https://t.co/MGxOiE9QXl #VasiliyZnamenskiy #Vasiliy #Znamenskiy #VZnamenskiy #Василий #Знаменский #ВасилийЗнаменский #ВасилийСерафимович pic.twitter.com/aDCL94Y9QC
— General Physics (@GeneralPhysics) December 7, 2016
PHY 110. General Physics. 4CRS. 3 HRS. 2 LAB HRS. This course serves as an introduction to Physics, especially for students who are not science-oriented. A selected number of basic physical ideas are carefully examined and interpreted non-mathematically. The relevance of the scientist and his/her work to the lives of non-scientists is continually examined. tutorstate@gmail.com
Tuesday, December 20, 2016
Monday, October 24, 2016
Two identical springs with spring constant k are connected to identical masses
Two identical #springs with spring #constant k are #connected to #identical #masses of #mass M, as shown in the figures. #Physics #problem pic.twitter.com/8ff4nK5pcn
— Vasiliy S Znamenskiy (@znamenski) October 24, 2016
Wednesday, 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
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℀℁℅℆
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℀℁℅℆
Sunday, October 2, 2016
How much does the spring compress?
A 2.0 - kg mass is
dropped 2.0 m above a spring with a spring constant 40.0 N/m. How
much does the spring compress?
Use g = 10 m/s².
Solution
Use g = 10 m/s².
Solution
Given Data:
m = 2.0kg
k = 40.0N/m
h = 2.0m
x = ?
m = 2.0kg
k = 40.0N/m
h = 2.0m
x = ?
Useful formulas:
Uₛ = ½kx²
Uₘ = mgh
Uₛ = ½kx²
Uₘ = mgh
Solution:
mg(h + x) = ½kx²
½kx² - mgx – mgh = 0
mg(h + x) = ½kx²
½kx² - mgx – mgh = 0
x² - (2mg/k)x –
(2mg/k)h = 0
x² – 2 (mg/k)x +
(mg/k)² = (mg/k)² + (2mg/k)h
(x - mg/k)² =
(mg/k)² + (2mg/k)h
x - mg/k = ±√
{(mg/k)² + 2(mg/k)h}
x = mg/k±√
{(mg/k)² + 2(mg/k)h}
Calculation:
mg/k = 2kg*10m/s ²
/ 40N/m = ½m
(mg/k)² = ¼m²
2(mg/k)h = 2·½m·2m
= 2m²
(mg/k)² + 2(mg/k)h
= ¼m² + 2m² = (⁹/₄)m²
√ {(mg/k)² +
2(mg/k)h} = (³/₂)m
x₁ = ½m + (³/₂)m
= 2m
x₂ = - ½m - (³/₂)m = - 1m
x₂ = - ½m - (³/₂)m = - 1m
Problem's
answer: 2.0 m
Monday, September 26, 2016
Chapter 1. Homework - Closed for Records. NOTHING Recorded
Chapter 1. Homework - (https://www.eztestonline.com/695230/14098613484607000.tp4) Closed for Records. NOTHING Recorded
Friday, September 16, 2016
Wednesday, August 31, 2016
Wednesday, July 6, 2016
Wednesday, June 8, 2016
Thursday, May 26, 2016
Saturday, May 21, 2016
Thursday, May 12, 2016
Monday, April 18, 2016
Thursday, April 7, 2016
Borough of Manhattan CC
Borough of Manhattan CC.
Professor:
University Physics-I Tu 11:00PM - 1:40PM Main Bldg N565
Office hour by app. Tu 2:00 PM - 3:00PM Main Bldg N687
University Physics-I We 11:00AM - 1:45PM Main Bldg N794
Office hour by app. We 2:00 PM - 3:00PM Main Bldg N687
General Physics (L) Th 11:00AM - 1:45PM Fiterman 206
Office hour by app. Th 2:00 PM - 3:00PM Main Bldg N687
Professor:
University Physics-I Tu 11:00PM - 1:40PM Main Bldg N565
Office hour by app. Tu 2:00 PM - 3:00PM Main Bldg N687
University Physics-I We 11:00AM - 1:45PM Main Bldg N794
Office hour by app. We 2:00 PM - 3:00PM Main Bldg N687
General Physics (L) Th 11:00AM - 1:45PM Fiterman 206
Office hour by app. Th 2:00 PM - 3:00PM Main Bldg N687
Wednesday, March 2, 2016
Using Newton's Laws: Friction, Circular Motion, Drag Forces
Using Newton's Laws: Friction, Circular Motion, Drag Forces
Using Newton's Laws:Friction, Circular Motion, Drag Forces |
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