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4 years ago

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What are the strengths and limitations of the doppler and transit methods? What kind of planets are easiest to detect with each

method?
help me please :( i will but brainliest

1 answer:

Arturiano [62]4 years ago

3 0

$\huge\mathfrak\red{✔Answer:-}$

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

Limitations: yield only planet's mass and orbital properties

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How do contact and non-contact forces affect the design of different modes of transportation?

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Contact and non contact forces are big factors to consider in designing of different modes of transportation because this factors are resistances for the mode of transportation. These contact and non- contact forces should be minimized in order for the energy requirement to be also minimum. But not the extent of risking the safety, for example a non contact force is the wieght, the should be optimum, safety of the design should not be compromised just to reduce weight, just like by removing essential parts, support just to remove weight is not good.

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4 years ago

As a rubber band and moves forward, which of the following is true

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3 years ago

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4 years ago

Read 2 more answers

A car is moving southwest at a velocity of 10 meters per second. Five seconds later, it's going 40 meters per second. What was t

Inga [223]

If "during this time" refers to the 5 second interval mentioned above, then the average acceleration is

$\bar a=\dfrac{v-v_0}{t-t_0}=\dfrac{40\,\frac{\mathrm m}{\mathrm s}-10\,\frac{\mathrm m}{\mathrm s}}{5\,\mathrm s-0\,\mathrm s}$

Notice that we took the start time to be the start of the 5 second interval and set that to $t_0=0$ . The starting velocity $v_0$ is the velocity measured at the start of the interval, and $v$ is the velocity measured at its end.

So the average velocity over these 5 seconds is

$\bar a=\dfrac{30\,\frac{\mathrm m}{\mathrm s}}{5\,\mathrm s}=6\,\dfrac{\mathrm m}{\mathrm s^2}$

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4 years ago

The intensity of sunlight reaching the earth is 1360 w/m2. Assuming all the sunligh is absorbed, what is the radiation pressure

krok68 [10]

(a) $1.15\cdot 10^9 N$

For an electromagnetic wave incident on a surface, the radiation pressure is given by (assuming all the radiation is absorbed)

$p=\frac{I}{c}$

where

I is the intensity

c is the speed of light

In this problem, $I=1360 W/m^2$ ; substituting this value, we find the radiation pressure:

$p=\frac{1360 W/m^2}{3\cdot 10^8 m/s}=4.5\cdot 10^{-6} Pa$

the force exerted on the Earth depends on the surface considered. Assuming that the sunlight hits half of the Earth's surface (the half illuminated by the Sun), we have to consider the area of a hemisphere, which is

$A=2pi R^2$

where

$R=6.37\cdot 10^6 m$

is the Earth's radius. Substituting,

$A=2\pi (6.37\cdot 10^6 m)^2=2.55\cdot 10^{14}m^2$

And so the force exerted by the sunlight is

$F=pA=(4.5\cdot 10^{-6} Pa)(2.55\cdot 10^{14} m^2)=1.15\cdot 10^9 N$

(b) $3.2\cdot 10^{-14}$

The gravitational force exerted by the Sun on the Earth is

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=1.99\cdot 10^{30}kg$ is the Sun's mass

$m=5.98\cdot 10^{24}kg$ is the Earth's mass

$r=1.49\cdot 10^{11} m$ is the distance between the Sun and the Earth

Substituting,

$F=(6.67\cdot 10^{-11} )\frac{(1.99\cdot 10^{30}kg)(5.98\cdot 10^{24} kg)}{(1.49\cdot 10^{11} m)^2}=3.58\cdot 10^{22}N$

And so, the radiation pressure force on Earth as a fraction of the sun's gravitational force on Earth is

$\frac{1.15\cdot 10^9 N}{3.58\cdot 10^{22}N}=3.2\cdot 10^{-14}$

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4 years ago