In the movement of the weight in vertical circle, using momentum balance, the largest tension is at the bottom of the circle. This is represented by:
<span>F = T - m g </span>
<span>T = F + m g
</span>F (centripetal) = mv^2/r
<span>= m v^2 / r + m g </span>
<span>m v^2 / r = T - m g </span>
<span>T= 0.5m * 100kgm/s^2 / 0.2kg - 9.81m/s^2 * 0.5m </span>
<span>T= 245 m^2/s^2 </span>
It would be a really bad idea to eat the snow because you obviously are trying to stay warm right? Well, the best thing to do is melt the snow. However, the process of melting the snow would have a few complications as well. But yes, the latter idea (drinking the snow) is a better idea (not the best).
Answer:
v= 2413.5 m/s
Explanation:
maximum change of speed of rocket
=(initial exhaust velocity)×ln [(initialmass/finalmass)]
let initial mass= m
final mass = m-m(4/5) = m/5
[since the 80% of mass which is fuel is exhausted]
V-0 = 1500 ln (1/0.2)
V= 1500×1.609 = 2413.5 m/s
therefore, its exhaust speed v= 2413.5 m/s
An activity that is relatively short in time <10 seconds and has few repetitions predominantly uses the ATP/PC energy system. The cellular respiration procedure that changes food energy into ATP which is a form of energy is largely reliant on oxygen obtainability. During exercise the source and request of oxygen obtainable to muscle is unnatural by period and strength and by the individual’s cardiorespiratory suitability level.
Steps of the ATP-PC system:
1. Primarily, ATP kept in the myosin cross-bridges which is microscopic contractile parts of muscle is broken down to issue energy for muscle shrinkage. This action consents the by-products of ATP breakdown which are the adenosine diphosphate and one single phosphate all on its own.
2. Phosphocreatine is then broken down by the enzyme creatine kinase into creatine and phosphate.
3. The energy free in the breakdown of PC permits ADP and Pi to rejoin creating more ATP. This newly made ATP can now be broken down to issue energy to fuel activity.
First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.
Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.
