Grant

Brian Cox visits the world's biggest vacuum chamber

What is the change in entropy of the bullet during this process?

A bullet of mass m slams into a concrete wall and melts on impact. If the temperature of the speeding bullet was T₁ and if the specific heat of bullet's material is c and its latent heat of fusion is L, what is the change in entropy of the bullet during this process? The melting point of lead is T₂.
Heating:
dS₁ = dQ₁/T
dQ₁ = cmdT
dS₁ = cmdT/T = cmd(lnT)
ΔS₁ = ∫cmd(lnT) = cm∫d(lnT) = cm(lnT₂-lnT₁) = cm ln(T₂/T₁)
ΔS₁ = cm ln(T₂/T₁)
Melting:
dS₂ = dQ₂/T₂
ΔS₂ = ΔQ₂/T₂ = Lm/T₂
ΔS = ΔS₁+ΔS₂ = cm ln(T₂/T₁)+Lm/T₂
A 16-gram lead bullet slams into a concrete wall and melts on impact. If the temperature of the speeding bullet was 25°C and if the specific heat of lead is 0.13 kJ/(kgK) and its latent heat of fusion is 23 kJ/kg, what is the change in entropy of the bullet during this process? The melting point of lead is 601K.
m=16g
T₁=25°C=25°+273.15=298.15K
T₂=601K
c=0.13 kJ/(kgK)
L=23 kJ/kg
ΔS=cm ln(T₂/T₁)+Lm/T₂
Calculation by Google calculator:
0.13kJ/(kg*K) * 16g * ln(601 / 298.15) in J/K =
1.4580763 J / K
23kJ/kg * 16g / 601K in J/K =
0.612312812 J / K
1.4580763 J / K + 0.612312812 J / K = 2.07038911 J / K
ΔS≈2.1J/K

Determine the Total Entropy Change of the Water

What is the diameter of the top section, d?

In the figure, water flows in the pipe as shown. What is the diameter of the top section, d?
d₁=5cm
v=4m/s
P=100kPa
P₂=31.5kPa
h=1.5m
d=?

vA=v₂A₂
vd₁²=v₂d²
v₂=vd₁²/d²
P+½ρv²=P₂+½ρv₂²+rgh
P+½ρv²=P₂+½ρ(vd₁²/d²)²+ρgh
P-P₂-ρgh=½ρv²d₁⁴/d⁴-½ρv²=½ρv²(d₁⁴/d⁴-1)
(P-P₂-ρgh)/(½ρv²)=d₁⁴/d⁴-1
(P-P₂-ρgh)/(½ρv²)+1=d₁⁴/d⁴
∜{(P-P₂-ρgh)/(½ρv²)+1} = d₁/d

d/d₁=∜{1/[(P-P₂-ρgh)/(½ρv²)+1]}
d  = d₁ ∜{1/[(P-P₂-ρgh)/(½ρv²)+1]}

Calculation

5cm*(1/( (100kPa-31.5kPa-1000kg/m^3*9.81m/s^2*1.5m)/(1000kg/m^3*(4m/s)^2)+1  ))^(1/4)=3.4598676 cm≈3.5cm

or
d₁⁴/d⁴-1= (P-P₂-ρgh)/(½ρv²)=
=(100kPa-31.5kPa-1000kg/m^3*9.81m/s^2*1.5m)/( 1000kg/m^3*(4m/s)^2)=
=3.3615625
d₁/d=(3.3615625+1)^(1/4)= 1.44514200657
d=d₁/∙1.44514200657= 5cm /1.44514200657=3.4598676 cm≈3.5cm

⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞√∛∜⨯∙ργ

What is the possible maximum frequency of this pendulum if the distance x can be varied?

