What is a question that an engineer might have about a can opener?

What is the correct answer?

pentagon [3]

Answer:

B) x^2+6x+8

Explanation:

x-4 | x^3+2x^2-16x-32

- x^3-4x^2 <-- (x-4)(x^2)

_________________

6x^2-16x-32

- 6x^2-24x <-- (x-4)(6x)

_________________

8x-32

- 8x-32 <- (x-4)(8)

___________________________

0 | x^2+6x+8

This means the answer is B) x^2+6x+8

3 0

2 years ago

At the race track, one race car starts its engine with a resulting intensity level of 104.0 dB at point P. Then 6 more cars star

dem82 [27]

To solve the problem it is necessary to apply the concepts related to sound intensity. The most common approach to sound intensity measurement is to use the decibel scale:

$\beta (dB) = 10log_{10}(\frac{I}{I_0})$

Where,

$I_0 = 1*10^{-12}$ is a reference intensity. It is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz.

I = Sound intensity

Our values are given by,

$\beta = 104dB$

$\#Autos = 7$

For each auto the intensity would be,

$104 = 10log\frac{I}{1*10^{-12}}$

$10.4= log_{10} (\frac{I}{10^{-12}})$

$10^{10.4}*10^{-12}=I$

$I = 0.02511W/m^2$

Therefore the sound intesity for the 7 autos is

$I= 7* 0.02511$

$I = 0.1748W/m^2$

The sound level for the 7 cars in dB is

$\beta (dB) = 10log_{10}(\frac{0.1748}{1*10^{-12}})$

$\beta (dB) = 112.42dB$

8 0

3 years ago

Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the

Scrat [10]

Answer: $1.8\°$

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d=3.4(10)^{-5}m$ is the width of the slit

$\lambda=540 nm=540(10)^{-9}m$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the second-order diffraction angle is given when $n=2$ , hence equation (1) becomes:

$dsin\theta_{2}=2\lambda$ (2)

Now we have to find the value of $\theta_{2}$ :

$sin\theta_{2}=\frac{2\lambda}{d}$ (3)

Then:

$\theta_{2}=arcsin(\frac{2\lambda}{d})$ (4)

$\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})$ (5)

Finally:

$\theta_{2}=1.8\°$ (6)

5 0

2 years ago

A very large sheet of insulating material has had an excess of electrons placed on it to a surface charge density of –3.00nC/m2

lys-0071 [83]

Answer: sheet of charge

Explanation:

a )

Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.

b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.

C ) Electric field which is almost uniform near the sheet of charge is equal t the following

E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²

E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²

= .3389 x 10³

= 338.9 V / m

spacing between 1 V

= 1 / 338.9 m

= 2.95 X 10⁻3 m

= 2.95 mm.

3 0

2 years ago

What other places in the world use hydro kinetic energy to generate electrical power?

Elina [12.6K]

Answer:

Alaska: Hydrokinetic Energy Campbell CR9000X used for in-stream hydrokinetic device evaluation. Marine hydrokinetic energy power generation is an emerging sector in the renewable energy portfolio. Hydrokinetic devices convert the energy of waves, tidal currents, ocean currents or river currents into electrical power.

3 0

3 years ago