Can u help me with this question

A rectangular lamina of uniform density is situated with opposite corners at (0,0) and (15,4). calculate its radii of gyration a

Sphinxa [80]

First, you have to find the moment of inertia along the x and y axes. Constant density is denoted as k.

$I_{x}= \int\limits^15_0\int\limits^4_0 {k y^{2} } \, dx dy= \frac{1}{3}k (15-4)^4=4880.33k$

$I_{y}= \int\limits^15_0\int\limits^4_0 {k x^{2} } \, dx dy= \frac{1}{3}k (15-4)^4=4880.33k$

Then, the radii of gyration for

x = √[I_x/m]
y = [I_y/m]

where m = k(15-4)² = 121k. Then,

x = y = [4880.33k/121k] = 40.33

I hope I was able to help you. Have a good day.

7 0

3 years ago

3/4+7/12<br>adding and subtracting fractions with value

lesantik [10]

Answer:It would be 4/3 or 1 1/3. You can put 3/4 to 9/12 and then add.

6 0

3 years ago

Help I’ve been working on this problem for about 2 days and I still don’t get it and please explain and show me how to do the wo

Bond [772]

the boiling point of jet fuel is 329 degrees F. Rounded to the nearest degree, what is the temperature in degrees Celcius? Use the Formula: C = 5/9(F - 32), where C represents degrees celcius, and F represents degrees fahrenheit.

A). 165 degrees C

B). 183 degrees C

C). 201 degrees C

D). 535 degrees C

Solve:

C = 5/9(F - 32)

Simplify both sides:

C = 5/9(329 - 32) = 0

Simplify:

5/9

Equation at end of step 1:

C - (5/9 * 297) = 0

C - 165 = 0

Solve single variable:

C - 165 = 0

Add 165 to both sides:

C = 165

Therefore, Your answer would be, Letter Choice (A), (C = 165 degrees C

Hope that helps!!!!! Have a great day!!!!! : )

4 0

3 years ago

What figures can be drawn in a circle with 2 perpendicular diameters?

nalin [4]

A little washer on wine cap

5 0

3 years ago

(a) Find the equation of the normal to the hyperbola x = 4t, y = 4/t at the point (8,2).

gtnhenbr [62]

The slope of the tangent line to the curve at (8, 2) is given by the derivative $\frac{dy}{dx}$ at that point. By the chain rule,

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Differentiate the given parametric equations with respect to $t$ :

$x = 4t \implies \dfrac{dx}{dt} = 4$

$y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}$

Then

$\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}$

We have $x=8$ and $y=2$ when $t=2$ , so the slope at the given point is $\frac{dy}{dx} = -\frac14$ .

The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation

$y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}$