Monday, October 17, 2016
Solve m₁g-t=m₁a₁, m₂g-½t=m₂a₂, m₃g-½t=m₃a₃, -2a₁=a₂+a₃, for t,a₁,a₂,a₃ where t is a tension of the cord, which is passing through the pulley A
Solve
m₁g-t=m₁a₁,
m₂g-½t=m₂a₂,
m₃g-½t=m₃a₃,
-2a₁=a₂+a₃,
for t,a₁,a₂,a₃
where t is a tension of the cord, which is passing through the pulley A
Friday, October 14, 2016
Saturday, October 1, 2016
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) ?
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)½t½
x(t)=(2P/m)½
(⅔t3/2)
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
Friday, September 30, 2016
Wednesday, September 21, 2016
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 traveled half the time necessary to reach Alpha Centauri?
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 traveled half the time necessary to reach Alpha
Centauri?
a(t)
= At²
v(0)
= 0
x(0)
= 0
D
= x(T)
D
/ 2 = x(τ)
v(τ)
/ v(½T) = ?
Solution
a(t)
= At²
v(t)
= ⅓At³
x(t)
= At⁴ / 12
D
= AT⁴ / 12
T⁴
= 12D / A
t⁴
= 12x / A
τ⁴
= 12(D / 2) / A
τ⁴
= 12D / A · (½)
= (½)T⁴
τ
= ⁴√(½)
· T
x(τ)
= Aτ⁴ / 12 =
A(⁴√(½)
· T)⁴
/ 12 = ½ AT⁴
/ 12 = ½D
v(½T)
= ⅓A(½T)³
= (½)³
· ⅓AT³
v(τ)
= v(⁴√(½)
· T)
= ⅓A(⁴√(½)
· T)³
= (⁴√(½))³
· ⅓AT³
v(τ)
/ v(½T)
= {(⁴√(½))³
· ⅓AT³}
/ {(½)³
· ⅓AT³}
= {(⁴√(½))³
} / {(½)³}
= (2 / ⁴√2)³
= 8 / ⁴√8
= (⁴√8)³
Tuesday, July 26, 2016
Vector Sum
#вектор #ВекторныеСвойства #сумма #ВекторнаяСумма #ВекторноеСложение #сложение #суммированиеhttps://t.co/1KHqrIq3lw pic.twitter.com/rWNKKr5vej
— TutorState.com (@TutorState) July 26, 2016
Задача на развитие интуитивного, образного понимания векторов и их свойств и, одновременно, на развитие глазомера. Сложите шесть изоброженных радиус-векторов. Найдите номер пятна, который достигает голова вектора-суммы. Начало вектора суммы, как и исходных векторов, расположено на пятне с номером 0?
Wednesday, July 13, 2016
Find functions with the following properties
Find functions with the following properties
Найдите функции, обладающие следующими свойствами.
For example f₁₁(x) = kx where k is a constant.
Например, f₁₁(x) = kx, где k - любая константа.
#function #properties #функции #свойства #constant #константа
Найдите функции, обладающие следующими свойствами.
For example f₁₁(x) = kx where k is a constant.
Например, f₁₁(x) = kx, где k - любая константа.
#function #properties #функции #свойства #constant #константа
Tuesday, June 28, 2016
Wednesday, June 8, 2016
Friday, May 13, 2016
Monday, April 25, 2016
Saturday, February 20, 2016
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