Answer:
The tension in the string when the speed increased is 134.53 N
Explanation:
Given;
Tension in the string, T = 120 N
initial speed of the transverse wave, v₁ = 170 m/s
final speed of the transverse wave, v₂ = 180 m/s
The wave speed is expressed as;

where;
μ represents mass per unit length

The new tension T₂ will be computed as;

Consequently, the tension in the string when the speed was increased is 134.53 N
The string vibrates in its third harmonic, where n = 3. The length of the string, l, measures 0.36 m. The frequency of the sound produced is f = 500 Hz. The speed of sound in air is 344 m/s. To find the speed of sound generated by the string in the third harmonic, we can apply the appropriate formula for frequency.
Answer:

Explanation:
The friction created between the tire and the ground generates thermal energy as force is applied during skidding.
The mentioned force relates to half the impact on the rear tire, resulting in a calculated normal force of,

The work executed is determined by the frictional force and the distance covered,

Where ![\mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)](https://tex.z-dn.net/?f=%20%5Cmu_k%20%5B%2F%20tex%5D%20is%20the%20coefficient%20of%20kinetic%20friction%3C%2Fp%3E%3Cp%3EN%20is%20the%20normal%20force%20previously%20found%20d%20is%20the%20distance%20traveled%2C%3C%2Fp%3E%3Cp%3EReplacing%2C%3C%2Fp%3E%3Cp%3E%5Btex%5DW_f%20%3D%20%280.80%29%28441%29%280.42%29)
The thermal energy produced from the work done is,

Answer:
29.4 N/m
0.1
Explanation:
a) The restoring force is described by: F_r = -k*x.
The gravitational force is given as: F_g = mg.
Where:
F_r = restoring force,
F_g = gravitational force,
g represents acceleration due to gravity,
and k is the force constant, while x1 and x2 are the displacements for the two masses.
Given:
m1 = 1.29 kg,
m2 = 0.3 kg,
x1 = -0.75 m,
x2 = -0.2 m,
g = 9.8 m/s².
Substituting the information yields:
F_r = F_g
-k*x1 + k*x2 = m1*g - m2*g.
Solving gives k = 29.4 N/m.
b) To find the unloaded length l:
l = x1 - (F_1/k).
Given:
m1 = 1.95kg, x1 = -0.75m.
Substituting in provides:
l = x1 - (F_1/k) = 0.1.