Hi! I believe your answer is decreasing. <u>An inclined plane makes work easier by decreasing the amount of effort force needed, but increases the distance</u>. I hope this helps you! Good luck and have a great day. ❤️✨
Answer:
You will hear the note E₆
Explanation:
We know that:
Your speed = 88m/s
Original frequency = 1,046 Hz
Sound speed = 340 m/s
The Doppler effect says that:

Where:
f = original frequency
f' = new frequency
v = velocity of the sound wave
v0 = your velocity
vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.
Replacing the values that we know in the equation we have:

This frequency is close to the note E₆ (1,318.5 Hz)
Generally, the length of the line will indicate how strong the force is. If you have two opposing forces and one is higher than the other, you would draw the line of the higher force visibly longer.
Complete question:
A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?
Answer:
The current in the circuit 7 ms later is 0.2499 A
Explanation:
Given;
Ideal inductor, L = 45-mH
Resistor, R = 60-Ω
Ideal voltage supply, V = 15-V
Initial current at t = 0 seconds:
I₀ = V/R
I₀ = 15/60 = 0.25 A
Time constant, is given as:
T = L/R
T = (45 x 10⁻³) / (60)
T = 7.5 x 10⁻⁴ s
Change in current with respect to time, is given as;

Current in the circuit after 7 ms later:
t = 7 ms = 7 x 10⁻³ s

Therefore, the current in the circuit 7 ms later is 0.2499 A