A 500 gram mass attached to a horizontal spring

Lumber has been manufactured in Texas since the early nineteenth century. In the beginning, these companies cut down all of the

salantis [7]

i need help toExplanation:

4 0

3 years ago

Read 2 more answers

The uniform slender bar AB has a mass of 6.4 kg and swings in a vertical plane about the pivot at A. If θ˙ = 2.7 rad/s when θ =

dolphi86 [110]

Answer:

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar, θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

∑ Mₐ = Iα

Where, Mₐ - the moment about A

α - angular acceleration

I - moment of inertia of the rod AB

$-mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha$

$\alpha=\frac{-3gcos\theta}{2l}$

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

$\vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

The acceleration at the center of the bar

$\vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}$

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}]$

Comparing the coefficients of i

$=-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}$

Comparing coefficients of j

$(\vec{a_{G}})_{y}=\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4}$

Net force on x direction

$F_{x}=(\vec{a_{G}})_{x}$

substituting the values

$F_{x}$ =1.5(14.58L+11.96)

Similarly net force on y direction

$F_{y}=(\vec{a_{G}})_{y}+mg$

= 3.2(2.97L - 157.03) + 62.72

Where L is the length of the bar AB

Therefore the net force,

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}$

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Substituting the value of L gives the force at pin A

8 0

3 years ago

It takes Serina 1.81 hours to drive to school. Her route is 36 km long. What is Serina's average speed on her drive to school?

AfilCa [17]

Answer:

60

Explanation:

because it shows he is moving at a low speed

4 0

3 years ago

I NEED HELP WITH ONE QUESTION!!

denis-greek [22]

Upon looking around a bit I found another brainly user who asked the same question. It was answered by a user named pinkfloyde and chosen by brainly as credible. this is what the user said:

"In designing electrical circuits, it is important to know the characteristics of circuits in order to know how to solve related problems that will arise whenever there will be problems in the circuits made. For example, series circuits makes the other lights go off whenever one bulb does not work. It would be easier to determine how to fix it because we know what type of circuit it is."

You should be able to use this and any other knowledge you know to come up with your desired answer

source: brainly.com/question/1269647

3 0

2 years ago

Which quantum numbers must be the same for the orbitals that they designate to be degenerate in a many-electron system?

stealth61 [152]

For the orbitals that designate to be degenerate in a many-electron system, the following quantum numbers should be equal:
1- principle quantum number (n) : this describes the most probable distance of the electron from the nucleus
2- orbital angular momentum quantum number (l) : this determines the shape of the orbital

8 0

3 years ago