https://t.co/MGxOiE9QXl #VasiliyZnamenskiy #Vasiliy #Znamenskiy #VZnamenskiy #Василий #Знаменский #ВасилийЗнаменский #ВасилийСерафимович pic.twitter.com/aDCL94Y9QC

— General Physics (@GeneralPhysics) December 7, 2016

# ☄ General Physics

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:

Useful Physics Formulas

Δ

Solution

Δ

Δ

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:

**= 1kg***m**y*₁ = 2m**U**₁ = 3J**y**₂ = 4m*U*₂ = ?Useful Physics Formulas

*=***U***mgh*Δ

*= Δ(***U****) =***mgh**Δ***mg***h*Solution

Δ

*= Δ(***U****) =***mgh**Δ***mg****=***h**(***mg****y**₂ -*y*₁)Δ

**=***U**₂ -***U****₁***U**₂ =***U****₁ +***U**(***mg****y**₂ -*y*₁)**₂ = 3J + 1kg · 10ᵐ/s² · (4m - 2m) = 3J + 20J = 23J***U*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

Subscribe to:
Posts (Atom)