Answer:
Change in electric potential energy ∆E = 365.72 kJ
Explanation:
Electric potential energy can be defined mathematically as:
E = kq1q2/r ....1
k = coulomb's constant = 9.0×10^9 N m^2/C^2
q1 = charge 1 = -2.1C
q2 = charge 2 = -5.0C
∆r = change in distance between the charges
r1 = 420km = 420000m
r2 = 160km = 160000m
From equation 1
∆E = kq1q2 (1/r2 -1/r1) ......2
Substituting the given values
∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)
∆E = 94.5 × 10^9 (3.87 × 10^-6) J
∆E = 365.72 × 10^3 J
∆E = 365.72 kJ
Answer:
D = 18000 kg/m3
V = 2.5*10{-7}m3
Explanation:
From the Archimedes principle,
Weight of fluid displaced = W_{air} - W_{water}
W_{air} = 4.5 gm
W_{water} = 4.25 gm
![W = [4.5 - 4.25]*9.81*10^{-3}](https://tex.z-dn.net/?f=W%20%3D%20%5B4.5%20-%204.25%5D%2A9.81%2A10%5E%7B-3%7D)
W = 2.4525*10{-3} N
D = 18000 kg/m3
b) object Volume can be obtained as ,
V = 2.5*10{-7}m3
Answer:
D. Half as great
Explanation:
Since we know that the friction force between the surface of crate and ground is given as
so here we know that
= friction coefficient between two surfaces which depends on the effective contact area between two surfaces
= normal force due to the object
So when we turn the object on another side such that the surface area is half then the friction coefficient will become also half
So here the friction force will also reduce to half
so correct answer will be
D. Half as great
Answer:
La única manera en que nuestro astronauta sería capaz de empujar la nave espacial en el espacio sin alejarse sería usar algo llamado "unidad de propulsión de astronauta". Supongamos que el astronauta está usando un SPK soviético, el sistema de cohetes mochila más poderoso jamás utilizado en el espacio.
Explanation:
