The mass m of the object = 5.25 kg
<h3>Further explanation</h3>
Given
k = spring constant = 3.5 N/cm
Δx= 30 cm - 15 cm = 15 cm
Required
the mass m
Solution
F=m.g
Hooke's Law
F = k.Δx

Answer:
The average acceleration of the bearings is 
Explanation:
Given that,
Height = 1.94 m
Bounced height = 1.48 m
Time interval 
Velocity of the ball bearing just before hitting the steel plate
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



Negative as it is directed downwards
After bounce back,
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



We need to calculate the average acceleration of the bearings while they are in contact with the plate
Using formula of acceleration

Put the value into the formula



Hence,The average acceleration of the bearings is 
In 2Cr2O7 there’s 2 items; Cr and O
Well you have to minus the 4.5 to 5.2 and the answer to that would be -11.5 and calculated that to be 4.5
Answer
given,
net charge = +2.00 μC
we know,
1 coulomb charge = 6.28 x 10¹⁸electrons
1 micro coulomb charge = 6.28 x 10¹⁸ x 10⁻⁶ electron
= 6.28 x 10¹² electrons
2.00 μC = 2 x 6.28 x 10¹² electrons
= 1.256 x 10¹³ electrons
since net charge is positive.
The number of protons should be 1.256 x 10¹³ more than electrons.
hence, +2.00 μC have 1.256 x 10¹³ more protons than electrons.