Answer:4
Explanation:
Given
![m_1=21\ gm](https://tex.z-dn.net/?f=m_1%3D21%5C%20gm)
![m_2=27\ gm](https://tex.z-dn.net/?f=m_2%3D27%5C%20gm)
Mass of stick is ![m=108\ gm](https://tex.z-dn.net/?f=m%3D108%5C%20gm)
Let T be the tension in the red string
Now if the red string is cut , suppose T is the tension in the blue rod immediately after cut
Therefore
![T=m_1g+m_2g](https://tex.z-dn.net/?f=T%3Dm_1g%2Bm_2g)
![T=(0.021+0.027)\times 10](https://tex.z-dn.net/?f=T%3D%280.021%2B0.027%29%5Ctimes%2010)
![T=0.48\ N](https://tex.z-dn.net/?f=T%3D0.48%5C%20N)
Answer:
b
Explanation:
she made a conclusion based off of her own observation, and isnt dumb enough to call a weather station
Answer:
i can't sorry
Explanation:
I didn't really pay attention in that class
The period of a simple pendulum is given by:
![T=2 \pi \sqrt{ \frac{L}{g} }](https://tex.z-dn.net/?f=T%3D2%20%5Cpi%20%20%5Csqrt%7B%20%5Cfrac%7BL%7D%7Bg%7D%20%7D%20)
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
![g= \frac{4 \pi^2}{T^2}L](https://tex.z-dn.net/?f=g%3D%20%5Cfrac%7B4%20%5Cpi%5E2%7D%7BT%5E2%7DL%20)
(1)
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
![f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7BN%7D%7Bt%7D%3D%20%5Cfrac%7B441%7D%7B1090%20s%7D%3D0.40%20Hz%20%20)
And its period is the reciprocal of its frequency:
![T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s](https://tex.z-dn.net/?f=T%3D%20%5Cfrac%7B1%7D%7Bf%7D%3D%20%5Cfrac%7B1%7D%7B0.40%20Hz%7D%3D2.47%20s%20%20)
So now we can use eq.(1) to find the gravitational acceleration of the planet:
Answer:
Explanation:I don't say you must have to mark my ans as brainliest but if you think it has really helped u plz don't forget to thank me....