# COFFEE

Newest questions and responses by COFFEE-
## PHYSICS

The pressure and volume of an ideal monatomic gas change from A to B to C. From A to B, volume remains 0.400 and pressure rises to 4.00x10^5. From B to C, volume changes from 0.400 to 0.200 while pressure remains the same. There is a curved line between A

*asked on December 1, 2015* -
## Surface Area

[Given] Fiber Linear Density = 1 denier = 1 g/9000m Fiber Density = 1.14 g/cm^3 [Find..] Fiber surface area in cm^2/g Assume that the fiber strand is a uniform cylinder [Answer] Surface area = 3,150 cm^2/g ....... how do i get to the answer? my professor

*asked on September 14, 2007* -
## calculus - power series ASAP please :)

using power series, integrate & evaluate to 4 dec. places integral from 0 to 1: sin x^2 dx i'm REALLY stuck on this. and i need help asap.. what is the inverse of "sin x^2" so that i could have it in a fraction that will fit the power series equation? and

*asked on July 31, 2007* -
## calculus - interval of convergence

infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n) .. my work so far. i used the ratio test = lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] | .. now my question is: was it ok for me to add "+1" to "n+2" to become

*asked on July 30, 2007* -
## calculus - ratio test

infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] | = lim (n->infinity) | ((e^n)(e^1)(n!)) / ((n+1)(n!)(e^n)) |

*asked on July 30, 2007* -
## calculus - derivatives

can you please find the first 5 derivatives for: f(x) = (0.5e^x)-(0.5e^-x) f'(x) = ? f''(x) = ? f'''(x) = ? f''''(x) = ? f'''''(x) = ? thanks :) f(x) = (0.5e^x)-(0.5e^-x) f'(x) = 0.5 e^x + 0.5 e^-x f''(x) = 0.5 e^x - 0.5 e^-x f'''(x) = 0.5 e^x + 0.5 e^-x

*asked on July 30, 2007* -
## calculus - interval of convergence

infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n) .. my work so far. i used the ratio test = lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] | .. now my question is: was it ok for me to add "+1" to "n+2" to become

*asked on July 29, 2007* -
## calculus - ratio test

Posted by COFFEE on Sunday, July 29, 2007 at 6:32pm. infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] | = lim

*asked on July 29, 2007* -
## Calculus - ratio test

infinity of the summation n=1: (e^n)/(n!) [using the ratio test] my work so far: = lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] | = lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] | = lim (n->infinity) | ((e^n)(e^1)(n!)) / ((n+1)(n!)(e^n)) |

*asked on July 29, 2007* -
## Calculus - Taylor #2

Find the Taylor series for f(x) centered at the given value of 'a'. (Assume that 'f' has a power series expansion. Do not show that Rn(x)-->0.) f(x) = x3, a = -1 and what i've done so far: f (x) = x^3 f ' (x) = 3x^2 f '' (x) = 6x^1 f ''' (x) = 6x f (-1) =

*asked on July 28, 2007* -
## Calculus - Taylor

could you please help me with solving this problem? #1) Find the Taylor polynomial Tn(x) for the function 'f' at the number 'a'. f(x) = sqrt(3+x^2) ; a=1; n=2; my work so far: f (x) = sqrt(3+x^2) = (3+x^2)^(1/2) f ' (x) = (1/2)(3+x^2)^(-1/2) f '' (x) =

*asked on July 28, 2007* -
## Calculus - series

I'm getting this answer wrong, can someone please help show me what i'm missing?? thank you :) Infinity of the summation n=0: [(-1)^n pi^(2n)] / [6^(2n) (2n)!] this is my work: [(-1^0) pi^(2*0)] / [6^(2*0) (2*0)!] + [(-1^1) pi^(2*1)] / [6^(2*1) (2*1)!] +

*asked on July 28, 2007* -
## Math

i'm a bit stuck with this.. 145/18 = x + x^2 what does x equal to? multiply each term by 18, then re-arrange to get 18x^2 + 18x - 145 = 0 This quadratic does not factor, so use the quadratic formula to get your two answers.

*asked on July 27, 2007* -
## Math/Physics

Please check my work below and comment. A tank initially contains 80 gallons of fresh water. A 10% acid solution flows into the tank at the rate of 3 gallons per minute. The well-stirred mixture flows out of the tank at the rate of 3 gallons per minute.

