Covalent bonds form when electrons are shared between atoms and are attracted by the nuclei of both atoms. In pure covalent bonds, the electrons are shared equally. In polar covalent bonds, the electrons are shared unequally, as one atom exerts a stronger force of attraction on the electrons than the other.
With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity
d= 250s
t= 17s
a= d/t
=

= 4.7
Answer:
Work done by the machine (W) = 500 × 1.5 = 750 J
Work supplied to the machine (W) = 100 × 10 = 1000 J
Here, work supplied to the machine is input work = 1000 J
Answer:
V_{a} - V_{b} = 89.3
Explanation:
The electric potential is defined by
= - ∫ E .ds
In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.
V_{b} - V_{a} = - ∫ E ds
We substitute
V_{b} - V_{a} = - ∫ (α + β/ y²) dy
We integrate
V_{b} - V_{a} = - α y + β / y
We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m
V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)
V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33
V_{b} - V_{a} = - 89.3 V
As they ask us the reverse case
V_{b} - V_{a} = - V_{b} - V_{a}
V_{a} - V_{b} = 89.3
Answer:
160 kg
12 m/s
Explanation:
= Mass of first car = 120 kg
= Mass of second car
= Initial Velocity of first car = 14 m/s
= Initial Velocity of second car = 0 m/s
= Final Velocity of first car = -2 m/s
= Final Velocity of second car
For perfectly elastic collision

Applying in the next equation


Mass of second car = 160 kg
Velocity of second car = 12 m/s