Answer:
what is at? is it at or as?
Answer:
Work done, W = 5534.53 J
Explanation:
It is given that,
Force acting on the piano, F = 6157 N
It is pushed up a distance of 2.41 m friction less plank.
Let W is the work done in sliding the piano up the plank at a slow constant rate. It is given by :

Since,
(in vertical direction)

W = 5534.53 J
So, the work done in sliding the piano up the plank is 5534.53 J. Hence, this is the required solution.
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Answer:
When have passed 3.9[s], since James threw the ball.
Explanation:
First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.
We'll use the kinematics equations to find these two unknowns.
![y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\](https://tex.z-dn.net/?f=y%3Dy_%7B0%7D%20%2Bv_%7B0%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cy%3D%20elevation%20%5Bm%5D%5C%5Cy_%7B0%7D%3Dinitial%20height%20%5Bm%5D%5C%5Cv_%7B0%7D%3D%20initial%20velocity%20%5Bm%2Fs%5D%20%3D41.67%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20passed%20%5Bs%5D%5C%5Cg%3D%20gravity%20%5Bm%2Fs%5E2%5D%3D9.81%5Bm%2Fs%5E2%5D%5C%5CNow%20replacing%3A%5C%5Cy%3D0%2B41.67%20%2A%282%29-%5Cfrac%7B1%7D%7B2%7D%20%2A%289.81%29%2A%282%29%5E%7B2%7D%20%5C%5C%5C%5Cy%3D63.72%5Bm%5D%5C%5C)
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can find the velocity after 2 seconds.
![v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bo%7D%20%2Bg%2At%5C%5Creplacing%3A%5C%5Cv_%7Bf%7D%20%3D41.67-%289.81%29%2A%282%29%5C%5C%5C%5Cv_%7Bf%7D%3D22.05%5Bm%2Fs%5D)
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

As we can see the equation is based on Time (t).
Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t
![y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%20%2Bv_%7Bo%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cv_%7Bo%7D%20%3D55.56%5Bm%2Fs%5D%20%3D%20initial%20velocity%5C%5Cy_%7Bo%7D%20%3D0%5Bm%5D%5C%5Cnow%20replacing%5C%5C63.72%20%2B22.05%20%2At-%284.905%29%2At%5E%7B2%7D%20%3D0%20%2B55.56%20%2At-%284.905%29%2At%5E%7B2%7D%20%5C%5C63.72%20%2B22.05%20%2At%20%3D0%20%2B55.56%20%2At%5C%5C63.72%20%3D%2033.51%2At%5C%5Ct%3D1.9%5Bs%5D)
Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.
Time = 2 + 1.9 = 3.9[s]