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2 years ago

The order of magnitude of 1001 is

1 answer:

8 0

Order of magnitude is usually written as 10 to the nth power. The n represents the order of magnitude. If you raise a number by one order of magnitude, you are basically multiplying that number by 10. If you decrease a number by one order of magnitude, you are basically multiplying that number by 0.1.

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3 years ago

How do you calculate distance using deceleration and time?

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3 years ago

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

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2 years ago

The order of magnitude of 1001 is​

The order of magnitude of 1001 is