(a) The tension the musician must stretch it is 147.82 N.
(b) The percent increase in tension is needed to increase the frequency is 26%.
<h3>Tension in the string</h3>
v = √T/μ
where;
- v is speed of the wave
- T is tension
- μ is mass per unit length = 0.0144 kg / 0.6 m = 0.024 kg/m
v = Fλ
in fundamental mode, v = F(2L)
v = 2FL
v = 2 x 65.4 x 0.6 = 78.48 m/s
v = √T/μ
v² = T/μ
T = μv²
T = 0.024 x (78.48)²
T = 147.82 N
<h3>When the frequency is 73.4 Hz;</h3>
v = 2FL = 2 x 73.4 x 0.6 = 88.08 m/s
T = μv²
T = (0.02)(88.08)²
T = 186.19 N
<h3>Increase in the tension</h3>
= (186.19 - 147.82)/(147.82)
= 0.26
= 0.26 x 100%
= 26 %
Thus, the tension the musician must stretch it is 147.82 N.
The percent increase in tension is needed to increase the frequency is 26%.
Learn more about tension here: brainly.com/question/24994188
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Explanation:
Efficiency is a way of describing the amount of useful output a process or machine can generate as a percentage of the input required to make it go. In other words, it compares how much energy is used to do work versus how much is lost or wasted to the environment. The more efficient the machine, the less energy wasted.
For example, if a heat engine is able to turn 75 percent of the fuel it receives into motion, while 25 percent is lost as heat in the process, it would be 75 percent efficient. Out of the original 100 percent of the fuel, 75 percent was output as useful work.
the equation:
energy efficiency =useful output energy/total input energy
As the box is moving with a constant velocity, the two forces acting on the box are canceling each other.
Then friction force = 80 Newtons but in the opposite direction.
Friction force = Mu * Normal force exerted by ground = Mu * weight of box
So we find Mu.
Mu = coefficient of friction between box and horizontal surface
= Force of friction / weight = 80 / 50 * 9.81 = 0.163
When an identical box is placed on top, the force of friction is
= Mu * total weight = 0.163 * (50+50) * 9.81 = 159.9 Newtons
Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
= - gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
Answer:
sorry about the other person but its b
Explanation:
