The distance from observer A of intensity of sound 59 db is 28.64 m and the distance from observer B of intensity of sound 83 db is 11.36m
Explanation:
Let's solve this problem in parts
let's start by finding the intensity of the sound in each observer
observer A β = 59 db
β = 
where I₀ = 
W / m²
= 
W / m²
Similarly for Observer b 
now we use that the emitted power that is constant is the intensity over the area of the sphere where the sound is distributed
P = I A
therefore for the two observers
the area of a sphere is
we substitute the above formula, we get
Let us call the distance from the observer A be to stereo speaker = x, so the distance from the observer B to the stereo speaker = 40- x; we substitute
after solving the above equation we get x = 28.64 m
This is the distance of observer A
similarly The distance from observer B is 35 - x
= 40 - 28.64
= 11.36m
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Answer:
Imagine you have an orange. This is your imaginary Earth. When you look at it in any direction, you won’t be able to see all sides of it. But when you peel the orange, flatten and stretch it out, you can begin to see everything.
Similarly, a map projection is a method by which cartographers translate a sphere or globe into a two-dimensional representation. In other words, a map projection systematically renders a 3D ellipsoid (or spheroid) of Earth to a 2D map surface.
Explanation:
formula for wavelength = speed/frequency
So 1500/200 = 7.5 meters
<span>We never really used the acronym "IMA", or ideal mechanical advantage, but I'm assuming you are trying to increase the leverage and ease the effort. If so, the answer is false. You want larger movement on the effort side, and smaller movement on the resistant side of the fulcrum.</span>
Answer:
h >5/2r
Explanation:
This problem involves the application of the concepts of force and the work-energy theorem.
The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.
Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)
So from newton's second law of motion,
W – N = mv²/r
N = normal force = 0
W = mg
mg = ma = mv²/r
mg = mv²/r
v²= rg
v = √(rg)
The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop
So
ΔPE = ΔKE = 1/2mv²
The height at the roller coaster starts is usually higher than the top of the loop by design. So
ΔPE =mgh - mg×2r = mg(h – 2r)
2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.
In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.
So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.
ΔPE > ΔKE
mg(h–2r) > 1/2mv²
g(h–2r) > 1/2(√(rg))²
g(h–2r) > 1/2×rg
h–2r > 1/2×r
h > 2r + 1/2r
h > 5/2r
