Find Ratio of Powers of Emitted Radiation

Two spheres are made from the same material. The radius of the second sphere is twice less than the radius of the first sphere, but the absolute temperature of the second sphere is twice greater than the absolute temperature of the first sphere. Find p₂/p₁, the ratio of the powers of emitted radiations of the second (p₂), and the first (p₁) spheres.

Deadline for the online homework quizzes is December 12, Monday, last day of classes, 12 PM

Deadline for the online homework quizzes is December 12, Monday, last day of classes, 12 PM.

Carnot Heat Engine

Find the Change in Entropy

A physical body with a constant mass and a constant heat capacity has the change of entropy ΔS₁₂ when the temperature of the body changes from the initial temperature T₁ to the final temperature T₂. If this body changes the temperature from the same initial temperature T₁ to the new final temperature T₃, find the change in entropy for this body in terms of ΔS₁₂, T₁, T₂, and T₃.

Find the heat Q₄₋₂₋₁, which is added to the gas in the process 4-2-1

In the process of taking a gas from state 1 to state 4 along the curved path the heat leaves the system is Q₁₋₄ and the work done on the system is W₁₋₄.

When the gas is taken along the path 4-2-1, the work done by the gas is W₄₋₂₋₁. Find the heat Q₄₋₂₋₁, which is added to the gas in the process 4-2-1, in terms of Q₁₋₄, W₁₋₄, W₄₋₂₋₁

Heat as Energy Transfer

Latent Heat

Physics Problem

What fraction of the original amount of air in the pressure vessel has been added into the pressure vessel?

A pressure vessel initially is filled with air at the temperature T₁ and the gauge pressure P₁. Finally air in the pressure vessel has the temperature T₂ and the gauge pressure P₂. What fraction of the original amount of air in the pressure vessel has been added into the pressure vessel, or removed from the pressure vessel? The pressure vessel has a constant volume.

Find the root mean square velocity of molecules with the mass m₂

A mixture of two gasses consists of molecules with the masses m₁ and m₂. Molecules of mass m₁ have the root mean square velocity v₁. Find the root mean square velocity of molecules with the mass m₂ in terms of m₁, m₂, and v₁

What Fraction of the Original Air Must Be Removed?

A tire is filled with air at the temperature T₁ to a gauge pressure of P₁. If the tire reaches a temperature T₂, what fraction of the original air must be removed if the original pressure P₁ is to be maintained?

Chapter 13. Fluids. Giancoli, Douglas C. Physics for Scientists and Engineers with Modern Physics - 4th Ed

Gravitational Potential Energy

#PhysicsProblem #Physics #problem #object #floor #gravitational #potential #energy #Howmuch #PotentialEnergy #AboveTheFloor #GivenData pic.twitter.com/JPMwGoTW69

— Physics I (@PHY210) October 12, 2016

A 1-kg object 2 m above the floor in a regular building at the Earth has a gravitational potential energy of 3 J.
How much gravitational potential energy does the object have when it is 4 m above the floor?

Given Data:
m = 1kg
y₁ = 2m
U₁ = 3J
y₂ = 4m
U₂ = ?

Useful Physics Formulas

U = mgh
ΔU = Δ(mgh) = mgΔh

Solution
ΔU = Δ(mgh) = mgΔh = mg(y₂ - y₁)
ΔU = U₂ - U₁
U₂ = U₁ + mg(y₂ - y₁)

U₂ = 3J + 1kg · 10ᵐ/s² · (4m - 2m) = 3J + 20J = 23J

Problem's answer is 23 J.

Δ·⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ⁱⁿᴬᴭᴮᴯᴱᴲᴳᴴᴵᴶᴷᴸᴹᴺᴻᴼᴽᴾᴿᵀᵁᵂᵃᵄᵅᵆᵇᵉᵊᵋᵌᵍᵏᵐᵑᵒᵓᵖᵗᵘᵚᵛᵝᵞᵟᵠᵡ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛᵢᵣᵤᵥᵦᵧᵨᵩᵪ½↉⅓⅔¼¾⅕⅖⅗⅘⅙⅚⅐⅛⅜⅝⅞⅑⅒⅟℀℁℅℆

BMCC Students, Please Repeat Your Submission for the Quiz 2!

