Without an atmosphere, the equatorial curve would show minimum daily values on the solstices in June when the sub-solar point is located at 23.5°N and in December when the sub-solar point is at 23.5°S latitude.
Explanation:
At the sub-solar point, the sun strikes directly at the surface with an angle of 90 degrees at a given point.
Solistice refers to that point in time when the sun’s zenith is located at the farthest point from the equator.
During summer solistice on June 21, the sun’s zenith reaches northernmost point, sub-solar point is fixed at 23.5°S Tropic of Cancer making the earth tilt 23.4 degrees
During winter soliscitse on December 21, the sub-solar point is fixed at) Tropic of Capricorn.
Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .
Given:
Amount of heat produced = 100 kcal per hour
Let's find the rate of energy production in joules.
We know that:
1 calorie = 4.184 Joules
1 kcal = 4.184 Joules
To find the rate of energy production in Joules, we have:
Therefore, the rate of energy production in joules is 418.4 kJ/h which is equivalent to 418400 Joules
ANSWER:
418.4 kJ/h
