Complete Question
The compete question is shown on the first uploaded question
Answer:
The speed is
Explanation:
From the question we are told that
The distance of separation is d = 4.00 m
The distance of the listener to the center between the speakers is I = 5.00 m
The change in the distance of the speaker is by 
The frequency of both speakers is 
Generally the distance of the listener to the first speaker is mathematically represented as
![L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}](https://tex.z-dn.net/?f=L_1%20%20%3D%20%20%5Csqrt%7Bl%5E2%20%2B%20%5B%5Cfrac%7Bd%7D%7B2%7D%20%5D%5E2%7D)
![L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}](https://tex.z-dn.net/?f=L_1%20%20%3D%20%20%5Csqrt%7B5%5E2%20%2B%20%5B%5Cfrac%7B4%7D%7B2%7D%20%5D%5E2%7D)

Generally the distance of the listener to second speaker at its new position is
![L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}](https://tex.z-dn.net/?f=L_2%20%20%3D%20%20%5Csqrt%7Bl%5E2%20%2B%20%5B%5Cfrac%7Bd%7D%7B2%7D%20%5D%5E2%20%2B%20k%7D)
![L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}](https://tex.z-dn.net/?f=L_2%20%20%3D%20%20%5Csqrt%7B5%5E2%20%2B%20%5B%5Cfrac%7B4%7D%7B2%7D%20%5D%5E2%20%2B%200.6%7D)
Generally the path difference between the speakers is mathematically represented as

Here
is the wavelength which is mathematically represented as

=> 
=>
=>
Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves
=>
=>
Answer:

(we need the mass of the astronaut A)
Explanation:
We can solve this by using the conservation law of the linear momentum P. First we need to represent every mass as a particle. Also we can simplify this system of particles by considering only the astronaut A with an initial speed
of 0 m/s and a mass
and the IMAX camera with an initial speed
of 7.5 m/s and a mass
of 15.0 kg.
The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

By the law of conservation we know that
For
(final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).
So:


Answer:
acceleration = 2.4525 m/s²
Explanation:
Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²
Tension in the rope = T
Sol: m2 > m1
i) for downward motion of m2:
m2 a = m2 g - T
5 a = 5 × 9.81 m/s² - T
⇒ T = 49.05 m/s² - 5 a Eqn (a)
ii) for upward motion of m1
m a = T - m1 g
3 a = T - 3 × 9.8 m/s²
⇒ T = 3 a + 29.43 m/s² Eqn (b)
Equating Eqn (a) and(b)
49.05 m/s² - 5 a = T = 3 a + 29.43 m/s²
49.05 m/s² - 29.43 m/s² = 3 a + 5 a
19.62 m/s² = 8 a
⇒ a = 2.4525 m/s²