Respuesta :
Answer:
The time period on Titan is 2.67 hours.
Explanation:
Given that,
Acceleration = 1.37 m/s²
Time on earth = 1 hour
Period of oscillation on earth
[tex]T_{e}=2\pi\sqrt{\dfrac{L}{g_{e}}}[/tex]
Where, L = length of the pendulum
The length of pendulum will be the same on the Titan.
The period of oscillation on Titan
[tex]T_{t}=2\pi\sqrt{\dfrac{L}{g_{t}}}[/tex]
We need to calculate the time period on Titan
[tex]\dfrac{T_{t}}{T_{e}}=\dfrac{2\pi\sqrt{\dfrac{L}{g_{t}}}}{2\pi\sqrt{\dfrac{L}{g_{e}}}}[/tex]
[tex]\dfrac{T_{t}}{T_{e}}=\sqrt{\dfrac{g_{e}}{g_{t}}}[/tex]
[tex]T_{t}=T_{e}\sqrt{\dfrac{g_{e}}{g_{t}}}[/tex]
Put the value into the formula
[tex]T_{t}=1\times\sqrt{\dfrac{9.8}{1.37}}[/tex]
[tex]T_{t}=2.67\ hours[/tex]
Hence, The time period on Titan is 2.67 hours.
Answer:
[tex]T_t =2.67\ hours[/tex]
Explanation:
given,
acceleration due to gravity in titan = 1.37 m/s²
period of oscillation on earth
[tex]T_e = 2\pi\sqrt{\dfrac{L}{g_e}}[/tex]
L is length of the pendulum
Period of oscillation on Tital
[tex]T_t = 2\pi\sqrt{\dfrac{L}{g_t}}[/tex]
dividing the above equation
[tex]\dfrac{T_t}{T_e} = \sqrt{\dfrac{g_e}{g_m}}[/tex]
time taken by hour hand for one revolution is equal to 1 hr
[tex]T_t= T_e\sqrt{\dfrac{g_e}{g_m}}[/tex]
[tex]T_t= 1 \sqrt{\dfrac{9.8}{1.37}}[/tex]
[tex]T_t =2.67\ hours[/tex]
