Answer:
The car C has KE = 100, PE = 0
Explanation:
The principle of conservation of energy states that although energy can be transformed from one form to another, the total energy of the given system remains unchanged.
The energy that a body possesses due to its motion or position is known as mechanical energy. There are two kinds of mechanical energy: kinetic energy, KE and potential energy, PE.
Kinetic energy is the energy that a body possesses due to its motion.
Potential energy is the energy a body possesses due to its position.
From the principle of conservation of energy, kinetic energy can be transformed into potential energy and vice versa, but in all cases the energy is conserved or constant.
In the diagram above, the cars at various positions of rest or motion are transforming the various forms of mechanical energy, but the total energy is conserved at every point. At the point A, energy is all potential, at B, it is partly potential partly kinetic energy, However, at the point C, all the potential energy has been converted to kinetic energy. At D, some of the kinetic energy has been converted to potential energy as the car climbs up the hill.
Therefore, the car C has KE = 100, PE = 0
Answer:
mu=12Tm^2
Explanation:
the magnetic moment mu of a single loop is given by:
![\mu = I A B](https://tex.z-dn.net/?f=%5Cmu%20%3D%20I%20A%20B)
where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:
![\mu=(0.5A)(4m*2m)(3T)=12Tm^2](https://tex.z-dn.net/?f=%5Cmu%3D%280.5A%29%284m%2A2m%29%283T%29%3D12Tm%5E2)
hope this helps!!
To solve this problem we will make a graph that allows us to understand the components acting on the body. In this way we will have the centripetal Force and the Force by gravity generating a total component. If we take both forces and get the trigonometric ratio of the tangent we would have the angle is,
![T_x = nsinA = \frac{mv^2}{r}](https://tex.z-dn.net/?f=T_x%20%3D%20nsinA%20%3D%20%5Cfrac%7Bmv%5E2%7D%7Br%7D)
![T_y = ncosA = mg](https://tex.z-dn.net/?f=T_y%20%3D%20ncosA%20%3D%20mg)
Dividing both.
![tan A = \frac{v^2}{rg}](https://tex.z-dn.net/?f=tan%20A%20%3D%20%5Cfrac%7Bv%5E2%7D%7Brg%7D)
![tan A = \frac{11.7^2}{50*9.8}](https://tex.z-dn.net/?f=tan%20A%20%3D%20%5Cfrac%7B11.7%5E2%7D%7B50%2A9.8%7D)
![A = tan^{-1} (0.279367)](https://tex.z-dn.net/?f=A%20%3D%20tan%5E%7B-1%7D%20%280.279367%29)
![A = 15.608\°](https://tex.z-dn.net/?f=A%20%3D%2015.608%5C%C2%B0)
Therefore the angle that should the curve be banked is 15.608°
Answer:
Explanation:
It would actually be A. 30 , as each hour of ascension (i am not sure about the correct terminology) equals 15 .