A) The power delivered to the lines is
And the voltage at which the lines work is
Since the power delivered is the product between the voltage and the current:
We can find the current flowing in the lines:
b) The voltage change along each line can be found by using Ohm's law:
c) The power wasted as heat along each line is given by:
And since we have 2 lines, the total power wasted as heat in both lines is
The bigger one as the luminosity has a direct exponential relationship with R of the star so the star which has more surface area will be more luminous than the smaller one
The momentum change of the running back is - 664.2 kg m/s or 664 west.
<u>Explanation:</u>
Momentum is defined as the change in velocity of any object along with its mass. So mathematically, momentum can be derived using the product of mass with the change in velocity.
As here mass is given as 82 kg and the initial velocity was 5.6 m/s and final velocity is 2.5 m/s.
Initial Momentum = 
Final Momentum = 
Momentum change = Final Momentum - Initial Momentum
Momentum change = - 205 - 459.2 = - 664.2 kg m/s or 664 west
Thus, the changing momentum is -664.2 kg m/s. The negative sign indicates that the momentum is acting in the opposite direction on changing in the direction of velocity.
Answer:
1.28 m, 14 m/s
Explanation:
At the maximum height, the velocity is 0.
Given:
a = -9.8 m/s²
v₀ = 5.00 m/s
v = 0 m/s
x₀ = 0 m
Find:
x
v² = v₀² + 2a(x - x₀)
(0 m/s)² = (5.00 m/s)² + 2(-9.8 m/s²) (x - 0 m)
x = 1.28 m
The maximum speed is at the bottom of the well.
Given:
a = -9.8 m/s²
v₀ = 5.00 m/s
x₀ = 0 m
x = -8.5 m
Find:
v
v² = v₀² + 2a(x - x₀)
v² = (5.00 m/s)² + 2(-9.8 m/s²) (-8.5 m - 0 m)
v = -13.8 m/s
Rounded to 2 sig-figs, the maximum speed is 14 m/s.
