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Ulleksa [173]
3 years ago
14

The first few steps in deriving the quadratic formula are shown. Which best explains why is not added to the left side of the eq

uation in the last step shown in the table? The term is added to the right side of the equation, so it needs to be subtracted from the left side of the equation to balance the sides of the equation. The distributive property needs to be applied to determine the value to add to the left side of the equation to balance the sides of the equation. The term needs to be converted so it has a common denominator before adding it to the left side of the equation to balance the equation. The square root of the term needs to be found before adding the term to the left side of the equation to balance the sides of the equation.
Physics
2 answers:
bija089 [108]3 years ago
6 0
The term b^2 / 4a^2 is not added to the left side of the equation, because the term that was added to the right was not either b^2 / 4 a^2.

As you can see the ther b^2 / 4a^2 that appears in the last step of the table is inside a parenthesis, which is preceded by factor a.


Then, you need to apply the distributive property to know the term that you are really adding to the right side,  i.e. you need to mulitply b^2 / 4a^2 * a which is b^2 / 4a.


That means that you are really adding b^2 / 4a to the right, so that is the same that you have to add to the left, which is what the last step of the table shows.


That situation is reflected by the statement "<span>The distributive property needs to be applied to determine the value to add to the left side of the equation to balance the sides of the equation".</span>  That is the answer.
Flauer [41]3 years ago
6 0

Answer:B

Explanation:

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Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

8 0
2 years ago
After the pendulum is dropped, determine the height at which the kinetic energy is equal to the potential energy.
Fofino [41]
0.5 m v² = m g h
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3 years ago
For thermal equilibrium at temperature Tan appropriate measure of energy is kT where k is Boltzmann's constant. Convert the foll
Schach [20]

Answer:

1 cm⁻¹ =1.44K  1 ev = 1.16 10⁴ K

Explanation:

The relationship between temperature and thermal energy is

     E = K T

The relationship of the speed of light

    c =λ f = f / ν          1/λ= ν

The Planck equation is

          E = h f

Let's start the transformations

     c = f λ = f / ν        

     f = c ν

     E = h f

     E = h c ν

     E = KT

     h c ν = K T

     T = h c ν  / K =( h c / K) ν

Let's replace the constants

     h = 6.63 10⁻³⁴ J s

     c = 3 10⁸ m / s

     K = 1.38  10⁻²³ J / K

 

     v = 1 cm-1 (100 cm / 1 m) = 10² m-1

   

     T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²

     A = h c / K = 1,441 10⁻²

     T =  1.44K

     ν = 103 cm⁻¹ = 103 10² m

     T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²

     T = 148K

1 Rydberg = 1.097 10 7 m

As we saw at the beginning the λ=1 / v

     T = (h c / K) 1 /λ

     T = 1,441 10⁻²  1 / 1,097 10⁷

     T = 1.3 10⁻⁹ K

    E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J

    E = KT

    T = E/K

    T = 1.6 10⁻¹⁹ /1.38 10⁻²³

    T = 1.16 10⁴ K

3 0
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An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance
marshall27 [118]

Answer:

E = \frac{\lambda}{2\pi \epsilon_0 r}

Explanation:

Let the linear charge density of the charged wire is given as

\frac{q}{L} = \lambda

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

\int E. dA = \frac{q}{\epsilon_0}

here we have

E \int dA = \frac{\lambda L}{\epsilon_0}

E. 2\pi r L = \frac{\lambda L}{\epsilon_0}

so we have

E = \frac{\lambda}{2\pi \epsilon_0 r}

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for a given type of wave in a given medium a larger frequency means a smaller wavelength

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3 years ago
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