BMCC Students, Please repeat your submission for the quiz 2! Unfortunately for technical reasons your previous submissions of the quiz 2 were not recorded.

Physics Problem - A Spring is Fixed Vertically to the Floor

A #spring is fixed #vertically to the floor.
A 10-kg #box is then #pressed down such that the spring is #compressed https://t.co/Z5fLcbq2LX

— Physics I (@PHY210) October 5, 2016

The potential energy of a 10-kg mass is given by ...

The potential energy is given by the function U(x) = kx⁻¹, where k is a constant, and variable x is the position, the coordinate. (a) Derive the function of the coordinate x, which describes the force produced by this potential energy.

F(x) = - d U(x) / dx

F(x) = - d kx⁻¹ / dx = kx⁻²

F(x) = kx⁻²

(b) Let this potential energy is obtained by the body with the mass of 10kg. The constant is known: k = -3.98 × 10¹⁵ Jm.
Find the force on this body when it is located atx = 6.38 × 10⁶ m?

F(x) = kx⁻² = -3.98 × 10¹⁵ Jm / ( 6.38 × 10⁶ m)²

The code for the Google calculator: (-3.98E15J*m) / ( 6.38E6m )^2

Google calculator's result: -97.7781272 newtons

Problem's answer: F = -97.8 N or 97.8 N towards x = 0

Constant Power

What is the power output of an engine when a car of mass M accelerates with the constant power from rest to a final speed V over a distance d on flat, level ground?

Ignore energy lost due to friction and air resistance.
Derive a formula for the power P in terms of variables: the mass M, the final speed V, and the distance d.
P=P(M,V,d) ?

v=v(t) The speed is a function of time.

Pt = ½m v(t)² Because of the energy conservation law.

v(t)²=2tP/m=(2P/m)t

v(t)=(2P/m)^½t^½

x(t)=(2P/m)^½ (⅔t^3/2)

x(t)²=(2P/m) (⅔)² t³

t=v(t)² / (2P/m)

t³=v(t)⁶ / (2P/m)³

x(t)²=(2P/m) (⅔)² t³=(2P/m) (⅔)² v(t)⁶ / (2P/m)³

x(t)²=(⅔)² v(t)⁶ / (2P/m) ²

x(t)=(⅔) v(t)³ / (2P/m)

2P/m = (⅔) v(t)³ / x(t)

P = ½(⅔) v(t)³m / x(t)

P = ⅓ v(t)³m / x(t) = ⅓ V³m / d

Physics Problem - Find Average Power

What is the #average #power #output of an #engine when a #car of #mass M #accelerates #uniformly (a=const) from #rest to a final #speed V... pic.twitter.com/iqz40MQpOh

— Physics I (@PHY210) October 1, 2016

Physics Problem - Work, Kinetic Energy

Problem from Quiz 5

A train with the mass M accelerates uniformly from the rest to the speed V when passing through a distance d.
What is its kinetic energy
when it passing through a distance d₁ (d₁ < d)?

Given Data: M = 15,000 kg

Vₒ = 0

V = 15 m / s

d = 150 m

d₁ = 75 m

K₁ = ?

