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2 years ago

Obtain the zeroes of polynomial

Physics

1 answer:

morpeh [17]2 years ago

3 0

I assume you meant to say

$f(x)=2x^4+3x^3-5x^2-9x-3$

Given that x = √3 and x = -√3 are roots of f(x), this means that both x - √3 and x + √3, and hence their product x ² - 3, divides f(x) exactly and leaves no remainder.

Carry out the division:

$\dfrac{2x^4+3x^3-5x^2-9x-3}{x^2-3} = 2x^2+3x+1$

To compute the quotient:

* 2x ⁴ = 2x ² • x ², and 2x ² (x ² - 3) = 2x ⁴ - 6x ²

Subtract this from the numerator to get a first remainder of

(2x ⁴ + 3x ³ - 5x ² - 9x - 3) - (2x ⁴ - 6x ²) = 3x ³ + x ² - 9x - 3

* 3x ³ = 3x • x ², and 3x (x ² - 3) = 3x ³ - 9x

Subtract this from the remainder to get a new remainder of

(3x ³ + x ² - 9x - 3) - (3x ³ - 9x) = x ² - 3

This last remainder is exactly divisible by x ² - 3, so we're left with 1. Putting everything together gives us the quotient,

2x ² + 3x + 1

Factoring this result is easy:

2x ² + 3x + 1 = (2x + 1) (x + 1)

which has roots at x = -1/2 and x = -1, and these re the remaining zeroes of f(x).

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astra-53 [7]

Answer: It will remain the same

Explanation:

According to Coulomb's Law:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them".

Mathematically this law is written as:

$F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}$ is the electrostatic force

$K$ is the Coulomb's constant

$q_{1}=q_{2}=q$ are the electric charges , which in this case have the same positive charge

$d$ is the separation distance between the charges

Rewritting we have:

$F_{E}=K\frac{q^{2}}{d^{2}}$ (2)

Now, if the first charge is doubled:

$q_{1}=2q_$

And the second is reduced to a half:

$q_{2}=\frac{1}{2}q$

We will have the following:

$F_{E}=K\frac{(2q)(\frac{1}{2}q)}{d^{2}}$ (3)

$F_{E}=K\frac{q^{2}}{d^{2}}$ (4)

As we can see equation (4) is equal to equation (2), this means the force of repulsion between both charges will remain the same

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3 years ago

Read 2 more answers

During wich two time intervals does the particle undergo equal displacement

sweet-ann [11.9K]

Basically, you want to take the integral of each interval and compare them. The two intervals with the same integral represent equal displacement of the particle. And since delta(x) is always 2, all you have to do is average the initial and final velocities of each interval and multiply by two to find total displacement.

Hope it helped.

Edit to show calculations:

2 * [ (0 + 10)/2 ] = 10 for interval AB
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2 years ago

Plz help ASAP I'll mark as brainliest

gogolik [260]

Hi there!

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

We can begin by looking at the given graph.

When the spring force = 4N, the total length of the spring is 35 cm.

Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:

35 - 30 = 5 cm.

5.1.

If the spring elongates by 10 cm, the total length of the spring is:

30 + 10 = 40 cm

According to the graph, a length of 40 cm corresponds to a force of 8N.

5.2.

We can solve for the weight of the ball using the following:

W (weight) = m (mass) · acceleration due to gravity (10N/kg)

Using a summation of forces:

∑F = T - W

The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

0 = T - W

T = W = 8 N

5.3.

0 = T - Mg

T = Mg

Use the prior value of T and gravity to solve:

8 = 10M

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