Answer:
a)906.5 Nm^2/C
b) 0
c) 742.56132 N•m^2/C
Explanation:
a) The plane is parallel to the yz-plane.
We know that
flux ∅= EAcosθ
3.7×1000×0.350×0.700=906.5 N•m^2/C
(b) The plane is parallel to the xy-plane.
here theta = 90 degree
therefore,
0 N•m^2/C
(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.
therefore, applying the flux formula we get
3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C
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.
Answer: The trip takes 
Explanation:
Velocity 
is the variation of the position of a body (distance traveled 
) with time 
:
In this case, the car travels a distance 
at a velocity 
and we need to find the time it takes the trip.
Isolating :
Finally:
<span> We're given that x=25 when t=2: </span>
<span>25 = 3 + 12(2) + (1/2)a(2)^2 </span>
<span>Thus a = -1 cm/sec^2</span>
Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.
Mass of the gold (m) = 0.072 kg
Temperature change (ΔT) = 90 - 20 = 70 degree Celsius
Specific heat capacity of the gold (c) = 136 J/kg C
Heat supplied = m × c × ΔT
Heat supplied = 0.072 × 136 × 70
Heat supplied = 685.44 Joules
Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules
