Answer:
(a). The acceleration due to gravity at the surface of Titania is 0.37 m/s².
(b). The average density of Titania is 1656.47 kg/m³
Explanation:
Given that,
Radius of Titania
Mass of Titania
We need to calculate the acceleration due to gravity at the surface of Titania
Using formula of the acceleration due to gravity on earth
The acceleration due to gravity on Titania
Put the value into the formula
The acceleration due to gravity at the surface of Titania is 0.37 m/s².
(b). We assume Titania is a sphere
The average density of the earth is 5500 kg/m³.
We need to calculate the average density of Titania
Using formula of density
The average density of Titania is 1656.47 kg/m³
Hence, (a). The acceleration due to gravity at the surface of Titania is 0.37 m/s².
(b). The average density of Titania is 1656.47 kg/m³
Mujhe nhi pata sorry. Lekin mai bhut logo ko jaanti hoon jo iss savaal ka jawaab de sakte hai.
Answer:
4.74 x 10⁷ yr
Explanation:
m = Rate of conversion of mass to energy by sun = 4 x 10⁹ kg/s
t = total time taken to convert mass of earth to energy
M = mass of earth = 5.98 x 10²⁴ kg
Using the equation
M = m t
5.98 x 10²⁴ = (4 x 10⁹) t
t = 1.495 x 10¹⁵ sec
we know that, 1 yr = 3.154 x 10⁷ sec
t = (1.495 x 10¹⁵ sec) (3.154 x 10⁷ sec)⁻¹ yr
t = 4.74 x 10⁷ yr
Answer: F = 19.2 N
Explanation: Given that the initial Force = 3.6N
The formula involved is
F = GMm/r^2
Substitute the force F
3.6 = GMm/r^2
If one of the masses is tripled and the distance between the masses is quadrupled. We have
3.6 = (G × 3Mm)/(4r)^2
Where G will be constant.
3.6 = 3GMm/16r^2
Separate the fraction of number
3.6 = 3/16 × GMm/r^2
Make GMm/r^2 the subject of formula
(3.6 × 16)/3 = GMm/r^2
19.2 = GMm/r^2
Therefore, the new force of attraction is 19.2 N