Active Optics.
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Answer:
420J
Explanation:
Power is the time rate of change in energy. Power is the ratio of energy to time. The S.I unit of power is in watts.
Given that the flash lasts for 1/675 s, power output is 2.7 * 10⁵ W. Hence:
Power = Energy / time
Substituting:
2.7 * 10⁵ W = Energy / (1/675)
Energy = 2.7 * 10⁵ W * 1/675 = 400J
Therefore the energy emitted as light is 400J.
Since the conversion of electric energy to light is 95% efficient, hence the energy stored as electrical energy is:
Energy(capacitor) = 5% of 400J + 400J = 0.05*400 + 400
Energy(capacitor) = 420J
Once all hydrogen is depleted the star begins to die
<span>6.20 m/s^2
The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be
9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N
Add in the atmospheric drag and you get
4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N
Now subtract that total drag from the thrust available.
1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N
So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So
3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2
Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2</span>
The perimeter of ΔWXY is : ( D ) 14.5 cm
<u>Calculating the </u><u>perimeter </u><u>of ΔWXY</u>
QR = WY / 2
RS = XW / 2
QS = XY / 2
Given that : QR = 2.93 cm , RS = 2.04 cm, QS = 2.28 cm
Therefore
Perimeter of ΔWXY = ∑ WY + XW + XY
= 2SR + 2QS + 2QR
= 2(2.04) + 2(2.28) + 2(2.93)
= 14.5 cm
Hence we can conclude that the perimeter of ΔWXY = 14.5 cm
learn more about perimeter calculations : brainly.com/question/24744445
