Pls help me answer this :'(

Two forces with magnitudes 300 pounds and 450 pounds, with an angle of 60° between them, act upon an object. What is the approxi

kifflom [539]

Answer:

653.84 lbs.

Step-by-step explanation:

To make the computation simple, assume the 450 lb force acts along the x axis in the positive direction.

The x component of the 300 lb force is 300cos(60)=150 lbs.

The y component of the 300 lb force is 300sin(60)= $300*\frac{\sqrt{3} }{2} =259.81\\$ lbs.

The x component of the total force is the sum of the x components of the two forces.

Fx=150+450=600

Fy=259.81

The net force is obtained using the Pythagorean theorem, since the two force components are perpindicular.

F=Sqrt(Fx^2+Fy^2)=Sqrt(360000+67501.24)=Sqrt(427501)=653.84 lbs.

4 0

2 years ago

Read 2 more answers

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

18.25=4(1.5-3.75x)-.25(4x-5) wat is x?

sdas [7]

The answer is x= - 11/ 16
You can check this by plugging in x for -11/16

5 0

2 years ago

What is the answer to 6 3/8 -3/4- 1/2 =

zysi [14]

Answer: 5 1/8

Step-by-step explanation:

1. 6 3/8=51/8

2. 3/4=6/8; 1/2=4/8

3. 51/8-6/8-4/8=41/8

4. 41/8=5 1/8

3 0

2 years ago

Four ways on how save water

irina1246 [14]

Turn off the faucet if not in use
Take shorter showers
Water your plants only when it needs it
Don’t throw trash on the sea

6 0

2 years ago