I need help with this answer I believe it's a democracy

Suppose you monitor a large number (many thousands) of stars over a period of 3 years, searching for planets through the transit

nasty-shy [4]

Answer:

2. You must be able to precisely measure variations in the star's brightness with time.

5. As seen from Earth, the planet's orbit must be seen nearly edge–on (in the plane of our line-of-sight).

6. You must repeatedly obtain spectra of the star that the planet orbits.

Explanation:

The transit method is a very important and effective tool for discovering new exoplanets (the planets orbiting other stars out of the solar system). In this method the stars are observed for a long duration. When the exoplanet will cross in front of theses stars as seen from Earth, the brightness of the star will dip. To observe this dip following conditions must be met:

1. The orbit of the planet should be co-planar with the plane of our line of sight. Then only its transition can be observed.

2. The brightness of the star must be observed precisely as the period of transit can be less than a second as seen from Earth. Also the dip in brightness depends on the size of the planet. If the planet is not that big the intensity dip will be very less.

3. The spectrum of the star needs to be studied and observe during the transit and normally to find out the details about the planets.

4. Also, the orbital period should be less than the period of observation for the transit to occur at least once.

4 0

3 years ago

A driver in a car traveling at a speed of 21.8m/s sees a cat 101 m away on the road. How long will it take for the car to accele

kobusy [5.1K]

The acceleration of the car will be needed in order to calculate the time. It is important to consider that the final speed is equal to zero:

$v^2 = v_0^2 + 2ad\ \to\ a = \frac{-v_0^2}{2d} = -\frac{21.8^2\ m^2/s^2}{2\cdot 99\ m} = -2.4\frac{m}{s^2}$

We can clear time in the speed equation:

$v = v_0 + at\ \to\ t = \frac{-v_0}{a} = \frac{-21.8\ m/s}{-2.4\ m/s^2} = \bf 9.08\ s$

If you find some mistake in my English, please tell me know.

3 0

3 years ago

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0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a

klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

$(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}$

$(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }$

where $h_{fg$ is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

$V_{fg$ is specific volume ( 1.5 ft³ )

we substitute

$(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5})$ / $( (15+460)\frac{1.5}{0.5})$

$(\frac{dP}{dT} )_{sat } =$ 272.98 Ibf-ft²/R

Now,

$(\frac{dP}{dT} )_{sat } =$ $(\frac{P_2 - P_1}{T_2 - T_1})_{sat$

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

$T_2$ $= ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}$

$T_2 = 475 + 5.2751\\$

$T_2 =$ 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

3 0

3 years ago

758 j of heat are added to 0.750 kg of copper. how much does its temperature change?(unit=degrees c) PLEASE HELPPPPPPP MEEEEEE

DiKsa [7]

Answer:

$2.6^{\circ}C$

Explanation:

When a substance is supplied with a certain amount of heat energy, the temperature of the substance increases according to the equation

$Q=mC\Delta T$

where

m is the mass of the substance

Q is the amount of energy supplied

C is the specific heat of the substance

$\Delta T$ is the temperature change

In this problem:

Q = 758 J is the energy supplied

m = 0.750 kg is the mass of the sample

$C=385 J/kg^{\circ}C$ is the specific heat of copper

Re-arranging the equation, we can find the increase in temperature:

$\Delta T=\frac{Q}{mC}=\frac{758}{(0.750)(385)}=2.6^{\circ}C$

5 0

3 years ago

an automobile is traveling 65km/h the brakes decelerate it at a rate of -6.0 m/s^2 how long will it take to stop the car?

joja [24]

Explanation:

It is given that,

Initial speed of the automobile, u = 65 km/hr =

Final speed of the automobile, v = 0

Deceleration of the automobile, $a=-6\ m/s^2$

We need to find the distance covered by the car as it comes to rest. It can be calculated using third equation of motion as :

$v^2-u^2=2ax$

$a=\dfrac{v^2-u^2}{2x}$

$a=\dfrac{0-(18.05)^2}{2\times (-6)}$

$a=27.15\ m/s^2$

So, the acceleration of the car is $27.15\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

I need help with this answer I believe it's a democracy​

I need help with this answer I believe it's a democracy