Answer:
The distance is 3.1 m
Explanation:
The position vector of the fly relative to the corner of the wall is
r = (3.1, 0.5).
The distance of the fly from the corner will be calculated as the magnitude of the vector "r"
magnitude of vector 
Since the numbers to be added have only one decimal place 3.<u>1</u> and 0.<u>5</u>, the result of the sum will have to have one decimal place. The result of the square root will also have one decimal place.
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr
Answer: D. The elements have the same number of valence electrons
Explanation: The chemical reactivity of elements is governed by the valence electrons present in the element.
The elements present in the same group or vertical column have similar valence configurations and thus behave similarly in chemical reactions or have similar bonding properties.
For Example: Both fluorine and chlorine belong to same family or group and both have 7 electrons in their valence shell and thus accept single electron to attain noble gas configuration.
thus both would bond with a cation bearing a single positive charge.
Not sure... need help with it
The correct answer is C) towards the center of the circle.
Although the object is moving at a constant speed it is constantly accelerating due to the constant change in direction as it describes the circular path. This causes a constant change in velocity as velocity is a vector quantity.
For the object to maintain the circular path there has to be centripetal force acting on the object and this centripetal force is directed towards the center of the circle.
