Ask question

Ask question

All categories

3 years ago

8

particles of mass 2m and m are fixed at the extreme of a light string of lenght l.find the centre of mass and the moment of iner

tia of the system of particles from the lighter particle.

1 answer:

Sever21 [200]3 years ago

3 0

Centre of mass will be at a distance of (L / 3) from mass 2m.

moment of inertia = m(2L/3)^2 + (2m)(L/3)^2

You might be interested in

Algebra II<br>Which is the correct statement?

Nataly_w [17]

B. g(x) is translated 2 units left. it's inside so it's a horizontal translation, and it's +2 which means it's translated left

6 0

3 years ago

Help me answer this question.

neonofarm [45]

Answer:

<h2>8 units²</h2>

Step-by-step explanation:

Let side of the square = a

The , Area of square = a²

Now, Midpoint of Diagonal DB is E

And DE = 2 units

So, DB = 2 DE = 2 × 2 = 4 units

Now, using Pythagoras theorem in ∆ BCD

DB² = DC² + BC²

plug the values

${4}^{2} = {a}^{2} + {a}^{2}$

Collect like terms

${4}^{2} = 2 {a}^{2}$

Evaluate the power

$16 = 2 {a}^{2}$

Swipe the side of the equation

$2 {a}^{2} = 16$

Divide both sides of the equation by 2

$\frac{2 {a}^{2} }{ 2 } = \frac{16}{2}$

Calculate

${a}^{2} = 8$

Therefore, The area of the square is 8 sq.units.

Hope this helps..

Best regards!!

3 0

3 years ago

Find an equation of the line described below. Write the equation is slope-intercept from ( solved for y), when possible. Through

Sladkaya [172]

Answer:

The equation of the line that is passing through the point (5,4) and is parallel to x-axis would be $y=4$

Step-by-step explanation:

Given that the line passes through the point (5,4).

As the line is parallel to the x-axis, the slope of the line would be zero.

And we have point (5,4) from which the line is passing.

So, $x_1=5\ ,\ y_1=4\ and\ m=0$

The equation of the line passing through point $(x_1,y_1)$ is

$y-y_1=m(x-x_1)\\y-4=0(x-5)\\y-4=0\\y=4$

So, the equation of the line that is passing through the point (5,4) and is parallel to x-axis would be $y=4$

4 0

3 years ago

How does the graph of y=-3√2x-4 compare to the graph of y=-3√x-4?

kifflom [539]

<span>Assuming the graph is y=-3(√2x)-4 and y=-3√(x-4) the transformation would be:

</span><span>The graph is compressed horizontally by a factor of 2
x=1/2x'
</span>y=-3(√2x)-4
y=-3(√x')-4 <span>

</span><span>moved left 4
x=x'-4
</span>y=-3(√x)-4
y=-3(√x'-4)-4
<span>
moved down 4
y=y'-4
</span>y=-3(√x-4)-4
y'-4=-3(√x'-4)-4
y'=-3(√x'-4)-4 +4
y'=-3(√x'-4)

Answer: C. <span>The graph is compressed horizontally by a factor of 2, moved left 4, and moved down 4.
</span>

6 0

3 years ago

Read 2 more answers

What’s the correct answer for this?

sammy [17]

Answer:

(0,2)

Step-by-step explanation:

2:4 means one part is 2/(2+4)=1/3 of AB and the other part is 2/3 of AB

Add 1/3 of the distance from -2 to 4. (1/3)(4+2)=2. -2+2=0 The x coordinate is 0

Subtract 1/3 of the distance from 6 to -6, (1/3)6+6)=4 6-4=2 The y coordinate is 2

The point is (0,2)

3 0

3 years ago