Answer:
hello your question has some missing parts below is the complete question
and the missing diagram
The two speakers emit sound that is 180° out of phase and of a single frequency,ƒ, Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.
answer : 1316.2 hertz
Explanation:
The frequency that produce the maximum sound intensity can be calculated using the relation below
dsin ∅ = n <em>A</em>
where <em>A = </em>dsin ∅ / n when n = 1 . d = 0.800
<em>A</em> = 0.800 * ( 1 / 3.162 )
<em>A</em> = 0.253 m
speed of sound = 333 m/s
frequency = speed /<em> A</em>
<em>= </em>333 / 0.253 = 1316.2 hertz
The magnitude of the component of the box’s weight
perpendicular to the incline can be olve using the formula:
F = wcos(a)
Where F is the box’s weight perpendicular to the incline
W is the weight of the box
A is the angle of the incline
F = (46)cos(25)
F = 42 N
<u>Answer</u>:
pairing of the copper which has an electron affinity of 0.34 and the silver which has an electron affinity of 0.80 makes a strong battery.
<u>Explanation</u>:
All the Batteries of this world are made with two metals having different-different electron affinities. What do the phrase “electrons affinities ” mean and how do these affinities affect the voltage of the batteries?
The Electron affinities are the energy change that occurs when electrons are added to atoms. The greater the attraction of the atoms to the electrons the more energy would released. If an atom has high electron affinity, the electron will be harder to gain The greater is the difference in metal affinities, the greater the voltage. That means, if you pair Coppers with Silver, the difference between their electron affinities would be (0.80-0.34) or, 0.46 and You can make a strong battery.
Explain<span> why it is </span>not advisable to use small values<span> of incident ray in </span>performing experiment<span> on the</span>refraction through a glass prism<span>.</span>
Answer:
<h2>1.5 ohms</h2>
Explanation:
Power is expressed as P = V²/R
R = resistance
V = supplied voltage
Given P = 600W and V = 30V
R = V²/P
R = 30²/600
R = 900/600
R = 1.5ohms
magnitude of its resistance is 1.5ohms