Any object that is spherical in shape would best represent a true scale model of the shape of the Earth. Examples are ping pong balls, billiard balls, marble and other smooth spherical objects. The shape of the Earth is called the oblate spheroid. The "oblate" would refer to an oblong shape and "spheroid" would refer to an almost spherical shape. The earth has on almost spherical shape and has a slightly oblong appearance. The diameter from the South pole to the north pole was measured to have a value of 12714 km while the diameter of the equator is approximately 12756 km. As you can see, the values are not equal. This makes the earth not a perfect sphere.
The correct definition of a fracture is break in the bone
<u>Explanation:</u>
When nay injury results in the breaking or causing any cracking in the bones of any parts then this will lead to fracture. When the injury caused is near the ligament or tissue in which the bone is connected or attached then it will lead to an avulsion fractures. Thus this will lead to the pulling of bone form the original position thereby leading more pain in the spot associated with the fracture.
Sports people are the victims of this type of fracture. Fracture may occur anywhere mostly legs,hands,ankle,hip and elbow. sometimes it may be in finger, shoulder,knee,etc. The main symptoms that are associated with fracture includes, selling, inability in moving the fractured part or pain associated when trying to move that part, Loss of the affected part's function,etc.
M1 = 750Kg, v1 = 10m/s
m2 = 2500Kg , v2= 0 (because in problem say cuz that object don t move).
The momentum before colision is equal with the momentum after colision:
m1v1 + m2v2 = (m1+m2)v3 => v3 is the velocity after colison and that s u want to caluclate for your problem
=> m1v1 = (m1+m2)v3 => v3 = m1v1/(m1+m2) now u should do the math i think v3 prox 2,4 but not sure u should caculate
Its rays point away from the charge
Answer:
i) E = 269 [MJ] ii)v = 116 [m/s]
Explanation:
This is a problem that encompasses the work and principle of energy conservation.
In this way, we establish the equation for the principle of conservation and energy.
i)
![W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]](https://tex.z-dn.net/?f=W_%7B1-2%7D%3D%20%28F%2Ad%29%20-%20%28m%2Ag%2Ah%29%5C%5CW_%7B1-2%7D%3D%28500000%2A2.5%2A10%5E3%29-%2840000%2A9.81%2A2.5%2A10%5E3%29%5C%5CW_%7B1-2%7D%3D%20269%2A10%5E6%5BJ%5D%20or%20269%20%5BMJ%5D)
At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.
Er = 269*10^6[J]
ii ) With the energy calculated at the previous point, we can calculate the speed developed.
![E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]](https://tex.z-dn.net/?f=E_%7Bk2%7D%3D0.5%2Am%2Av%5E2%5C%5C269%2A10%5E6%3D0.5%2A40000%2Av%5E2%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B269%2A10%5E6%7D%7B0.5%2A40000%7D%20%7D%5C%5C%20v%3D116%5Bm%2Fs%5D)
