Express the standard form: (i) radius of the earth= 6 400 000 m (ii) radius of hydrogen atom = 0.000 000 000 0529

I) 6.4 × 10^(6) m

II) 5.29 × 10^(-11) m

Explanation:

Standard form is how we write very large number in such a way that it is easier to read.

Thus;

I) radius of the earth = 6 400 000 m

What we will now have is;

6.4 × 10^(6) m

II) radius of hydrogen atom = 0.000 000 000 0529 m

What we will have is;

5.29 × 10^(-11) m

i. 6.4 × 10⁶ m.

ii. 5.29 × 10⁻¹¹ m.

Explanation:

To express a number in standard form, we write it in the form a × 10ⁿ where 1 ≤ a < 10 and n is an integer.

(i) radius of the earth= 6 400 000 m

To express the radius of the earth = 6 400 000 m in standard form, we count the number of decimal places after the first digit. We have 6 decimal places after digit 6. So, we put a decimal point between digit 6 and 4. So, we have 6.4. Since we have 6 decimal places after the six digit, the exponent of our base which is 10 is 6.

So, writing 6 400 000 in the form in a × 10ⁿ, we have a = 6.4 and n = 6. So, in standard form we have 6.4 × 10⁶ m.

So, 6 400 000 =  6.4 × 10⁶ m in standard form.

(ii) radius of hydrogen atom = 0.000 000 000 0529 m

To express the radius of hydrogen atom = 0.000 000 000 0529 m in standard form, we count the number of decimal places after the decimal point including the first non-zero digit. We have 11 decimal places after the decimal point up to and including the first non-zero digit which is digit 5. So, we put a decimal point after the first non-zero digit,5. So, we have 5.29. Since we have 11 decimal places after the decimal point, the exponent of our base which is 10 is -11. The exponent is negative since the digit places are taken after the decimal point.

So, writing 0.000 000 000 0529 m in the form in a × 10ⁿ, we have a = 5.29 and n = -11. So, in standard form we have 5.29 × 10⁻¹¹ m.

So,  0.000 000 000 0529 m = 5.29 × 10⁻¹¹ m in standard form.

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1