*asked on July 13, 2007* -
## Math/Physics

I am given a damping constant of 20 dyne*sec/meter...do I need to convert this if the rest of my givens are, for mass = 2kg, k (spring constant) = 82 N/m. I am trying to find the equation of motion of a spring but cannot solve it until I know how to

*asked on July 13, 2007* -
## Math/Calculus

Please check my work and correct any errors/point out any errors. Thanks. Solve the initial-value problem using the method of undetermined coefficients. y''-4y=e^xcos(x), y(0)=1, y'(0)=2 r^2-4=0, r1=2, r2=-2 yc(x)=c1*e^2x+c2*e^-2x

*asked on July 12, 2007* -
## Math/Calculus

A series circuit contains a resistor with R = 24 ohms, an inductor with L = 2 H, a capacitor with C = 0.005 F, and a generator producing a voltage of E(t) = 12 sin(10t). The initial charge is Q = 0.001 C and the initial current is 0. (a) Find the charge at

*asked on July 12, 2007* -
## Math/Calculus

Solve the initial-value problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else? Thanks. y''+4y'+6y=0, y(0)=2, y'(0)=4 r^2+4r+6=0, r=(-4 +/- sqrt(16-4(1)(6))/2 r=-2 +/- sqrt(2)*i , alpha = -2, beta = 2(sqrt(2))

*asked on July 12, 2007* -
## Math/Calculus

Please take a look at my work below and provide a good critique: Solve the differential equation using the method of undetermined coefficients or variation of parameters. y'' - 3y' + 2y = sin(x) yc(x)= c1*e^2x+c2*e^x y"-3y'+2y=sin(x) r^2-3r+2=0

*asked on July 11, 2007* -
## Math/Calculus

A spring with a 4 kg mass has natural length 1 m and is maintained stretched to a length of 1.3 m by a force of 24.3 N. If the spring is compressed to a length of 0.8 m and then released with zero velocity, find the position of the mass at any time t. Here

*asked on July 11, 2007* -
## Math

A series circuit contains a resistor with R = 24 , an inductor with L = 2 H, a capacitor with C = 0.005 F, and a generator producing a voltage of E(t) = 12 sin(10t). The initial charge is Q = 0.001 C and the initial current is 0. Find the charge at time t.

*asked on July 10, 2007* -
## Calculus

Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2

*asked on July 10, 2007* -
## Calculus - Second Order Differential Equations

Solve the initial-value problem. y'' - 2y' + y = 0 , y(2) = 0 , y'(2) = 1 r^2-2r+1=0, r1=r2=1 y(x)=c1*e^x+c2*x*e^x y(2)=c1*e^2+c2*2*e^2=0 c1=-(2*c2*exp(2))/exp(2) c1=-2*c2 y'(x)=-2*c2*e^x+c2*e^x*(x-1) y'(2)=-2*c2*e^2+c2*e^2*(2-1)=1 c2(-2e^2+e^2)=1

*asked on July 10, 2007* -
## Calculus - Second Order Differential Equations

Posted by COFFEE on Monday, July 9, 2007 at 9:10pm. download mp3 free instrumental remix Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i,

*asked on July 10, 2007* -
## Calculus - Second Order Differential Equations

Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 y'(0)=4, c2=4

*asked on July 9, 2007* -
## Calculus - Second Order Differential Equations

Solve the boundary-value problem. y''+5y'-6y=0, y(0)=0, y(2)=1 r^2+5r-6=0, r1=1, r2=-6 y=c1*e^x + c2*e^-6x y(x)=c1*e^x+c2*e^-6x y'(x)=c1*e^x-6*c2*e^-6x y(0)=c1+c2=0, c1=-c2 y(2)=c1*e^2+c2*e^(-12)=1 -c2*e^2-6c2*e^(-12)=1 -c2(e^2-6*e^-12)=1

*asked on July 9, 2007* -
## calc check please?

