Answer:
The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m
Energy approach has been used to sole the problem.
The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring
The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved
Explanation:
The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.
As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .
x = compression of the spring = 0.89
Answer:
Nanotechnology refers to the branch of science and engineering devoted to designing, producing, and using structures, devices, and systems by manipulating atoms and molecules at nanoscale, i.e. having one or more dimensions of the order of 100 nanometres (100 millionth of a millimetre) or less.
Explanation:
Answer: The bottom of the ladder is moving at 3.464ft/sec
Explanation:
The question defines a right angle triangle. Therefore using pythagorean
h^2 + l^2 = 10^2 = 100 ...eq1
dh/dt = -2ft/sec
dl/ dt = ?
Taking derivatives of time in eq 1 on both sides
2hdh/dt + 2ldl/dt = 0 ....eq2
Putting l = 5ft in eq2
h^ + 5^2 = 100
h^2 = 25 = 100
h Sqrt(75)
h = 8.66 ft
Put h = 8.66ft in eq2
2 × 8.66 × (-2) + 2 ×5 dl/dt
dl/dt = 17.32 / 5
dl/dt = 3.464ft/sec
Answer:
Asch (1956) found that group size influenced whether subjects conformed. The bigger the majority group (no of confederates), the more people conformed, but only up to a certain point.
Explanation:
Answer:
sweeps out equal areas in equal times.
Explanation:
As we know that there is no torque due to Sun on the planets revolving about the sun
so we will have
![\tau_{net} = 0](https://tex.z-dn.net/?f=%5Ctau_%7Bnet%7D%20%3D%200)
now we have
![\frac{dL}{dt}= 0](https://tex.z-dn.net/?f=%5Cfrac%7BdL%7D%7Bdt%7D%3D%200)
now we also know that
![Area = \frac{1}{2}r^2d\theta](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7Dr%5E2d%5Ctheta)
so rate of change in area is given as
![\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7Dr%5E2%5Cfrac%7Bd%5Ctheta%7D%7Bdt%7D)
so we will have
![\frac{dA}{dt} = \frac{1}{2}r^2\omega](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7Dr%5E2%5Comega)
![\frac{dA}{dt} = \frac{L}{2m}](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%20%3D%20%5Cfrac%7BL%7D%7B2m%7D)
since angular momentum and mass is constant here so
all planets sweeps out equal areas in equal times.