Answer:
The tension in the string is 
.
Explanation:
For a string with tension 
and linear density 
carrying a transverse wave at speed 
it is true that
solving for we get:
Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is
And it's linear density (mass per unit length) is
Therefore, the tension in the cord is
or in micro newtons
Answer. dry friction- <span> resists relative lateral motion of two solid surfaces in contact.</span>
Answer:
t = 1.77 s
Explanation:
The equation of a traveling wave is
y = A sin [2π (x /λ -t /T)]
where A is the oscillation amplitude, λ the wavelength and T the period
the speed of the wave is constant and is given by
v = λ f
Where the frequency and period are related
f = 1 / T
we substitute
v = λ / T
let's develop the initial equation
y = A sin [(2π / λ) x - (2π / T) t +Ф]
where Ф is a phase constant given by the initial conditions
the equation given in the problem is
y = 5.26 sin (1.65 x - 4.64 t + 1.33)
if we compare the terms of the two equations
2π /λ = 1.65
λ = 2π / 1.65
λ = 3.81 m
2π / T = 4.64
T = 2π / 4.64
T = 1.35 s
we seek the speed of the wave
v = 3.81 / 1.35
v = 2.82 m / s
Since this speed is constant, we use the uniformly moving ratios
v = d / t
t = d / v
t = 5 / 2.82
t = 1.77 s
Answer:
527 Hz
Solution:
As per the question:
Beat frequency of the player, 
Frequency of the tuning fork, f = 523 Hz
Now,
The initial frequency can be calculated as:
when
when
But we know that as the length of the flute increases the frequency decreases
Hence, the initial frequency must be 527 Hz
Answer:
Earth attract the Moon with a force that is greater.
Explanation:
According to the law of gravitation, the gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Mathematically, F1 = Gm1m2/r²... 1
Let m1 be the mass of the earth and m2 be that of the moon
If the Earth is much more massive than is the Moon, the new force of attraction between them will become;
F2= G(2m1)m2/r²
F2 = 2Gm1m2/r² ... (2)
Dividing eqn 1 by 2 we have;
F1/F2 = (Gm1m2/r²)÷(2Gm1m2/r²)
F1/F2 = Gm1m2/r²×r²/2Gm1m2
F1/F2 = 1/2
F2=2F1
This shows that that the earth will attract the moon by a force 2times the initial force of the masses(i.e a much greater force)