Given the differential equation: dy/dx = y(1+x), y(0)=1, Use Euler's method with step size .1 to approximate y(.3). ... please check this for me! no one has responded to this question yet.. thanks. y' = y(1+x), y'(0) = 1(1+0)=1 ->the solution has slope 1

*asked on July 2, 2007* -
## calc: avg value

Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2]. and this is what i did.. please check for mistakes. thanks :D f(x) = x^2 sqrt(1+x^3), [0,2] f ave = (1/(b-a))*inegral of a to b for: f(x) dx f ave = (1/(2-0))*integral

*asked on July 2, 2007* -
## calc check: curve length

Find the length of the curve y=(1/(x^2)) from ( 1, 1 ) to ( 2, 1/4 ) [set up the problem only, don't integrate/evaluate] this is what i did.. let me know asap if i did it right.. y = (1/(x^2)) dy/dx = (-2/(x^3)) L = integral from a to b for:

*asked on July 2, 2007* -
## calc check: euler's method

Given the differential equation: dy/dx = y(1+x), y(0)=1, Use Euler's method with step size .1 to approximate y(.3). y' = y(1+x), y'(0) = 1(1+0)=1 ->the solution has slope 1 at the point (0,1); x0=0, y0=1, h=0.1, F(x,y)=y(1+x) y1=y0+h*F(x0,y0) y1=1 +

*asked on July 1, 2007* -
## calc check: average value

Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2]. and this is what i did.. please check for mistakes. thanks :D f(x) = x^2 sqrt(1+x^3), [0,2] f ave = (1/(b-a))*inegral of a to b for: f(x) dx f ave = (1/(2-0))*integral

*asked on June 30, 2007* -
## calc check: hooke's law

A force of 27N is required to maintain a spring stretched from its natural length of 12cm to a length of 15cm. How much work is done in stretching the spring from 15 to 25cm? and this is what i did.. please check to see if i did it correctly.. thanks :)

*asked on June 29, 2007* -
## calculus

Given the differential equation: dy/dx = y(1+x), y(0)=1, Use Euler's method with step size .1 to approximate y(.3). y' = y(1+x), y'(0) = 1(1+0)=1 ->the solution has slope 1 at the point (0,1); x0=0, y0=1, h=0.1, F(x,y)=y(1+x) y1=y0+h*F(x0,y0) y1=1 +

*asked on June 29, 2007* -
## Calculus - Hydrostatic Pressure

Please check my work: Find the hydrostatic pressure on one end of a water trough full of water, the end of which is a trapezoid with given dimensions: top of trapezoid = 20 feet, sides of trapezoid both = 8 feet, bottom of trapezoid = 12 feet. Depth of

*asked on June 26, 2007* -
## Calculus - Seperable Equations

Solve the separable differential equation (dy/dx)=y(1+x) for y and find the exact value for y(.3). dy/dx = y(1+x) dy/y = (1+x)dx Integral (dy/y) = Integral (1+x)dx ln (y) = x + (1/2)x^2 + C y = e^(x + (1/2)x^2 + C) y(0.3) = e^(0.345 + C) I am stuck here.

*asked on June 26, 2007* -
## Calculus

Given the differential equation: dy/dx = y(1+x), y(0)=1, Use Euler's method with step size .1 to approximate y(.3). y' = y(1+x), y'(0) = 1(1+0)=1 ->the solution has slope 1 at the point (0,1); x0=0, y0=1, h=0.1, F(x,y)=y(1+x) y1=y0+h*F(x0,y0) y1=1 +

*asked on June 26, 2007* -
## Calculus - Orthogonal Trajectories

Find the orthogonal trajectories of the family of curves: y = k*(e^-x) --------------- so k = y/(e^-x) differentiating we get: 1 = -k(e^-x)*(dx/dy) 1/(dx/dy) = -k(e^-x) dy/dx = -k(e^-x)...substituting for k: dy/dx = -(y/(e^-x))*(e^-x) dy/dx = -y

*asked on June 26, 2007* -
## Calculus - Center of Mass

Find the exact coordinates of the centroid given the curves: y = 1/x, y = 0, x = 1, x = 2. X = 1/Area*Integral from a to b: x*f(x)dx Y = 1/Area*Integral from a to b: [(1/2)*(f(x))^2]dx How do I find the area for this? Once I know that, is this the correct

*asked on June 26, 2007* -
## Calculus

Solve the differential equation. Let C represent an arbitrary constant. (Note: In this case, your answer willto have a negative sign in front of the arbitrary C.) (dz)/(dt) + e^(t+z) = 0 --------------- (dz/dt) + (e^t)(e^z) = 0 (dz/dt) = -(e^t)(e^z) dz =

