Suppose a piano tuner stretches a steel piano wire 7.5 mm. The wire was originally 0.975 mm in diameter, 1.45 m long, and has a
1 answer:
The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.
<h3>What is force?</h3>Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.
The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.
Given data;
Young modulus, E=2.10 × 10¹¹ N/m²
Cross-sectional area,A
Final length,
Initial length,
Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.
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1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)
2) The angular acceleration is C)
3) The frequency of the oscillation is C) 1.6 Hz
4) The period of vibration is B) 0.6 s
5) The diameter of the nozzle is A) 5.0 mm
6) The force that must be applied is B) 266.7 N
Explanation:
1)
The period of a simple pendulum is given by
where
T is the period
L is the length of the pendulum
g is the acceleration of gravity
From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is
B) III. only (the length of the pendulum)
2)
The angular acceleration of the rotating disc is given by the equation
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is the final angular velocity
is the initial angular velocity
t is the time elapsed
For the compact disc in this problem we have:
(since it starts from rest)
is the final angular velocity
t = 2 s
Substituting, we find
3)
For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation
where
a is the acceleration
x is the displacement
is the angular frequency of the system
For the oscillator in this problem, we have the following relationship
which implies that
And so
Also, the angular frequency is related to the frequency by
Therefore, the frequency of this simple harmonic oscillator is
4)
When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write
where
m is the mass
is the acceleration of gravity
k is the spring constant
x = 8.0 cm = 0.08 m is the stretching of the spring
We can re-arrange the equation as
The angular frequency of the spring is given by
And therefore, its period is
5)
According to the equation of continuity, the volume flow rate must remain constant, so we can write
where
is the cross-sectional area of the hose, with
being the radius of the hose
is the speed of the petrol in the hose
is the cross-sectional area of the nozzle, with
being the radius of the nozzle
is the speed in the nozzle
Solving for , we find the radius of the nozzle:
So, the diameter of the nozzle will be
6)
According to the Pascal principle, the pressure on the two pistons is the same, so we can write
where
is the force that must be applied to the small piston
is the area of the first piston, with
being its radius
is the force applied on the bigger piston (the weight of the car)
is the area of the bigger piston, with
being its radius
Solving for , we find
So, the closest answer is B) 266.7 N.
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