3 years ago

Which of the following statements is true?

Physics

1 answer:

maksim [4K]3 years ago

8 0

There are NO true statements on the list you provided.

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Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equ

sdas [7]

Answer:

$x_{3}=2.35$

Explanation:

Given $x^2-2x-1=0,x_1=2$

From Newton's method

$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$

$f(x)=x^2-2x-1$

$f'(x)=2x-2$

Now

$x_{2}=x_1-\dfrac{f(x_1)}{f'(x_1)}$

$f(x)=x^2-2x-1$

$f(2)=2^2-2\times 2-1$

f(x)=-1

$f'(2)=2x-2$

$f'(1)=2\times 2-2$

f'(1)=2

$x_{2}=2+\dfrac{1}{2}$

$x_{2}=2.5$

$x_{3}=x_2-\dfrac{f(x_2)}{f'(x_2)}$

$f(2.5)=2.5^2-2\times 2.5-1$

f(2.5)=0.45

$f'(2.5)=2\times 2.5-2$

$f'(2.5)=3$

$x_{3}=2.5-\dfrac{0.45}{3}$

$x_{3}=2.35$

8 0

2 years ago

If light intensity obeyed an inverse square law you would expect to find the intensity of light to decrease as the square of the

Reptile [31]

Inverse square law: $\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
where
$I_{1}$ is the intensity at distance 1
$I_{2}$ is the intensity at distance 2
$d_{1}$ is distance 1
$d_{2}$ is distance 2

The inverse squared law state that intensity decreases in inverse proportion to the distance squared. So if light obeyed that rule, it will decreases its intensity as the square of the distance increases.

We can conclude that the correct answer is: true.

4 0

3 years ago

Read 2 more answers

In order of fastest to slowest, how do sound waves travel through different mediums?

Nikitich [7]

Gases, liquids, solids

7 0

3 years ago

A man throws a ball vertically upward with an initial speed of 20m/s. What is the maximum height reached by the ball and how lon

Luba_88 [7]

Answer:

It's your first question!!!

Explanation:

Enjoy it!!!

8 0

2 years ago

A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of

Fudgin [204]

Answer:

Part a)

$v_f = v_x = 32.77 m/s$

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

$v_x = 40 cos35 = 32.77 m/s$

$v_y = 40 sin35 = 22.94 m/s$

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

$v_f = 32.77 m/s$