String the·o·ry
noun
a cosmological theory based on the existence of cosmic strings.
Answer:
hmax=81ft
Explanation:
Maximum height of the object is the highest vertical position along its trajectory.
The vertical velocity is equal to 0 (Vy = 0)

we isolate th (needed to reach the maximum height hmax)

The formula describing vertical distance is:

So, given y = hmax and t = th, we can join those two equations together:


if we launch a projectile from some initial height h all you need to do is add this initial elevation


Answer:
The water acts like a lubricant therefore has a smooth motion over the ice.
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]
Answer:
A. α = - 1.047 rad/s²
B. θ = 14.1 rad
C. θ = 2.24 rev
Explanation:
A.
We can use the first equation of motion to find the acceleration:
where,
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s
t = time = 3 s
α = angular acceleration = ?
Therefore,
<u>α = - 1.047 rad/s²</u>
B.
We can use the second equation of motion to find the angular distance:
<u>θ = 14.1 rad</u>
C.
θ = (14.1 rad)(1 rev/2π rad)
<u>θ = 2.24 rev</u>