Given that a hot air balloon lifts 50 meters vertically into the air and then comes back down.
The displacement is the distance covered in a specific direction.
When the balloon is going up, the displacement is positive. and when the balloon is coming down, the displacement is negative.
The total displacement = 50 - 50 = 0
The distance is a measurement of length between to different points or position.
For distance, there is no need to consider direction. There is no consideration for positive or negative signs
While the distance = 50 + 50 = 100 meters
Therefore, the correct answer is C
That is, The displacement is zero and the distance is 100 meters
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Answer:
I think the acceleration is 12m/s
The option that takes place as water cycles from the bottom of the pot toward the top is that A. thermal energy is transferred.
As the pot gets warmer and warmer, the heat flows everywhere inside the pot, ultimately reaching the top, and heating the water at the top as well. There is no chemical energy here, and molecules don't gain thermal energy, it is just transferred to the top of the pot.
Answer:
Explanation:
Given
Diameter of Pulley=10.4 cm
mass of Pulley(m)=2.3 kg
mass of book![(m_0)=1.7 kg](https://tex.z-dn.net/?f=%28m_0%29%3D1.7%20kg)
height(h)=1 m
time taken=0.64 s
![h=ut+frac{at^2}{2}](https://tex.z-dn.net/?f=h%3Dut%2Bfrac%7Bat%5E2%7D%7B2%7D)
![1=0+\frac{a(0.64)^2}{2}](https://tex.z-dn.net/?f=1%3D0%2B%5Cfrac%7Ba%280.64%29%5E2%7D%7B2%7D)
![a=4.88 m/s^2and [tex]a=\alpha r](https://tex.z-dn.net/?f=a%3D4.88%20m%2Fs%5E2%3C%2Fp%3E%3Cp%3Eand%20%5Btex%5Da%3D%5Calpha%20r)
where
is angular acceleration of pulley
![4.88=\alpha \times 5.2\times 10^{-2}](https://tex.z-dn.net/?f=4.88%3D%5Calpha%20%5Ctimes%205.2%5Ctimes%2010%5E%7B-2%7D)
![\alpha =93.84 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D93.84%20rad%2Fs%5E2)
And Tension in Rope
![T=m(g-a)](https://tex.z-dn.net/?f=T%3Dm%28g-a%29)
![T=1.7\times (9.8-4.88)](https://tex.z-dn.net/?f=T%3D1.7%5Ctimes%20%289.8-4.88%29)
T=8.364 N
and Tension will provide Torque
![T\times r=I\cdot \alpha](https://tex.z-dn.net/?f=T%5Ctimes%20r%3DI%5Ccdot%20%5Calpha%20)
![8.364\times 5.2\times 10^{-2}=I\times 93.84](https://tex.z-dn.net/?f=8.364%5Ctimes%205.2%5Ctimes%2010%5E%7B-2%7D%3DI%5Ctimes%2093.84)
![I=0.463\times 10^{-2} kg-m^2](https://tex.z-dn.net/?f=I%3D0.463%5Ctimes%2010%5E%7B-2%7D%20kg-m%5E2)
![I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2](https://tex.z-dn.net/?f=I_%7Boriginal%7D%3D%5Cfrac%7Bmr%5E2%7D%7B2%7D%3D0.31%5Ctimes%2010%5E%7B-2%7Dkg-m%5E2)
Thus mass is uniformly distributed or some more towards periphery of Pulley
To find out scientific notation, you want to make sure that number is less than 10. So do 5.000000, you don't rally need the zeros but I just want to make my point. So use 10^x meaning ten the whatever power adds zeros like 5.000000x10^6 meaning it is increasing it by six zeros moving it out of the decimals and letting become 5,000,000.