Answer:
The gravitational acceleration on the surface of Planet-X is 1.5 [tex]ms^{-2}[/tex]
Explanation:
Firstly, we are to list out the parameters given:
Mass (m) = 35.5 g, Length of string (L) = 133 cm = 1.33 m, t = 70.7 seconds for 12 oscillations, T = 70.7 ÷ 12 = 5.89 seconds
The formula for calculating the period of a simple pendulum (assuming the angle of deflection is lesser than 15º) is given by:
[tex]T=2\pi\sqrt{\frac{L}{g}}\\[/tex]
We want to calculate for the gravitational acceleration (g), hence, we have to make g the subject of the formula
[tex]g=4\pi^{2}\frac{L}{T^{2}}\\[/tex]
Substitute the parameters into the equation, we have:
g = [tex]4\pi^{2}[/tex] * 1.33 ÷ [tex]5.89^{2}[/tex] = 1.51
g = 1.5 [tex]ms^{-2}[/tex]
The gravitational acceleration on the surface of Planet-X is 1.5 [tex]ms^{-2}[/tex]
Answer:
the gravitational acceleration of the Xplanet is 1.344m/s^2
Explanation:
You can use the formula for the calculation of the frequency of a pendulum, in order to find an expression for the gravitational constant:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{l}}\\\\g=4\pi^2 f^2l[/tex]
Where you can notice that mass ob the object does not influence of the gravitatiolan acceleration. By the information of the question, you have the values of f and l. By replacing these values (with units of meter and seconds) you obtain:
[tex]f=\frac{12}{70.7}=0.16s^{-1}\\\\g=4\pi^2(0.16s^{-1})^2(1.33m)=1.344\frac{m}{s^2}[/tex]
Hence, the gravitational acceleration of the Xplante is 1.344m/s^2
