Answer:
T = 92.8 min
Explanation:
Given:
The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:
Find:
- How long does the International Space Station take to orbit the earth? Give an exact answer.
Solution:
- Using the the expression given we can extract the angular speed of the International Space Station orbit:
- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8
- We know that the relation between angular speed w and time period T of an orbit is related by:
T = 2*p / w
T = 2*p / (2*p / 92.8)
Hence, T = 92.8 min
Answer:
Explanation:
Work done = ∫Fdx
= ∫(cx-3.00x²) dx
[ c x² / 2 - 3 x³ / 3 ]₀²
= change in kinetic energy
= 11-20
= - 9 J
[ c x² / 2 - x³ ]₀² = - 9
c x 2² / 2 - 2³ = -9
2c - 8 = -9
2c = -1
c = - 1/2
The correct answer would be the first option. The process that would need more energy would be vaporizing 1 kg of saturated liquid water at a pressure of 1 atmosphere. This can be seen from the latent heat of vaporization of each system. For the saturated water at 1 atm, the latent heat is equal to 40.7 kJ per mole while, at 8 atm, the latent heat is equal to 36.4 kJ per mole. The latent heat of vaporization is the amount of heat needed in order to vaporize a specific amount of substance without any change in the temperature. As we can observe, more energy is needed by the liquid water at 1 atm.
The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.
<h3>What is cutoff frequency?</h3>
The work function is related to the frequency as
W0 = h x fo
where, fo = cutoff frequency and h is the Planck's constant
Given is the work function for magnesium is 3.70 eV.
fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴
fo = 8.93 x 10¹⁴ Hz.
Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.
Learn more about cutoff frequency.
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PLS HELP ME AS QUICK AS POSSIBLE,
THANKS :)) I'm a bit confused
Can you answer 1 and 2, then confirm 3 :))))