*asked on June 24, 2007* -
## Calculus

The tank shown is full of water. Given that water weighs 62.5 lb/ft3, find the work required to pump the water out of the tank. The tank shown is a hemisphere with r = 5 ft. The water is to be pumped out at the top. First I solved for ri, ri / (5 - xi) = 5

*asked on June 23, 2007* -
## Calculus

The tank shown is full of water. Given that water weighs 62.5 lb/ft3, find the work required to pump the water out of the tank. The tank shown is a hemisphere with r = 5 ft. The water is to be pumped out at the top. First I solved for ri, ri / (5 - xi) = 5

*asked on June 22, 2007* -
## Calculus

Use Euler's method with step size 0.2 to estimate y(1.4), where y(x) is the solution of the initial-value problem below. Give your answer correct to 4 decimal places. y' = x - xy y(1) = 0 h = 0.2 Since I am at y(1) = 0 and not y(0) = 0 would I just do this

*asked on June 22, 2007* -
## Calculus

Find the exact coordinates of the centroid. y = sqrt[x], y = 0, x = 9. -------------- Is this basically 1/4 of an oval/ellipse? If so then the area would be: pi*9*3, correct? So the X coordinate would equal: 1/Area * Integral from 0 to 9 of (x*f(x))*dx

*asked on June 22, 2007* -
## Calc: euler's method

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem below. Give your answer correct to 4 decimal places. y' = 1 - xy y(0) = 0 y(1) = ____ ? ... help, this is what i've done but got the wrong

*asked on June 20, 2007* -
## Calculus

Find the exact coordinates of the centroid. y = sqrt[x], y = 0, x = 9. -------------- Is this basically 1/4 of an oval/ellipse? If so then the area would be: pi*9*3, correct? So the X coordinate would equal: 1/Area * Integral from 0 to 9 of (x*f(x))*dx

*asked on June 18, 2007* -
## Calculus

The hemispherical tank shown is full of water. Given that water weighs 62.5 lb/ft3, find the work required to pump the water out of the tank. ---------- What is shown is just the tank (a hemisphere) with a radius of 5 ft. ---------- First I calculated the

*asked on June 18, 2007* -
## Calculus

Find the exact coordinates of the centroid. y = sqrt[x], y = 0, x = 9. -------------- Is this basically 1/4 of an oval/ellipse? If so then the area would be: pi*9*3, correct? So the X coordinate would equal: 1/Area * Integral from 0 to 9 of (x*f(x))*dx

*asked on June 17, 2007* -
## Calculus

The hemispherical tank shown is full of water. Given that water weighs 62.5 lb/ft3, find the work required to pump the water out of the tank. ---------- What is shown is just the tank (a hemisphere) with a radius of 5 ft. ---------- First I calculated the

*asked on June 17, 2007* -
## Calculus

A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by the following. y = 150 - (1/40)(x-50)^2 Find the distance traveled by the kite. y = 150 - (1/40)(x-50)^2 y = 150 -

*asked on June 15, 2007* -
## Calculus

A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by the following. y = 150 - (1/40)(x-50)^2 Find the distance traveled by the kite. y = 150 - (1/40)(x-50)^2 y = 150 -

*asked on June 13, 2007* -
## Calculus

A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by the following. y = 150 - (1/40)(x-50)^2 Find the distance traveled by the kite. y = 150 - (1/40)(x-50)^2 y = 150 -

*asked on June 13, 2007* -
## Calculus

Graph the curve and find its exact length. x = e^t + e^-t, y = 5 - 2t, from 0 to 3 Length = Integral from 0 to 3 of: Sqrt[(dx/dt)^2 + (dy/dt)^2] dx/dt = e^t - e^-t, correct? dy/dt = -t^2 - 5t, correct? So: Integral from 0 to 3 of Sqrt[(e^t - e^-t)^2 +

*asked on June 13, 2007* -
## calc: simpson's rule & arc length

i'm still getting this question wrong. please check for my errors: Use Simpson's Rule with n = 10 to estimate the arc length of the curve. y = tan x, 0

*asked on June 13, 2007* -
## calc: arc length

Posted by COFFEE on Monday, June 11, 2007 at 11:48pm. find the exact length of this curve: y = ( x^3/6 ) + ( 1/2x ) 1/2

*asked on June 12, 2007* -
## calc: arc length

find the exact length of this curve: y = ( x^3/6 ) + ( 1/2x ) 1/2

*asked on June 11, 2007* -
## Math/Euler's Method

Consider a cooling cup of coffee whose initial temperature is 205°. The room temperature is held at 70°. Suppose k = 1/16. Let y be the temperature, and y' its time derivative. ----------------------------------- I have the differential equation: y' =

*asked on June 7, 2007* -
## Math/Calculus

How would I integrate the following: (2x^2 + 5)/((x^2+1)(x^2+4))dx I think I would start with making it a sum of two partial fractions.

