energy (kinetic)
E_k = 1/2mv^2/ 1/2mV_li^2 + 1/2mV_{hi}^2 = 1/2mV_{lf}^2 + 1/2mV_{hf}^2
Where i and h stand for initial and final values.
We are already told the masses, \[M_l = 110kg\] \[M_h = 85kg\] and the final velocities \[V_{fi} = 8.5ms^{-1}\] and \[V_{ih} = 7.2ms^{-1} </span>
During the period of constant acceleration, the car's average speed is (1/2) (16 + 32) = 24 m/s.
At that average speed, it covers 240 meters in 10 seconds.
Different wavelengths of light are seen as colors