P2O5 is a covalent compound used to purify sugar. What is the name of this compound?
2 answers:
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a)
b) See plot attached
c) 10.0 m
d) 0.500 cm
Explanation:
a)
The position of the tip of the lever at time t is described by the equation:
(1)
The generic equation that describes a wave is
(2)
where
A is the amplitude of the wave
T is the period of the wave
t is the time
By comparing (1) and (2), we see that for the wave in this problem we have
Therefore, the period is
b)
The sketch of the profile of the wave until t = 4T is shown in attachment.
A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.
The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.
The maximum displacement of the wave is given by the value of y when , and from eq(1), we see that this is equal to
y = 0.500 cm
So, this is the maximum displacement represented in the sketch.
c)
When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:
where
is the wavelength of the wave
L is the length of the string
In this problem,
L = 5.00 m is the length of the string
Therefore, the wavelength is
d)
The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.
In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).
And by comparing it with the general equation of a wave:
In fact, the maximum displacement occurs when the sine part is equal to 1, so when
which means that
And therefore in this case,
So, this is the displacement.
Answer:
The speed of the student just before she lands, v₂ is approximately 8.225 m/s
Explanation:
The given parameters are;
The mass of the physic student, m = 43.0 kg
The height at which the student is standing, h = 12.0 m
The radius of the wheel, r = 0.300 m
The moment of inertia of the wheel, I = 9.60 kg·m²
The initial potential energy of the female student, P.E.₁ = m·g·h₁
Where;
m = 43.0 kg
g = The acceleration due to gravity ≈ 9.81 m/s²
h = 12.0 h
∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J
The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;
The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0
∴ K₁ = 0 + 0 = 0
The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0
Where;
h₂ = 0 m
The final kinetic energy, K₂, of the wheel and student is give as follows;
Where;
v₂ = The speed of the student just before she lands
ω₂ = The angular velocity of the wheel just before she lands
By the conservation of energy, we have;
P.E.₁ + K₁ = P.E.₂ + K₂
∴ m·g·h₁ + = m·g·h₂ +
Where;
ω₂ = v₂/r
∴ 5061.96 J + 0 = 0 +
5,061.96 J = 21.5 kg × v₂² + 53. kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²
v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²
v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s
The speed of the student just before she lands, v₂ ≈ 8.225 m/s.