The time for the echo to return is directly proportional to the distance. vw = fλ. In a given medium under fixed conditions, vw is constant, so that there is a relationship between f and λ; the higher the frequency, the smaller the wavelength.
Base on your question where a 14.8g of piece of Styrofoam carries a net charge of -0.742C and is suspended in equilibrium above the center of a large, horizontal sheet of plastic so the ask of the problem is to calculate the charge per unit area on the plastic sheet. The answer would be 21.96
Answer:
The force on one side of the plate is 3093529.3 N.
Explanation:
Given that,
Side of square plate = 9 m
Angle = 60°
Water weight density = 9800 N/m³
Length of small strip is


The area of strip is

We need to calculate the force on one side of the plate
Using formula of pressure


On integrating




Hence, The force on one side of the plate is 3093529.3 N.
Here we deal with a lever law. It states that product of force and distance from a fixed point on a lever is equal on both sides.
F₁*d₁ = F₂*d₂
By analysing this formula we can see that applying small force on a great length equals great force on a small length.
To remove nail we need to apply certain force. If we use F₁ for this required force we can see that on other side we need to apply certain force. If we have greater arm length we need smaller force. In a crowbar arm length along which we apply force is greater than length of our arm. This leads to a conclusion that we need smaller force when using crowbar. Depending on the length of a nail it is possible that we need to apply force that is greater than force required to remove nail.
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)