Rwrite a definition of a fraction equivalent

1 answer:

7 0

Multiplying the numerator and the denominator of a fraction by the same nonzero whole number will change that fraction into an equivalent fraction, but it will not change its value.

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In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53 × 10−1

tresset_1 [31]

For the answer to the question above,
a) Use F = k * q1 * q2 / d²
where k = 8.99e9 N·m²/C²
and q1 = -1.602e-19 C (electron)
and q2 = 1.602e-19 C (proton)
and d = distance between point charges = 0.53e-10 m
The negative result indicates "attraction".

b) Here, just use F = ma
where F was found above, and
m = mass of electron = 9.11e-31kg, if memory serves
a = radial acceleration

c) Now use a = v² / r
where a was found above
and r was given

d) period T = 2π / ω = 2πr / v
where v was found above
and r was given

7 0

3 years ago

Use pascal's triangle to expand the binomial (d-5y)^6 please show work

Ratling [72]

Pascals triangle to the 6th:
1 x^0
1 1 x^1
1 2 1 x^2
1 3 3 1 x^3
1 4 6 4 1 x^4
1 5 10 10 5 1 x^5
1 6 15 20 15 6 1 x^6

the problem is to the 6th power so your going to use the 6th row of pascals triangle (don't count the first row). these numbers represent the coefficients of the variables

1(d-5y)^6 + 6(d-5y)^5 + 15(d-5y)^4 + 20(d-5y)^3 + 15(d-5y)^2 + 6(d-5y) + 1
then simplify

4 0

3 years ago

Read 2 more answers

This trapezium is drawn on a centimetre grid. Find the area of the trapezium.

aivan3 [116]

Answer:

20 unit²

Step-by-step explanation:

A trapezium is given to us on the grid and we need to find out the area of the trapezium . In order to find the area , we need to find the measure of the parallel sides and the distance between the parallel sides.

From the grid :-

$\rm\implies Side_1 = 7 \ units$

$\rm\implies Side_2 = 3 \ units$

$\rm\implies \perp \ Distance =4 \ units$

Now here we got the two parallel sides of the trapezium and the distance between the two parallel sides. Now we can find the area as ,

$\rm\implies Area_{Trapezium}= \dfrac{1}{2}\times ( s_1 + s_2) \times \perp \ Distance \\\\\rm\implies Area = \dfrac{1}{2} \times ( 7 + 3 ) \times 4 \ unit^2 \\\\\rm\implies Area = \dfrac{1}{2} \times ( 10) \times 4 \ unit^2 \\\\\rm\implies\boxed{\rm Area = 20 \ unit^2}$