<span>In Ionic type of bonding, electrons are lost (more
protons than electrons and positive charge) or gained (more electrons than
protons, still a negative charge) by atoms, and the atoms are held together by
electrical attraction in the process. Covalent bondings are the sharing of electrons
as well as partial bondings. Covalent bondings’ electrons have the same charges
thus, there is no gaining or losing electrons in the process of sharing. Strong
bondings are applicable only to Hydrogen (H) atoms. </span>
Answer:
Explanation:
a ) work done by gravitational force
= mg sinθ ( d + .21)
Potential energy stored in compressed spring
= 1/2 k x²
= .5 x 431 x ( .21 )²
= 9.5
According to conservation of energy
mg sinθ ( d + .21) = 9.5
3.2 x 9.8 x sin 30( d + .21 ) = 9.5
d = 40 cm
b )
As long as mg sin30 is greater than kx ( restoring force ) , there will be acceleration in the block.
mg sin30 = kx
3.2 x 9.8 x .5 = 431 x
x = 3.63 cm
When there is compression of 3.63 cm in the spring , block will have maximum velocity. there after its speed will start decreasing.
Answer:
Explanation:
Sam mass=75kg
Height is 50m
20° frictionless slope
Horizontal force on Sam is 200N
According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.
Therefore
Wg - Ww =∆K.E
Note initial the body was at rest at top of the slope.
Then, ∆K.E is K.E(final) - K.E(initial)
K.E Is given as ½mv²
Since initial velocity is zero then, K.E(initial ) is zero
Therefore, ∆K.E=½mVf²
Wg is work done by gravity and it is given by using P.E formulas
Wg=mgh
Wg=75×9.8×50
Wg=36750J
Ww is work done by wind and it's is given by using formulae for work
Work=force × distance
Ww=horizontal force × horizontal distance
Using Trig.
TanX=opposite/adjacent
Tan20=h/x
x=h/tan20
x=50/tan20
x=137.37m
Then,
Ww=F×x
Ww=200×137.37
We=27474J
Now applying the formula
Wg - Ww =∆K.E
36750 - 27474 =½×75×Vf²
9276=37.5Vf²
Vf²=9275/37.5
Vf²= 247.36
Vf=√247.36
Vf=15.73m/s
As per the question a frog jumps 5 m towards east.
Frog again jumps 2 m north.
Let the displacement along east is denoted by vector A and the displacement towards north is denoted as vector B.
Hence magnitude of A = 5 m
Magnitude of B = 2 m
We are asked to calculate the total displacement.
Here the angle between them is 90 degree as A is towards east and B is towards north.
As per parallelogram law of vector addition,the magnitude of total displacement [R] will be-


[cos90= 0]

[ans]