Respuesta :
Answer:
Explanation:
Given
mass of rock [tex]m=15\ kg[/tex]
time taken by rock [tex]t=1.75\ s[/tex]
Rock is dropped from the same height suppose h on Enceladus surface
time taken on Enceladus surface [tex]t_2=18.6[/tex]
Using Equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
where
y=displacement
u=initial velocity
a=acceleration
t=time
For Earth surface
here initial velocity u=0
[tex]y=0+\frac{1}{2}a_{earth}t^2[/tex]
[tex]y=0.5a_{earth}(1.75)^2 -----1[/tex]
For Enceladus surface
[tex]y=0+0.5a_{Enceladus}(18.6)^2----2[/tex]
Divide 1 and 2 we get
[tex]a_{Enceladus}(18.6)^2=a_{earth}(1.75)^2[/tex]
[tex]a_{Enceladus}=\frac{1.75^2}{18.6^2}\times g[/tex]
[tex]a_{Enceladus}=0.008852\times 9.8[/tex]
[tex]a_{Enceladus}=0.086\ m/s^2[/tex]
The acceleration due to gravity on Enceladus is 0.086 m/s²
Gravitational Acceleration:
The acceleration generated on an object due to the force of gravity is called gravitational acceleration. For different celestial bodies, the value of gravitation acceleration is different. It also varies on earth with the variation of height.
For the case of the earth:
The height from which the rock is dropped is given by:
[tex]h=ut+\frac{1}{2} gt^2[/tex]
here, u is the initial speed = 0
[tex]h=\frac{1}{2}gt^2\\\\h=0.5\times9.8\times(1.75)^2\\\\h=15m[/tex]
Let the acceleration due to gravity on Saturn's satellite be g', so if we apply the same equation with g':
[tex]h=\frac{1}{2}g't'^2\;\;(t'=time\;taken \;on\;Enceladus)\\\\ \frac{15\times2}{(18.6)^2}=g'\\\\g'=0.086\;m/s^2[/tex]
Learn more about gravitational acceleration:
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