(a) A submarine descends to a depth of 70 m below the surface of water. The density of the water is 1050 kg/m3. Atmospheric pres
1 answer:
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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
Answer:
1.07 x 10⁻⁸N
Explanation:
Given parameters:
Mass 1 = 200kg
Mass 2 = 500kg
Distance of separation = 25m
Unknown:
Gravitational attraction between the two bodies = ?
Solution:
To solve this problem, we use the equation of the universal gravitation;
F =
G is the universal gravitation constant = 6.67 x 10⁻¹¹Nm²kg⁻²
r is the distance
Now insert the parameters and solve;
F = = 1.07 x 10⁻⁸N
This question is incomplete
Complete Question
Three equal point charges are held in place as shown in the figure below
If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.
A) F1=2F2
B) F1=3F2
C) F1=4F2
D) F1=9F2
Answer:
D) F1=9F2
Explanation:
We are told in the question that there are three equal point charges.
q, Q1, Q2 ,
q = Q1 = Q2
From the diagram we see the distance between the points d
q to Q1 = d
Q1 to Q2 = nd
Assuming n = 2
= 2 × d = 2d
Sum of the two distances = d + 2d = 3d
F1 is the force on q due to Q1 and
F2 is the force on q due to Q2,
Since we have 3 equal point charges and a total sum of distance which is 3d
Hence,
F1 = 9F2
