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Mandarinka [93]

3 years ago

6

Give an example of a situation where you could use a cost/ benefit analysis. Be sure to explain at least one cost and one benefi

t for a given choice.

1 answer:

evablogger [386]3 years ago

5 0

Answer: An example is where my uncle built a house with around 2 million and he sold it for 3,260,023 million

Explanation: Don't really think that help, but if it did then you're welcome.

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A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r

kaheart [24]

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

$\Phi = \frac{2\pi \delta}{\lambda}$

Where

$\delta =$ Horizontal distance between two points

$\lambda =$ Wavelength

From our values we have,

$\lambda = 500nm = 5*10^{-6}m$

$\theta = 1\°$

The horizontal distance between this two points would be given for

$\delta = dsin\theta$

Therefore using the equation we have

$\Phi = \frac{2\pi \delta}{\lambda}$

$\Phi = \frac{2\pi(dsin\theta)}{\lambda}$

$\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}$

$\Phi= 1.096 rad \approx = 1.1 rad$

Therefore the correct answer is C.

6 0

3 years ago

When can a theory be modified if a new type of technology allows for new observations that raise new questions?

olga nikolaevna [1]

Answer:

id say B, but not sure

Explanation:

4 0

3 years ago

Read 2 more answers

When a wave passes from one medium to another, its _________ remains constant.

Mashcka [7]

The frequency will not change

3 0

3 years ago

ale4655 [162]

m = 43.2 kg

Explanation:

volume of sphere = (4/3)pi(r)^3

= (4/3)(3.14)(2 m)^3

= 33.5 m^3

density = mass/volume

or solving for mass m,

m = (density)×(volume)

= (1.29 kg/m^3)(33.5 m^3)

= 43.2 kg

3 0

3 years ago

At what height h above the ground does the projectile have a speed of 0.5v?

maw [93]

Answer:

$h=\dfrac{3v^2}{8g}$

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

$v=u+at$

$v=u-gt$

Initial speed of the projectile is v and final speed is 0.5 v.

$0.5v=v-gt$

$t=\dfrac{v}{2g}$

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

$h=vt-\dfrac{1}{2}gt^2$

$h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2$

$h=\dfrac{3v^2}{8g}$

So, the height of the projectile above the ground is $\dfrac{3v^2}{8g}$ . Hence, this is the required solution.

6 0

3 years ago