Answer is 3. Volume= mass divided by density
Answer:
(a) The ratio of the pressure amplitude of the waves is 43.21
(b) The ratio of the intensities of the waves is 0.000535
Explanation:
Given;
density of gas,
= 2.27 kg/m³
density of liquid,
= 972 kg/m³
speed of sound in gas,
= 376 m/s
speed of sound in liquid,
= 1640 m/s
The of the sound wave is given by;

Where;
is the pressure amplitude

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

Answer:
a) W = 46.8 J and b) v = 3.84 m/s
Explanation:
The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy
W = ΔK =
-K₀
a) work is the scalar product of force by distance
W = F . d
Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.
W = F d cos θ
W = 39.0 1.20 cos 0
W = 46.8 J
b) zero initial kinetic language because the package is stopped
W -
=
-K₀
W - fr d= ½ m v² - 0
W - μ N d = ½ m v
on the horizontal surface using Newton's second law
N-W = 0
N = W = mg
W - μ mg d = ½ m v
v² = (W -μ mg d) 2/m
v = √(W -μ mg d) 2/m
v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3
]
v = √(31.63 2/4.3)
v = 3.84 m/s
Answer:
<h2>73 kg</h2>
Explanation:
The mass of the object can be found by using the formula

f is the force
a is the acceleration
From the question we have

We have the final answer as
<h3>73 kg</h3>
Hope this helps you