Answer:
To calculate displacement, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your displacement is 0. In physics, displacement is represented by Δs.
Explanation:
hope helps
Answer:
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Steps 1 and 2)
The variables are W = work, P = power, and t = time. In this case, W = 9514 joules and P = 347 watts.
The goal is to solve for the unknown time t.
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Step 3)
Since we want to solve for the time, and we have known W and P values, we use the equation t = W/P
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Step 4)
t = W/P
t = 9514/347
t = 27.4178674351586
t = 27.4 seconds
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Step 5)
The lawn mower ran for about 27.4 seconds. I rounded to three sig figs because this was the lower amount of sig figs when comparing 9514 and 347.
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Note: we don't use the mass at all
Answer:
a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]
Explanation:
a)
The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.
![E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%3D%2060%5Bkg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20%3D%2010%20%5Bm%5D%5C%5CE_%7Bp%7D%3D60%2A9.81%2A10%5C%5CE_%7Bp%7D%3D5886%5BJ%5D)
b)
Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.
![E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20E_%7Bk%7D%20%5C%5C5886%20%3D0.5%2Am%2Av%5E%7B2%7D%20%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B5886%7D%7B0.5%2A60%7D%20%7D%5C%5Cv%20%3D%2014%5Bm%2Fs%5D)
c)
The work is equal to
W = 5886 [J]
d)
We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.
![v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%3D%20v_%7Bo%7D%20%5E%7B2%7D-2%2Aa%2Ad%5C%5Cwhere%3A%5C%5Cd%20%3D%202.5%5Bm%5D%5C%5Cv_%7Bf%7D%3D0%5C%5Cv_%7Bo%7D%20%3D14%5Bm%2Fs%5D%5C%5CTherefore%5C%5Ca%20%3D%20%5Cfrac%7B14%5E%7B2%7D%20%7D%7B2%2A2.5%7D%20%5C%5Ca%20%3D%2039.2%5Bm%2Fs%5E2%5D)
By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.
m*g - F = m*a
F = m*a - m*g
F = (60*39.2) - (60*9.81)
F = 1763.4 [N]
Average speed = (1/2) (beginning speed + ending speed)
= (1/2) ( 13 m/s + 30 m/s )
= (1/2) ( 43 m/s )
= 21.5 m/s