# Idowu taye Ebenezer

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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:

$$-4 + 4 + \text{S} = 0 + 4$$

$$\text{S} = +4$$