*asked on May 29, 2007* -
## Math/Calculus

How would I integrate the following by parts: Integral of: (x^2)(sin (ax))dx, where a is any constant. Just like you did x^2 exp(x) below. Also partial integration is not the easiest way to do this integral. You can also use this method. Evaluate first:

*asked on May 28, 2007* -
## Math/Calculus

How would I evaluate the following integral by using integration by parts? Integral of: (t^3)(e^x)? You mean (x^3)(e^x)? x^3 exp(x) dx = x^3 d[exp(x)] = d[x^3 exp(x)] - exp(x) d[x^3] = d[x^3 exp(x)] - 3 x^2 exp(x) dx So, if you integrate this you get x^3

*asked on May 28, 2007* -
## Math/Calculus

How would I solve the following integral with the substitution rule? Integral of: [(x^3)*(1-x^4)^5]dx Put 1-x^4 = y Then -4x^3 dx = dy Integral is then becomes: Integral of -1/4 y^5 dy ok, thanks a lot! I got it now.

*asked on May 28, 2007* -
## Math/Calculus #2

Integrate: 1/(x-sqrt(x+2) dx I came up with: (2/3)(2*ln((sqrt(x+2))-2)+ln((sqrt(x+2))-1)) but it keeps coming back the wrong answer even though I integrated correctly. Is there a way to simplify this answer, and if so, how? I found: Ln[x-sqrt(x+2)] +

*asked on May 27, 2007* -
## Math/Calculus

Integrate: (2x^2+5)/((x^2+1)(x^2+4)) I came up with: (tan^-1)(x)-(1/2)((tan^-1)(2/x)) but it keeps coming back the wrong answer even though I integrated correctly. Is there a way to simplify this answer, and if so, how? Your answer is correct, but I think

*asked on May 27, 2007* -
## physics - doppler effect

Two identical tuning forks can oscillate at 329.6 Hz. A person is located somewhere on the line between them. The speed of sound in the air is 343 m/s. Calculate the beat frequency as measured by this individual under the following conditions. (a) the

*asked on April 21, 2007* -
## physics - sound level

For two sounds whose sound levels differ by 69 dB, find the ratios (greater value / smaller value) of the following values. (a) the intensities Intensity Final/Intensity Initial = log^-1 (69 / 10) = 7.9x10^6 (b) the pressure amplitudes (c) the particle

*asked on April 21, 2007* -
## physics - sound level

The source of a sound wave has a power of 2.50 µW. Assume it is a point source. (a) What is the intensity 6.70 m away? I used I = Power / 4*pi*r^2 and found I to be 4.43x10^-9 W/m^2 (b) What is the sound level at that distance? Sound level = 10 dB*log

*asked on April 21, 2007* -
## physics - waves

Two sinusoidal waves, identical except for phase, travel in the same direction along a string producing a net wave y'(x, t) = (1.5 mm) sin(29x - 4.0t + 0.960 rad), with x in meters and t in seconds. (a) What is the wavelength of the two waves? I found the

*asked on April 15, 2007* -
## physics - SHM

Calculate the speed of the pulse from the following: y(x,t) = 2/((x - 3t)^2 + 1) Well the speed of the pulse is given by: y(x,t) = f (x - vt) for a pulse traveling to the right and y(x,t) = f (x + vt) for a pulse traveling to the left but in this case the

*asked on April 14, 2007* -
## physics - SHM

Two sinusoidal waves with identical wavelengths and amplitudes travel in opposite directions along a string with a speed of 11 cm/s. If the time interval between instants when the string is flat is 0.33 s, what is the wavelength of the waves? wavelength =

