The derived unit for density are g/cm3.<br><br> True<br> False

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

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

$f'(2.5)=3$

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

So

$x_{3}=2.35$

8 0

3 years ago

Natalie accelerates her skateboard along a straight path from 4 m/s to 0 m/s in 25.0 s. Find the acceleration.

Lorico [155]

Its o.16 m/s2....really

7 0

3 years ago

The linear impulse delivered by the hit of a boxer is 202 N · s during the 0.244 s of contact. What is the magnitude of the aver

zlopas [31]

Answer: Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

Explanation: Impulse is defined as the force acting on an object for a short period or interval of time.

Mathematically it is given by the relation:

Impulse = Force $\times$ Time

According to the numerical values given in the question, I = 202 Ns and T = 0.244 s

So, Force F = $\frac{Impulse}{Time}$ = $\frac{202}{0.244}$ = 827.86 N

Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

7 0

3 years ago

You are in your car at rest when the traffic light turns green. You place your coffee cup on the horizontal dash and hit the gas

umka21 [38]

Answer:

(d) Negative.

Explanation:

let's test each at a time.

(a) It can't be 0, because cup would slide back other wise.

(b) Positive, well if forward is positive, than the work done against the forward acceleration must be negative , so it can't be positive.

(c) Equal to non-conservative work done by the car's engine.

well no, because work done by car's engine dosen't go all of it into getting car to move, so it can't be that.

(d) negative, this look like it, because work that friction does must be nagative to counteract positive thrust of car which is positive and in forward direction.

(d) this can't be true.

So the answer is (d) negative.

3 0

3 years ago

The robot arm is elevating and extending simultaneously. At a given instant, θ = 30°, ˙ θ = 10 deg / s = constant θ˙=10 deg/s=co

motikmotik

Explanation:

The position vector r:

$\overrightarrow{r(t)}=lcos\theta\hat{i}+lsin\theta\hat{j}$

The velocity vector v:

$\overrightarrow{v(t)}=\overrightarrow{\frac{dr}{dt}}=\dot{l}cos\theta-lsin\theta\dot{\theta}\hat{i}+\dot{l}sin\theta+lcos\theta\dot{\theta}\hat{j}$

The acceleration vector a:

$\overrightarrow{a(t)}}=cos\theta(\ddot{l}-l\dot{\theta}^2)-sin\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})\hat{i}+cos\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})+sin\theta(\ddot{l}-l\dot{\theta}^2)\hat{j}$

$\overrightarrow{v(t)}=0.13\hat{i}+0.18\hat{j}$

$\overrightarrow{a(t)}}=-0.3\hat{i}-0.1\hat{j}$

5 0

3 years ago

The derived unit for density are g/cm3. True False