Solution:

K(kinetic energy) = Kₒ(kinetic energy initial) + W(work) = W

W = Fx

K = W = Fd = ½MV²

F = ½MV² / d

K₁ = W₁ = Fd₁ = ½MV² d₁ / d

K₁ = ½(15,000 kg)(15 m/s)² (75m) / (150m) = 843750 J

Google Calculator Code:

0.5*(15000 kg)*(15 m/s)^2*(75m) / (150m)

Answer: 840 kJ

System of Two Pulleys and Two Masses

#Physics #PhysicsProblem #PhysicalProblem #GeneralPhysics #PhysicsI #Physics1 #CalculusBasedPhysics #CalculusBasedPhysicsI #CalculusBasedPhysics1

A photo posted by Vasiliy S. Znamenskiy (@znamenski) on Sep 28, 2016 at 12:21pm PDT

Physics Problem

pic.twitter.com/3tokEWsytw

— Physics I (@PHY210) September 12, 2016

2 Describing Motion: Kinematics in One Dimension

For students PHYS 1441 (NYC College of Technology) E474 [54392]: pic.twitter.com/Ta5cpIU7vK

— Physics I (@PHY210) September 1, 2016

Introduction

Online HomeWork Quizzes

Online HomeWork Quizzes:

PHYS 1441 (NYC College of Technology) E474 [54392] - www.eztestonline.com/695230/index15.tpx

PHY 215 (Borough of Manhattan Community College) 122L [29642] - www.eztestonline.com/695230/index12.tpx


Do Self-Registration!

PDF Versions of Textbook

In general, all exams in my class are open-book exams. It means students may use paper textbooks during exams. But I don't give permission to use in exams of any electronic devices except simple calculators. Students may use PDF versions of the textbook during lectures, laboratory lessons, or at home.

Linear Uniform Motion

Physics I

Problem

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

‪#‎физика‬ ‪#‎Physics‬ ‪#‎exam‬ ‪#‎экзамен‬ ‪#‎подготовка‬ ‪#‎preporation‬ ‪#‎FinalExam‬‪#‎examenation‬ ‪#‎ИтоговыйЭкзамен‬ ‪#‎задачи‬ ‪#‎problems‬ ‪#‎PhysicsProblem‬‪#‎PhysicalProblems‬ ‪#‎ЗадачиПоФизике‬
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

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The figure shows the position of a car (black circles) at one-second intervals. What is the acceleration at the time T = 4 s?

The figure shows the position of a car (black circles) at one-second intervals. What is the acceleration at the time T = 4 s?

Solution:
t=T-1s
x=v₀t+at²/2
40m=v₀∙4s+a∙(4s)²/2
13m=v₀∙1s+a∙(1s)²/2=(v₀+a∙1s/2)∙1s
v₀=13m/s-a∙1s/2
40m=(13m/s-a∙1s/2)∙4s+a∙(4s)²/2 = 52m-2s²∙a+8s²∙a = 52m+6s²∙a
6s²∙a=-12m
a=-2m/s²

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

The potential energy of a mass m=10kg is given by U=-3.98 × 10¹⁵Jm⋅ r⁻¹

The potential energy of a mass m=10kg is given by U=-3.98 × 10¹⁵Jm⋅ r⁻¹ (potential energy decreases as r approaches zero). The force on this mass when it is located at r=6.38 × 10⁶ m is _________

Solution:
F = -dU/dr=
= -d( -3.98 × 10¹⁵⋅ r⁻¹)/dr=
=  3.98 × 10⁻¹⁵⋅ d(r⁻¹)/dr =
= -3.98 × 10⁻¹⁵⋅ r⁻² =
= -3.98 × 10¹⁵/ (6.38 × 10⁶)²

Calculation on the Google Calculator: -3.98E15/ (6.38E6)^2=-97.7781271804

|F|=97.8N, the direction is toward the origin (r=0).

⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞

A rocket, speeding along toward Alpha Centauri, has an acceleration a(t) = At².

