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3 years ago

Dexter uses a compass to walk 140m due east. What is his distance and displacement?

1 answer:

3 0

The distance is 140 m and the displacement is 140 m E. Displacement can change though. For example, if Dexter walked 140 m east then walked 40 meters west the displacement would be 100 m E. Displacement is just how far away an object is from the start.

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Estimate the length of the neptunian year using the fact that the earth is 1.50Ã108km from the sun on average.

Ksenya-84 [330]

Answer:

Explanation:

We know that,

Neptune is 4.5×10^9 km from the sun

And given that,

Earth is 1.5×10^8km from sun

Then,

Let P be the orbital period and

Let a be the semi-major axis

Using Keplers third law

Then, the relation between the orbital period and the semi major axis is

P² ∝ a³

Then,

P² = ka³

P²/a³ = k

So,

P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³

Period of earth P(earth) =1year

Semi major axis of earth is

a(earth) = 1.5×10^8km

The semi major axis of Neptune is

a (Neptune) = 4.5×10^9km

So,

P(E)²/a(E)³ = P(N)² / a(N)³

1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³

Cross multiply

P(N)² = (4.5×10^9)³ / (1.5×10^8)³

P(N)² = 27000

P(N) =√27000

P(N) = 164.32years

The period of Neptune is 164.32years

6 0

4 years ago

A thin beam of light of wavelength 50 μm (in the infrared portion of the em spectrum) goes through

Brut [27]

Solution :

Given :

Wavelength of the thin beam of light, λ = 50 μm

Distance of the screen from the slit, D = 3.00 m

Width of the fringe, Δy = ±8.24 mm

Therefore, width of the slit is given by :

$d=\frac{n \lambda D}{\Delta y}$

$=\frac{2 \times 50 \times 10^{-9} \times 3}{2 \times 8.24 \times 10^{-3}}$

= 0.000018203 m

= 0.0182 mm

= 0.018 mm

The intensity of light is given by :

$I=I_0\left(\frac{\sin \beta /2}{\beta/ 2}\right)^2$ , where $\beta=\frac{2 \pi D \sin \theta}{\lambda}$

$I=I_0\left(\frac{\sin \frac{\pi d \sin \theta}{\lambda}}{\frac{\pi d \sin \theta}{\lambda}}\right)^2$

$I=I_0\left(\frac{\sin \frac{\pi d y}{D\lambda}}{\frac{\pi d y}{D\lambda}}\right)^2$

Now, $\frac{dy}{D \lambda} = \frac{8.24 \times 10^{-3}\times 0.018 \times 10^{-3}}{4 \times 50\times 10^{-9}\times 4}$

= 0.1854

≈ 0.18

$I=I_0\left(\frac{\sin 0.56}{0.56}\right)^2$

$=I_0 \times 0.81$

= 2 x0.81

$= 1.62 \ W/m^2$

6 0

3 years ago

Think about the questions below. You will be reporting the answers in your summary. Write your summary in the essay box below.

GrogVix [38]

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3 years ago

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What is the wavelength of a wave?

zubka84 [21]

Definition: Wavelength can be defined as the distance between two successive crests or troughs of a wave.

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3 years ago