Answer: Jupiter's mass
Explanation:
From Kepler's third law:
where T is the orbital period of a satellite, a is the average distance of the satellite from the Planet, M is the mass of the planet, G is the gravitational constant.
If the average distance of one of Jupiter's moons to Jupiter and its orbital period around Jupiter is given then mass of the Jupiter can be found:
We know that 1 minute= 60 seconds (or 1 min= 60 s).
10 min* (60 s/ 1 min)* (2.0 m/ 1 s)= 1,200 m.
(Note that the units cancel out so you get the answer)
The final answer is 1,200 m.
Hope this helps~
Answer:
ill get back to this question once i get the answer
Answer: 
Explanation:
Given
Volume of air 
Temperature of air 
Increase in temperature 
Specific heat for diatomic gas is 
Energy required to increase the temperature is
![\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}]](https://tex.z-dn.net/?f=%5CRightarrow%20Q%3DnC_pdT%5C%5C%5C%5C%5CRightarrow%20Q%3Dn%5Ctimes%20%5Cdfrac%7B7R%7D%7B2%7D%5Ctimes%20%5CDelta%20T%5C%5C%5C%5C%5CRightarrow%20Q%3D%5Cdfrac%7B7%7D%7B2%7DnR%5CDelta%20T%5C%5C%5C%5C%5CRightarrow%20Q%3D%5Cdfrac%7B7%7D%7B2%7D%5Ctimes%20%5Cdfrac%7BPV%7D%7BT%7D%5Ctimes%20%5CDelta%20T%5Cquad%20%5B%5Ctext%7Busing%20PV%3DnRT%7D%5D)
Insert the values
Question
A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g
Part (b) what is the angle of banking of the highway? Give your answer in degrees
Answer:
a. Equation of Tangent
tan(θ) = v²/rg
b. Angle of the banking highway
θ = 0.087°
Explanation:
Given
Radius of the curve = r = 330m
Acceleration of gravity = g = 9.8m/s²
Velocity = v = 8km/h = 8 * 1000/3600
v = 2.22 m/s
a . Write an equation for the tangent of the highway's angle of banking
The Angle is calculated by
tan(θ) = v²/rg
θ = tan-1(v²/rg)
b.
Part (b) what is the angle of banking of the highway? Give your answer in degrees
θ = tan-1(v²/rg)
Substituting the values of v,g and r
θ = tan-1(2.22²/(330 * 9.8)
θ = tan-1(0.001523933209647)
θ = 0.087314873580116°
θ = 0.087°