A rocket, speeding along toward Alpha Centauri, has an acceleration
a(t) = At².
Assume that the rocket began at rest at the Earth (x = 0) at t = 0. Assuming it simply travels in a straight line from Earth to Alpha Centauri (and beyond), what is the ratio of the speed of the rocket when it has covered half the distance to the star to its speed when it has travelled half the time necessary to reach Alpha Centauri?
Solution:
T - total time necessary to reach Alpha Centauri
D - distance to the star
v - speed
v₁=v(D/2)
v₂=v(T/2)
v(t)=∫At²dt=⅓At³
x(t)=∫v(t)dt=⅓At³dt=⅓¼At⁴
D=AT⁴/12
D/2=AT⁴/24
D/2=At⁴/12;   At⁴/12=AT⁴/24;   t⁴=T⁴/2;  t⁴/T⁴=½;   t/T=∜½
v₁/v₂ = v(t)/v(T/2) = {⅓At³} / {⅓A(T/2)³}
=t³ / (T/2)³ =2³ (t/T)³ = 2³ ⨯ (∜½)³ = (2∜½)³ = (∜16⨯∜½)³ = (∜8)³=4∜2
This result is not among the proposed answers to choose from. Check again algebraic transformations.

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

What is the location of the center of mass of the rod?

A long, thin rod lies along the x-axis.
One end of the rod is located at x = 2.00 m and the other end of the rod is located at x = 6.00 m.
The linear mass density of the rod is given by ax³+b ,
where a = 0.100 kg/m⁴ and b = 0.200 kg/m.
What is the location of the center of mass of the rod?
xₘ={∫x(ax³+b)dx}/{∫(ax³+b)dx}={∫(ax⁴+bx)dx}/{∫(ax³+b)dx}=
=(⅕a(x₂⁵-x₁⁵)+½b(x₂²-x₁²)) / (¼a(x₂⁴-x₁⁴)+b(x₂-x₁))=
={(⅕0.1(6⁵-2⁵)+½0.2(6²-2²))}/{(¼0.1(6⁴-2⁴)+0.2(6-2))}
Calculation:
(0.1*(6^5-2^5)/5+0.2*(6^2-2^2)/2)/(0.1*(6^4-2^4)/4+0.2(6-2))
=4.81951219512
xₘ≈4.82

⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ∫≈Δ¼½⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞

What is the distance of the resulting plate's center of mass from the origin?

A thin rectangular plate of uniform density σ₁ = 5.94 kg/m² has a length a = 0.740 m and a width of b = 0.330 m. The lower left hand corner is placed at the origin, (x, y) = (0, 0). A circular hole is cut in the plate of radius r = 0.064 m with center at (x, y) = (0.084 m, 0.084 m) and replaced with a circular disk with the same radius composed of another material of the same thickness with uniform density σ₂ = 1.67 kg/m². What is the distance of the resulting plate's center of mass from the origin? (in m)

xₘ=(σ₁ab a/2+πr²(σ₂-σ₁) x) / (σ₁ab a/2+πr²(σ₂-σ₁) x)
yₘ=(σ₁ab a/2+πr²(σ₂-σ₁) y) / (σ₁ab a/2+πr²(σ₂-σ₁) y)
d²=xₘ²+yₘ²
xₘ=(5.94*0.74*0.33*0.74/2 + (pi)*0.064^2*(1.67-5.94)*0.084)/(5.94*0.74*0.33+ (pi)*0.064^2*(1.67-5.94))=0.381260099
yₘ=(5.94*0.74*0.33*0.33/2 + (pi)*0.064^2*(1.67-5.94)*0.084)/(5.94*0.74*0.33+ (pi)*0.064^2*(1.67-5.94))=0.381260099=0.16818904902
d=sqrt(0.381260099^2+0.16818904902^2)=0.41670951429≈0.417 (m)

⁰¹²³⁴⁵⁶⁷⁸⁹ⁱ⁺⁻⁼⁽⁾ⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ
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Center of Mass

Find the y-coordinate of the center of mass of a 10-cm square plate whose density (in kg/m³) is ρ = 4.22⋅10⁴kg/m⁶⋅xy². The lower left-hand side of the plate is at the origin and the upper right-hand corner is at (10 cm, 10 cm). The plate is 0.30 cm thick.

(∫y⋅y²dy) / (∫y²dy) = ¾y = ¾⋅10cm = 7.5cm