Answer:
The tension in the rope is 41.38 N.
Explanation:
Given that,
Mass of bucket of water = 14.0 kg
Diameter of cylinder = 0.260 m
Mass of cylinder = 12.1 kg
Distance = 10.7 m
Suppose we need to find that,
What is the tension in the rope while the bucket is falling
We need to calculate the acceleration
Using relation of torque
![\tau=F\times r](https://tex.z-dn.net/?f=%5Ctau%3DF%5Ctimes%20r)
![I\times\alpha=F\times r](https://tex.z-dn.net/?f=I%5Ctimes%5Calpha%3DF%5Ctimes%20r)
Where, I = moment of inertia
= angular acceleration
![\dfrac{Mr^2}{2}\times\dfrac{a}{r}=F\times r](https://tex.z-dn.net/?f=%5Cdfrac%7BMr%5E2%7D%7B2%7D%5Ctimes%5Cdfrac%7Ba%7D%7Br%7D%3DF%5Ctimes%20r)
...(I)
Here, F = tension
The force is
...(II)
Where, F = tension
a = acceleration
From equation (I) and (II)
![\dfrac{M}{2}a=m(g-a)](https://tex.z-dn.net/?f=%5Cdfrac%7BM%7D%7B2%7Da%3Dm%28g-a%29)
![a=\dfrac{g}{1+\dfrac{M}{2m}}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bg%7D%7B1%2B%5Cdfrac%7BM%7D%7B2m%7D%7D)
Put the value into the formula
![a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B9.8%7D%7B1%2B%5Cdfrac%7B12.1%7D%7B2%5Ctimes14.0%7D%7D)
![a=6.84\ m/s^2](https://tex.z-dn.net/?f=a%3D6.84%5C%20m%2Fs%5E2)
We need to calculate the tension in the rope
Using equation (I)
![F=\dfrac{M}{2}a](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7BM%7D%7B2%7Da)
Put the value into the formula
![F=\dfrac{12.1}{2}\times6.84](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B12.1%7D%7B2%7D%5Ctimes6.84)
![F=41.38\ N](https://tex.z-dn.net/?f=F%3D41.38%5C%20N)
Hence, The tension in the rope is 41.38 N.
It is a conglomeration of rock that collected together over time, and it rotating in a spinning ball that is orbiting a sun that is in the middle of our solar system.
You should look at a picture of the earth from a satellite.<span />
Answer:
8.87 m/s^2
Is the same for both planets
Explanation:
Hello!
The surface gravity can be calculated from Newton's Law of Gravitation and Newton's Second Law :
ma = F =G Mm/r^2
Solving for a:
a = G M/r^2
And the surface graity g = a(R), that is, the surface gravity is equal to the acceleration evaluated at the radius of the planet:
g = G M/R^2
Since G is a constant, we need to evaluate M/R^2 for both to know in which planet the surface gravity is the geratest:
M_u/R_u^2 = 1.323 x 10^11 kg/m^2
M_v/R_v^2 = 1.323 x 10^11 kg/m^2
It turns out that the surface gravity in both planets is the same! which is:
g = G M_u/R_u^2
= ( 6.67408 × 10-11 m^3 / (kg s^2) ) *( 1.323 x 10^11 kg/m^2)
= 8.87 m/s^2
*as you can check on google*
You would feel the same weigth in both planets, however you wil feel lighter in these planets than in earth.
Usain Bolt I think, sorry if it’s not right