*asked on April 14, 2007* -
## physics

The angle of the pendulum is given by θ = θmcos(ωt + φ), where ω = 3.24 rad/s. If at t = 0, θ = 1 rad and dθ/dt = -0.9 rad/s, what are φ and θm? So if I substitute in omega and t=0 I have θ = θmcos(φ). How do I solve for phi and omega center of

*asked on April 12, 2007* -
## Physics check

A performer, seated on a trapeze, is swinging back and forth with a period of 9.55 s. If she stands up, thus raising the center of mass of the trapeze + performer system by 20.0 cm, what will be the new period of the system? Treat trapeze + performer as a

*asked on April 12, 2007* -
## Physics - Pendulums

A uniform circular disk whose radius R is 32.0 cm is suspended as a physical pendulum from a point on its rim. (a) What is its period of oscillation? __ s (b) At what radial distance r < R is there a point of suspension that gives the same period? __ cm in

*asked on April 12, 2007* -
## Physics - SHM

An oscillating block-spring system has a mechanical energy of 1.00 J, an amplitude of 11.2 cm, and a maximum speed of 1.08 m/s. (a) Find the spring constant. ___ N/m (b) Find the mass of the block. ___ kg (c) Find the frequency of oscillation. ___ Hz .. im

*asked on April 12, 2007* -
## Physics - angular acceleration

An object rotates about a fixed axis, and th angular position of a reference line on the object is given by THETA(t)=0.4e^2t, where THETA is in radians, and t is in seconds. [a.] what is the object's angular acceleration at t = 2 s? ..this is my work so

*asked on April 8, 2007* -
## Physics - Conservation of Angular Momentum

a man is standing on the center of a platform that is rotating without friction. his arms are outstretched holding a brick in each hand. the rotational inertia of the system consists of the man, bricks, and platform about the central vertical axis of the

*asked on April 8, 2007* -
## Physics - Angular Momentum

When the angular momentum changes, the 'change' in the angular momentum vector (ie. dL) is ____. [a.] perpendicular to the torque vector. [b.] parallel to the angular momentum vector. [c.] parallel to the torque vector. .. im confused on this one.. i think

*asked on April 8, 2007* -
## Physics Phase Constant

Figure A [[which i tried to recreate below]] is a partial graph of the position function x(t) for a simple harmonic oscillator with an angular frequency of 1.1 rad/s. [a] ............x(cm).............. .................5-|-.................

*asked on April 4, 2007* -
## Physics Phase Constant

..im really stuck on this. can someone please explain? ------- Figure A [[which i tried to recreate below]] is a partial graph of the position function x(t) for a simple harmonic oscillator with an angular frequency of 1.1 rad/s. [a].....x(cm).......

*asked on April 4, 2007* -
## Physics SHM

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of 4 Hz. (a) What is the spring constant of each

*asked on April 4, 2007* -
## Physics - Torque

A particle is acted on by two torques about the origin: T1 has a magnitude of 4.3 N*m and is directed in the positive direction of the x axis, and T2 has a magnitude of 3.2 N*m and is directed in the negative direction of the y axis. What are the magnitude

*asked on March 31, 2007* -
## Physics - Torque

A particle is acted on by two torques about the origin: T1 has a magnitude of 4.3 N*m and is directed in the positive direction of the x axis, and T2 has a magnitude of 3.2 N*m and is directed in the negative direction of the y axis. What are the magnitude

*asked on March 31, 2007* -
## Re: Physics

Posted by COFFEE on Friday, March 30, 2007 at 4:25am. A 1.1 kg particle-like object moves in a plane with velocity components Vx = 30 m/s and Vy = 90 m/s as it passes through the point with (x, y) coordinates of (3.0, -4.0) m. (a) What is its angular

*asked on March 31, 2007* -
## Re: Physics (check)

Posted by COFFEE on Friday, March 30, 2007 at 4:16am. A sanding disk with rotational inertia 1.1 x 10^-3 kg*m^2 is attached to an electric drill whose motor delivers a torque of 6 Nm about the central axis of the disk. What are the following values about

*asked on March 31, 2007* -
## Physics

A 1.1 kg particle-like object moves in a plane with velocity components Vx = 30 m/s and Vy = 90 m/s as it passes through the point with (x, y) coordinates of (3.0, -4.0) m. (a) What is its angular momentum relative to the origin at this moment? _____

