Explanation:
In uniform gravity it is the same as the centre of mass. For regular shaped bodies it lies at the centre of the that particular body. Hence for a cylinder centre of gravity lies at the midpoint of the axis of the cylinder.
Answer:
4.3 g/ cm^3
Explanation:
Mass / Volume = density . so 2150/500= 4.3 g/cm^3
R=10
F=R/2
F=10/2=5
F=-5(CONCAVE MIRROR)
U=-8(CONCAVE MIRROR)
HEIGHT OF OBJECT=1.5
V=?
HEIGHT OF IMAGE=?
I/F=1/U+1/V
-I/5=-1/8-1/V
-1/V=-1/5+1/8
-1/V=-8+5/40
-1/V=-3/40
1/V=3/40
V=40/3
HEIGHT OF IMAGE/HEIGHT OF OBJECT =-V/U
HEIGHT OF IMAGE=40/3*1/-8*15/10
=-20/8
=-2.5
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)
An atom in which is the number of electrons that surrounds the nucleus