Answer:8.28 km
Explanation:
Given
First it drifts
2.5 km

Secondly it drifts
4.70 km

After that it drifted along east direction 5.1 km

After that it drifts
7.2 km

After that it drifts
2.8 km

=
+![\left [ 2.5sin45-4.7sin60-7.2sin55+2.8sin5\right ]\hat{j}](https://tex.z-dn.net/?f=%5Cleft%20%5B%202.5sin45-4.7sin60-7.2sin55%2B2.8sin5%5Cright%20%5D%5Chat%7Bj%7D)


for direction

south of east
Sharing of 4 electrons between two atoms results in two double bonds. This can be see in the case of oxygen molecule (O2)
Atomic number of O = 8
Electron configuration of O = 1s²2s²2p⁴
Valence electron configuration: 2s²2p⁴
When 2 O atoms combine they share 4 electrons to form 2 double bonds. In addition, there are two lone pairs on each O atom.
Structural formula: O=O
Answer:
a)The electric Field will be zero at the point between the sheets
b)
c)
Explanation:
Let
be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.
Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

The Field will be away from the sheet and perpendicular to it.
a) The Electric Field between them

b)The Electric Field to the right of the sheets

c)The Electric Field to the left of the sheets

Answer:
- R = ( 4.831 m , 1.469 m )
- Direction of R relative to the x axis= 16°54'33'
Explanation:
Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

where
its the magnitude and θ.
So, for our vectors, we will have:


and


Now, we can take the sum of the vectors




This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem





To find the direction, we can use




As we are in the first quadrant, this is relative to the x axis.
Answer:
Create
Broken
Explanation:
Bond formation or creation requires the use of energy. Energy is used during bond formation between chemical species. The energy is required for the reaction to occur.
- When bonds are broken, energy is released
- Bond breaking process is a procedure that releases energy.
- This energy makes them able to react.