A pendulum consists of a sphere of radius R tied to a light rod so that the center of mass of the sphere is located at a distance x from the pivot point. What is the possible maximum frequency of this pendulum if the distance x can be varied?
#Physics #problem #PhysicsProblem #PhysicalProblem #pendulum #sphere #radius #light #rod #center #mass #distance #pivot #point #maximum #frequency #lightrod #centerofmass

Ideal Simple Pendulum

Hints for Quiz 15

Formulae:    v= λ f;    μ=m/l;    v²=Fₜ/μ; P=½μω²A²v;   T=1/f;    ω=2πf; v²=v₀²+2ad; d=v₀t+at²; L =½mλ, m=1,2,... (fixed at both ends);    L =¼λ+½mλ   m=0,1,...(fixed at one end),

v= λ f          (v Wave speed, λ wavelength, f frequency)
μ=m/l          (μ linear density, m mass, l cord length)
v²=Fₜ/μ        (v speed of transverse waves in a cord,  Fₜ tension force)
P=½μω²A²v (P power transported by transverse waves in a string, cord, rope)
v²=v₀²+2ad; d=v₀t+at² (a acceleration, d displacement, t time)
Standing waves with the wavelength λ in a string with the length L:  L =½mλ, m=1,2,... (strings are fixed at both ends)
L =¼λ+½mλ,   m=0,1,...(strings are fixed at only one end)

1. A wire of uniform linear mass density hangs from the ceiling. It takes the time t for a pulse to travel the length of the wire. How long is the wire?
Use: μ=m/l;  F=mg; v²=v₀²+2ad; d=v₀t+at²

2. A stretched string with a mass m and a length l is placed under a tension F. How much power must be supplied to the string to generate traveling waves that have a frequency f and an amplitude A?
Use: μ=m/l;  v²=Fₜ/μ; P=½μω²A²v; v= λ f.

3.  A string of mass m and length l can carry a wave with the power P.  The amplitude of the wave is A.  What is the period of oscillation if the tension applied to the string is F?
Use: μ=m/l;  v²=Fₜ/μ; P=½μω²A²v; T=1/f; v= λ f.

4. A string of mass m and length l can carry a wave with the power P. When a tension of F is applied to the string it oscillates with a period T. What is the amplitude of the wave?
Use: μ=m/l;  v²=Fₜ/μ; P=½μω²A²v; T=1/f.

5. In the figure, point A is below the ceiling on the distance h. Determine how much longer it will take for a pulse to travel along wire 1 than it would along wire 2. Both wires are made of the same material and have identical cross sections.
Use:   μ=m/l,    v²=Fₜ/μ; F=mg.

6. A guitar string with a mass m and a length l is attached to a guitar at two points separated by the distance x. What tension must the guitar string have so that its fundamental note is f?
Use L =½mλ, m=1,2,...; v= λ f; μ=m/l;  v²=Fₜ/μ;

7. Two adjacent frequencies of transverse standing waves on a string held fixed at both ends are f1 and f2. What is the fundamental frequency of the transverse standing waves on this string?
Use: L =½mλ, m=1,2,...;

8.  A wave of amplitude A, wavelength λ, and frequency f.  What is its speed?
Use: v= λ f
⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞√∛∜⨯∙ργ

Tautochrone curve - брахистохроны и таутохроны

What is the magnitude of the normal force N the support exerts on the object?

The 2-dimensional object of art shown in the figure is held in display at a museum by support S and rope R. Support S is on the floor and rope R is attached to the ceiling. The coefficient of static friction between the object and support S is 0.5. If the angle, θ has its maximum value such that the object remains in static equilibrium on the support, what is the magnitude of the normal force N the support exerts on the object? The object's mass is 25 kg and the distances from object's center of mass, CM, to its contact points with the support and the rope are s = 35 cm and r = 55 cm, respectively.
μ=0.5; m=25kg; s=35cm; r=55cm
fₛ≤μN
fₛ=μN (maximum)
fₛ=Tsinθ
N+Tcosθ=mg
N∙s∙sin30°+fₛ∙s∙cos30°=rTcosθ
N∙s∙sin30°+μN∙s∙cos30°=r(mg-N)
N∙s∙sin30°+μN∙s∙cos30°=rmg-rN
N∙s∙sin30°+μN∙s∙cos30°+rN=rmg
N∙(s∙sin30°+μ∙s∙cos30°+r)=rmg
N∙=rmg/(s∙sin30°+μ∙s∙cos30°+r)
Calculation: 55*25*9.81/(35*sin(30degree)+0.5*35*cos(30degree)+55cm)=153.88376691
N=154 N

What is the length of vector A-B?

