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3 years ago

The period of a satellite in a circular orbit as the distance from the central object increases.

Physics

2 answers:

vovangra [49]3 years ago

5 0

Answer:

Increases

Explanation:

The period of a satellite in a circular orbit increases as the distance from the central object increases. It makes the most sense. Correct on Edge!

kow [346]3 years ago

4 0

Answer:

increases

Explanation:

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A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc

balandron [24]

Answer:

Approximately $28^{\circ}$ .

Explanation:

The refractive index of the air $n_{\text{air}}$ is approximately $1.00$ .

Let $n_\text{glass}$ denote the refractive index of the glass block, and let $\theta _{\text{glass}}$ denote the angle of refraction in the glass. Let $\theta_\text{air}$ denote the angle at which the light enters the glass block from the air.

By Snell's Law:

$n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}})$ .

Rearrange the Snell's Law equation to obtain:

$\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}$ .

Hence:

$\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}$ .

In other words, the angle of refraction in the glass would be approximately $28^{\circ}$ .

7 0

2 years ago

You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen

Nikolay [14]

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂ = d₁ / d₂

But

λ₁ < λ₂

d₁ < d₂

Therefore no of lines per unit length of grating A will be more because

no of lines per unit length ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

λ₁ D/ d₂

As λ₁ < λ₂

λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

5 0

3 years ago

A hydrocarbon compound z decolourises bromine liquid

Leno4ka [110]

Answer:

...

Explanation:

................................................

8 0

3 years ago

What change does this cause concerning weather?

Vika [28.1K]

Answer:

More extreme weather.

Explanation:

The Conveyor Belt of tides functions on a local and global level to spread out the cold and hot temperature differences on the planet. It is a delicate but important process that is easily disrupted, which causes it to slow down. And when it slows down, all those temperature differences will become more concentrated, causing colder places to be colder and hotter places to be hotter, ultimately leading to more extreme weather events as these cold and hot spots collide more violently than before.

Here's a picture I found on it:

6 0

3 years ago

Read 2 more answers

A resistor, inductor, and a battery are arranged in a circuit. The circuit has an inductance of L = 1 H and a resistance of 1.4

shtirl [24]

Answer:

$\tau \approx 7.14 \times 10^{-4}s \approx0.714ms$

Explanation:

In a LC circuit The time constant τ is the time necessary for 60% of the total current (maximum current), pass through the inductor after a direct voltage source has been connected to it. The time constant can be calculated as follows:

$\tau =\frac{L}{R}$

Therefore, the time needed for the current to reach a fraction f = 0.6(60%) of its maximum value is:

$\tau =\frac{1}{1.4\times 10^{3}} =7.142857143 \times 10^{-4} \approx7.14 \times 10^{-4}s$

8 0

4 years ago