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3 years ago

5

One way to search for planets around other stars is the doppler technique. another way uses transits of planets around other sta

rs. how does the transit technique work?

1 answer:

katovenus [111]3 years ago

3 0

The transit method requires watching the light output of a star over long periods of time. A transit occurs when the planet crosses in front of its star from earths point of view. Since there is a small object (the planet) now blocking some of the star, it appears to dim a little bit for a while until the planet passes. If we are in a position where that occurs regularly (most paths of planets do not happen to be on the line of sight between earth and their star) we can deduce the period of orbit. From the amount of dimming and the period you can estimate the mass

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Find the interest. A 105-day note for $14,200 at 8.5% interest.

mr Goodwill [35]

<h3><u>Answer</u>;</h3>

$347.22

<h3><u>Explanation</u>;</h3>

Principal = $14,200

Rate = 8.5%

Time = 105 days = 105/365

Interest = Principal x Rate x Time

Interest = 14,200 x 0.085 x 105/365

Interest = 347.219

= $347.22

7 0

3 years ago

Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take

KiRa [710]

Answer:

$\frac{V_{e}}{V_{h}}=0.428*10^{2}$

Explanation:

From conservation of energy states that

$K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}$

5 0

3 years ago

Light travels _____________ in a material with a higher index of refraction.

Aleks [24]

<span>Light travels Slower in a material with a higher index of refraction.
because </span><span>Water has a higher index of refraction than air and it travels slower (bends) in water
high index mean slow travel
hope it helps</span>

4 0

3 years ago

Read 2 more answers

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

8 0

3 years ago

The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on

Neporo4naja [7]

Answer:

162.8 K

Explanation:

initial current = io

final current, i = io/8

Let the potential difference is V.

coefficient of resistivity, α = 43 x 10^-3 /K

Let the resistance is R and the final resistance is Ro.

The resistance varies with temperature

R = Ro ( 1 + α ΔT)

V/i = V/io (1 + α ΔT )

8 = 1 + 43 x 10^-3 x ΔT

7 = 43 x 10^-3 x ΔT

ΔT = 162.8 K

Thus, the rise in temperature is 162.8 K.

5 0

3 years ago