Answer:
Velocity = 20.3 [m/s]
Explanation:
This is a typical problem of energy conservation, where potential energy is converted to kinetic energy. We must first find the potential energy. In this way, we will choose as a reference point or point where the potential energy is zero when the carrriage rolls down 21 [m] from the top of the hill.
![E_{p} =m*g*h\\ where:\\m = mass = 25[kg]\\g = gravity = 9.81 [m/s^2]\\h = elevation = 21 [m]\\E_{p} =potential energy [J]\\E_{p} =25*9.81*21=5150[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5C%20where%3A%5C%5Cm%20%3D%20mass%20%3D%2025%5Bkg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%20%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20%3D%2021%20%5Bm%5D%5C%5CE_%7Bp%7D%20%3Dpotential%20energy%20%5BJ%5D%5C%5CE_%7Bp%7D%20%3D25%2A9.81%2A21%3D5150%5BJ%5D)
Now this will be the same energy transformed into kinetic energy, therefore:
![E_{p}=E_{k} = 5150[J]\\E_{k} =0.5*m*v^{2} \\where:\\v=velocity [m/s]\\v=\sqrt{\frac{E_{k}}{0.5*25} } \\v=20.3[m/s]](https://tex.z-dn.net/?f=E_%7Bp%7D%3DE_%7Bk%7D%20%3D%205150%5BJ%5D%5C%5CE_%7Bk%7D%20%3D0.5%2Am%2Av%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cv%3Dvelocity%20%5Bm%2Fs%5D%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7BE_%7Bk%7D%7D%7B0.5%2A25%7D%20%7D%20%5C%5Cv%3D20.3%5Bm%2Fs%5D)
Capillarity "Capillarity causes the part of the surface of a liquid in contact with a solid, to be either elevated above (e.g. water), or depressed below (e.g. mercury), the rest of the surface. This trait is named for the behaviour of liquids in capillary tubes placed perpendicular to the surface. The forces operating within capillarity are cohesion, adhesion, and surface tension."
Answer:
True
Explanation:
True\
I'm not sure but i think it is true la
Answer:
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,
Explanation:
When a satellite is in orbit the most important force is the docking of gravity with the Earth
F = m a
where the acceleration is centripetal and F is the force of universal attraction
centripetal acceleration is
a = v² / r
F = m v² / r
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.
The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.