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Elenna [48]
3 years ago
14

Choose the option that best answers the question. The hypotenuse of a right triangle is 16 ft longer than the length of the shor

ter leg. If the area of this triangle is exactly 120 ft2, what is the length of the hypotenuse in feet?
Physics
1 answer:
FrozenT [24]3 years ago
3 0

Answer: The hypotenuse is 26 ft.

Step-by-step explanation: Let's call the shorter leg of the triangle x. The hypotenuse would be x + 16 and the other leg is y.

As the area is 120, we know that x.y/2 = 120 (because it's a right triangle).

This way, y = 240/x

we know that in a right triangle, we have a² = b² + c²

(x + 16)² = x² + y²

(x + 16)² = x² + (240/2)²

x² + 32x + 256 = x² + 57600/x²

32x + 256 = 57600/x²

32x³ + 256x² = 57600 (÷ 32)

x³ + 8x² = 1800

x = 10

∴

x + 16 = 26

y = 240/10 = 24

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Read 2 more answers
Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha
Westkost [7]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, A_p = 180 cm^2

- The charge on each plate, q = 17 * 10^-^6 C

- Permittivity of free space, e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}

- The radius for the flux region, r = 3.3 cm

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

                             A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ = \frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

                         E+ = E- = \frac{sigma}{2*e_o} \\\\

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

                        E_n_e_t = (E+)  + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C}  \\

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C

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