Vector A points along the y-axis with length 5.0, vector B makes an angle of 55° with respect to the x-axis and has a length of 7.0. What is the length of vector A-B?
Solution:
Aₓ=0
Ay=5
Bₓ=7cos55°
By=7sin55°
(A-B)ₓ=-7cos55°
(A-B)y=5-7sin55°
|(A-B)| = √{(-7cos55°)²+(5-7sin55°)²}
Calculation in Google calculator
sqrt((7*cos(55degree))^2+(5-7*sin(55degrees))^2)=4.0815875465
|(A-B)|≈4.1

Trigonometrical Solution:
In trigonometry, the law of cosines (also known as the cosine formula or cosine rule) relates the lengths of the sides of a triangle to the cosine of one of its angles. The law of cosines states
c² = a² + b² - 2ab∙cos γ
where γ denotes the angle contained between sides of lengths a and b and opposite the side of length c.
The vector (A-B) geometrically looks as the third side of the triangle built by two vectors A and B.
|(A-B)|=√{A+B-2|A||B|cos(γ)},
where |A|=5, |B|=7, γ is the angle between vector, γ=90°-55°=35
Calculation:
sqrt(7^2+5^2-2*7*5*cos(35degree))=4.0815875465
|(A-B)|≈4.1

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⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞√∛∜⨯∙ργ

5 Fun Physics Phenomena

An object is launched up a 30-degree slope at an angle of 20 degrees above the incline of the slope and lands a distance L= 3.12 m up the slope. What is the object's initial speed?

An object is launched up a 30-degree slope at an angle of 20 degrees above the incline of the slope and lands a distance L= 3.12 m up the slope.  What is the object's initial speed?
ϴ₁=30°
ϴ₂=20°
L= 3.12 m
v₀=?
v₀ cos(ϴ₁+ϴ₂) ∙t =L∙cos(ϴ₁)
v₀ sin(ϴ₁+ϴ₂) ∙t - ½gt² = L∙sin(ϴ₁)
t ={L∙cos(ϴ₁) / (v₀ cos(ϴ₁+ϴ₂))}

v₀ sin(ϴ₁+ϴ₂) ∙{L∙cos(ϴ₁) / (v₀ cos(ϴ₁+ϴ₂))} - ½g{L∙cos(ϴ₁) / (v₀ cos(ϴ₁+ϴ₂))}² = L∙sin(ϴ₁)

sin(ϴ₁+ϴ₂) ∙L∙cos(ϴ₁) / cos(ϴ₁+ϴ₂) - ½gL²∙cos²(ϴ₁) / (v₀² cos²(ϴ₁+ϴ₂)) = L∙sin(ϴ₁)

sin(ϴ₁+ϴ₂) ∙cos(ϴ₁) / cos(ϴ₁+ϴ₂) - ½gL∙cos²(ϴ₁) / (v₀² cos²(ϴ₁+ϴ₂)) = sin(ϴ₁)

½gL∙cos²(ϴ₁) / (v₀² cos²(ϴ₁+ϴ₂))=sin(ϴ₁+ϴ₂) ∙cos(ϴ₁)/cos(ϴ₁+ϴ₂)-sin(ϴ₁)

½gL∙cos²(ϴ₁) / (v₀² cos²(ϴ₁+ϴ₂))=
=(sin(ϴ₁+ϴ₂) ∙cos(ϴ₁) -cos(ϴ₁+ϴ₂)∙sin(ϴ₁))/cos(ϴ₁+ϴ₂)= sin(ϴ₂)/cos(ϴ₁+ϴ₂)

½gL∙cos²(ϴ₁) / (v₀² cos²(ϴ₁+ϴ₂))=sin(ϴ₂)/cos(ϴ₁+ϴ₂)

v₀² cos²(ϴ₁+ϴ₂) / (½gL∙cos²(ϴ₁))   =  cos(ϴ₁+ϴ₂)/sin(ϴ₂)
v₀²  / cos²(ϴ₁)   =  gL/(2sin(ϴ₂)cos(ϴ₁+ϴ₂))

v₀/cos(ϴ₁)   =  √{gL/(2sin(ϴ₂)cos(ϴ₁+ϴ₂))}
v₀   =  cos(ϴ₁)√{gL/(2sin(ϴ₂)cos(ϴ₁+ϴ₂))}
Calculation:

cos(30deg)*sqrt(9.81m/s^2*3.12m/(2*sin(20deg)*cos(30deg+20deg)))=
=7.22549898 m/s
v₀ ≈ 7.23 m/s

°⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞√∛∜⨯∙ρϴ