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Idowu Taye provided a contribution 2 years ago

Question 26 | Physics Past Questions

Let's start from K.E = P.E:

\(\text{mgh} = \frac{1}{2}\text{mv}^2\)

\(\text{mgh} = \cfrac{\text{mv}^2}{2}\)

Cross multiply:

It gives us:

\(2\text{mgh} = \text{mv}^2\)

Divide both sides by m:

It gives:

\(2\text{gh} = \text{v}^2\)

To get the value of h, divide both sides by 2g

The formula to use will become:

\(\text{h} = \cfrac{\text{v}^2}{2\text{g}}\)

Then \(\text{v} = 5\text{ms}^{-1}\)

g = 10. Put this into the above formula:

\(\text{h} = \cfrac{5^2}{2 \times 10}\)

\(\text{h} = \cfrac{25}{20}\)

\(\text{h} = 1.3\text{m}\)


Idowu Taye provided a contribution 2 years ago

Question 6 | Physics Past Questions

Because bata particles emit -1 and this is just balancing equation.

Idowu Taye provided a contribution 2 years ago

Question 25 | Chemistry Past Questions

\(\mathrm{H_2SO_3}\)

\((+1 \times 2) + \text{S} + (-2 \times 3) = 0\)

\(2 + \text{S} - 6 = 0\)

\(-4 + \text{S} = 0\)

TO GET S ADD +4 TO THE BOTH SIDES:

\(-4 + 4 + \text{S} = 0 + 4\)

\(\text{S} = +4\)