Answer: Speeding up the orbital speed of earth so it escapes the sun require the greater energy.
Explanation: To find the answer, we need to know more about the Orbital and escape velocities.
<h3>
What is Orbital and Escape velocity?</h3>
- Orbital velocity can be defined as the minimum velocity required to put the satellite in its orbit around the earth.
- The expression for orbital velocity near to the surface of earth will be,
- Escape velocity can be defined as the minimum velocity with which a body must be projected from the surface of earth, so that it escapes from the gravitational field of earth.
- The expression for orbital velocity will be,
- If we want to get into the sun, we want to slow down almost completely, so that your speed relative to the sun became almost zero.
- We need about twice the raw speed to go to the sun than to leave the sun.
Thus, we can conclude that, the speeding up the orbital speed of earth so it escapes the sun require the greater energy.
Learn more about orbital and escape velocity here:
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Answer:
Explanation:
Given that,
An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.
We know that,
1 THz = 10¹² Hz
So,
f = 4.39 × 10¹² Hz
We need to find the wavelength of the infrared radiation.
We know that,
So, the wavelength of the infrared radiation is 
.
Answer:
The average speed of the blood in the capillaries is 0.047 cm/s.
Explanation:
Given;
radius of the aorta, r₁ = 1 cm
speed of blood, v₁ = 30 cm/s
Area of the aorta, A₁ = πr₁² = π(1)² = 3.142 cm²
Area of the capillaries, A₂ = 2000 cm²
let the average speed of the blood in the capillaries = v₂
Apply continuity equation to determine the average speed of the blood in the capillaries.
A₁v₁ = A₂v₂
v₂ = (A₁v₁) / (A₂)
v₂ = (3.142 x 30) / (2000)
v₂ = 0.047 cm/s
Therefore, the average speed of the blood in the capillaries is 0.047 cm/s.
The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.
