Tuesday

November 24, 2015
Posted by **akha** on Sunday, October 13, 2013 at 8:37pm.

Express the answer of the following questions in terms of some or all of the variables c_1, r , m , g, v_x , v_y , and u (enter C_1 for , v_x for and v_y for ). Enter e^(-z) for exp(-z) (the exponential function of argument -z).

(a) What is component of the acceleration in the direction as a function of the component of the velocity in the direction ? express your answer in terms of v_x , c_1 , r , g , m and u as needed: What is the acceleration in the direction as a function of the component of the velocity in the direction ? express your answer in terms of v_y , c_1 , r , g, m and as needed: Using your result from part (a), find an expression for the horizontal component of the ball's velocity as a function of time? Using your result from part (b), find an expression for the vertical component of the ball's velocity as a function of time t. How long does it take for the vertical speed to reach 99 of its maximum value.What value does the horizontal component of the ball's velocity approach as t becomes infinitely large?What value does the vertical component of the ball's velocity approach as t becomes infinitely large?

- Physics pls anyone can help me? its urgent -
**Anonymous**, Monday, October 14, 2013 at 3:26pma)(-C_1*r*v_x)/m

b)g-(C_1*r*v_y)/m

c)u*e^(-(C_1*r*t)/m)

d)((m*g)/(C_1*r))-((m*g)/(C_1*r))*e^(-(C_1*r*t)/m)

e)4.6*(m/(C_1*r))

f)0

g)(m*g)/(C_1*r)

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**Anonymous**, Tuesday, October 15, 2013 at 7:38amU r awesome...

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**Anonymous**, Tuesday, October 15, 2013 at 3:19pmYou are welcome, did you get number 6?

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**ss01**, Tuesday, October 15, 2013 at 10:02pmq6 (5*R)/2

did anyone got vertical spring?

- Physics pls anyone can help me? its urgent -
**Anonymous**, Tuesday, October 15, 2013 at 10:31pmDarn, I said 6, I meant 5 the half loop. Vertical spring is a) .4785

b) .72

c) 1.1785

d) 1.812

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**ss01**, Tuesday, October 15, 2013 at 11:01pmfor me q6 is full loop and q5 is half loop

i did not got that one i am on half way

your answers are showing wrong for me.

do you have the steps?

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**Anonymous**, Wednesday, October 16, 2013 at 12:01amYou mean the one that starts d1=.4785? I took the general equation, Acos(wt+p), used the initial conditions to solve for the constants. Then I used that to get the answers. At least that is what I remember. I don't recall all the details I'll see if I can find the details.

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**ss01**, Wednesday, October 16, 2013 at 12:07amyup

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**Anonymous**, Wednesday, October 16, 2013 at 12:35amI'm looking for my notes. The first one comes from the equation ma=-kx. then

(ma/k)=-x = (4*9.81)/82=x=.4785

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**Anonymous**, Wednesday, October 16, 2013 at 12:38amThe second one comes from

f=(1/2pi)sqrt(k/m)

=(1/2pi)*sqrt(82/4)=.72

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**Anonymous**, Wednesday, October 16, 2013 at 12:41amfor the other two I think I got

x(t)=.4cos(4.53t+pi) and then set v(t)=0 to get the time and then used that to get the velocity. And I used v(tm)=0 to get the time when it reaches the highest point and plugged that value back into x(t).

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**ss01**, Wednesday, October 16, 2013 at 1:25amthnx..

and what about spring block question?

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**Anonymous**, Wednesday, October 16, 2013 at 2:02amI got 3, don't ask me how I got it. I'm not sure.

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**ss01**, Wednesday, October 16, 2013 at 2:32amjust tell the answer

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**Anonymous**, Wednesday, October 16, 2013 at 3:12amThe answer is "3"

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**Anonymous**, Wednesday, October 16, 2013 at 3:12amHey are you in quantum also?

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**ss01**, Wednesday, October 16, 2013 at 5:51amyes