*asked on March 30, 2007* -
## Physics

A sanding disk with rotational inertia 1.1 x 10^-3 kg*m^2 is attached to an electric drill whose motor delivers a torque of 6 Nm about the central axis of the disk. What are the following values about the central axis at the instant the torque has been

*asked on March 30, 2007* -
## Physics - KE

In Figure 11-32 (which shows a ball at the top of an incline, at the bottom of the incline a loop begins with radius R and Q a point on the loop lined up with the center of the loop), a solid brass ball of mass m and radius r will roll without slipping

*asked on March 27, 2007* -
## Physics - KE

In Figure 11-32 (which shows a ball at the top of an incline, at the bottom of the incline a loop begins with radius R and Q a point on the loop lined up with the center of the loop), a solid brass ball of mass m and radius r will roll without slipping

*asked on March 27, 2007* -
## Physics - KE/inertia

The rigid body shown in Figure 10-66 (it is a triangle with the upper mass = to M and the bottom two corners/masses equal to 2M. The bottom side length equals .70 cm with point P in the middle of the side and the other two sides equal .55 cm)consists of

*asked on March 27, 2007* -
## Physics - KE/Inertia

The rigid body shown in Figure 10-66 (it is a triangle with the upper mass = to M and the bottom two corners/masses equal to 2M. The bottom side length equals .70 cm with point P in the middle of the side and the other two sides equal .55 cm)consists of

*asked on March 26, 2007* -
## Physics - KE/rotation

In Figure 11-32 (which shows a ball at the top of an incline, at the bottom of the incline a loop begins with radius R and Q a point on the loop lined up with the center of the loop), a solid brass ball of mass m and radius r will roll without slipping

*asked on March 25, 2007* -
## Physics - KE/Inertia

The rigid body shown in Figure 10-66 (it is a triangle with the upper mass = to M and the bottom two corners/masses equal to 2M. The bottom side length equals .70 cm with point P in the middle of the side and the other two sides equal .55 cm) consists of

*asked on March 25, 2007* -
## Re: PHYSICS

"Relative" is an important word. Block L of mass mL = 1.90 kg and block R of mass mR = 0.500 kg are held in place with a compressed spring between them. When the blocks are released, the spring sends them sliding across a frictionless floor. (The spring

*asked on February 27, 2007* -
## PHYSICS, still cant get it

A man (weighing 915 N) stands on a long railroad flatcar (weighing 2805 N) as it rolls at 18.0 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 40.00 m/s relative to

*asked on February 27, 2007* -
## Re: PHYSICS

A stationary block explodes into two pieces L and R that slide across a frictionless floor and then into regions with friction, where they stop. Piece L, with a mass of 2.0 kg, encounters a coefficient of kinetic friction µL = 0.40 and slides to a stop in

*asked on February 27, 2007* -
## Re: PHYSICS

"Relative" is an important word. Block L of mass mL = 1.90 kg and block R of mass mR = 0.500 kg are held in place with a compressed spring between them. When the blocks are released, the spring sends them sliding across a frictionless floor. (The spring

*asked on February 27, 2007* -
## Re: PHYSICS

A man (weighing 915 N) stands on a long railroad flatcar (weighing 2805 N) as it rolls at 18.0 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 40.00 m/s relative to

*asked on February 27, 2007* -
## RE: PHYSICS

A man (weighing 915 N) stands on a long railroad flatcar (weighing 2805 N) as it rolls at 18.0 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 40.00 m/s relative to

*asked on February 26, 2007* -
## RE: PHYSICS

A stationary block explodes into two pieces L and R that slide across a frictionless floor and then into regions with friction, where they stop. Piece L, with a mass of 2.0 kg, encounters a coefficient of kinetic friction µL = 0.40 and slides to a stop in

*asked on February 26, 2007* -
## RE: PHYSICS

"Relative" is an important word. Block L of mass mL = 1.90 kg and block R of mass mR = 0.500 kg are held in place with a compressed spring between them. When the blocks are released, the spring sends them sliding across a frictionless floor. (The spring

*asked on February 26, 2007* -
## Physics

*asked on February 25, 2007* -
## Physics

A stationary block explodes into two pieces L and R that slide across a frictionless floor and then into regions with friction, where they stop. Piece L, with a mass of 2.0 kg, encounters a coefficient of kinetic friction µL = 0.40 and slides to a stop in

*asked on February 25, 2007*

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