In the given question, one important information for getting to the actual solution is not given and that is the atmospheric pressure. To find the approximate absolute pressure, it is needed to add the value of atmospheric pressure with the gage pressure.
Atmospheric pressure = 100 kPa
Then
Absolute pressure = 156 + 100 kPa
= 256 KPa.
Answer:
By nature, laws of Physics are stated facts which have been deduced and derived based on empirical observations. Simply put, the world around us works in a certain way, and physical laws are a way of classifying that “working.”
Answer:
![\rho = 12580.7 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%2012580.7%20kg%2Fm%5E3)
Explanation:
As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet
So here we will have
![F = \frac{GMm}{(r + h)^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7BGMm%7D%7B%28r%20%2B%20h%29%5E2%7D)
here we have
![F =\frac {mv^2}{(r+ h)}](https://tex.z-dn.net/?f=F%20%3D%5Cfrac%20%7Bmv%5E2%7D%7B%28r%2B%20h%29%7D)
![\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E2%7D%7Br%20%2B%20h%7D%20%3D%20%5Cfrac%7BGMm%7D%7B%28r%20%2B%20h%29%5E2%7D)
here we have
![v = \sqrt{\frac{GM}{(r + h)}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7BGM%7D%7B%28r%20%2B%20h%29%7D%7D)
now we can find time period as
![T = \frac{2\pi (r + h)}{v}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%20%28r%20%2B%20h%29%7D%7Bv%7D)
![T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%20%282.05%20%5Ctimes%2010%5E6%20%2B%20310%20%5Ctimes%2010%5E3%29%7D%7B%5Csqrt%7B%5Cfrac%7BGM%7D%7B%28r%20%2B%20h%29%7D%7D%7D)
![1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}](https://tex.z-dn.net/?f=1.15%20%5Ctimes%203600%20%3D%20%5Cfrac%7B2%5Cpi%20%282.05%20%5Ctimes%2010%5E6%20%2B%20310%20%5Ctimes%2010%5E3%29%7D%7B%5Csqrt%7B%5Cfrac%7B%286.67%20%5Ctimes%2010%5E%7B-11%7D%29%28M%29%7D%7B%282.05%20%5Ctimes%2010%5E6%20%2B%20310%20%5Ctimes%2010%5E3%29%7D%7D%7D)
![M = 4.54 \times 10^{23} kg](https://tex.z-dn.net/?f=M%20%3D%204.54%20%5Ctimes%2010%5E%7B23%7D%20kg)
Now the density is given as
![\rho = \frac{M}{\frac{4}{3}\pi r^3}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7BM%7D%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3%7D)
![\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7B4.54%20%5Ctimes%2010%5E%7B23%7D%7D%7B%5Cfrac%7B4%7D%5B3%7D%5Cpi%282.05%20%5Ctimes%2010%5E6%29%5E3%7D)
![\rho = 12580.7 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%2012580.7%20kg%2Fm%5E3)
Answers:
a) ![0.5 m/s^{2}](https://tex.z-dn.net/?f=0.5%20m%2Fs%5E%7B2%7D)
b) ![1.5 N](https://tex.z-dn.net/?f=1.5%20N)
Explanation:
a) The centripetal acceleration
of an object moving in a uniform circular motion is given by the following equation:
Where:
is the angular velocity of the ball
is the radius of the circular motion, which is equal to the length of the string
Then:
This is the centripetal acceleration of the ball
b) On the other hand, in this circular motion there is a force (centripetal force
) that is directed towards the center and is equal to the tension (
) in the string:
![F=T=m. a_{c}](https://tex.z-dn.net/?f=F%3DT%3Dm.%20a_%7Bc%7D)
Where
is the mass of the ball
Hence:
![T=(3 kg)(0.5 m/s^{2})](https://tex.z-dn.net/?f=T%3D%283%20kg%29%280.5%20m%2Fs%5E%7B2%7D%29)
This is the tension in the string
Answer:
Friction is a force that holds back the movement of a sliding object.
Explanation:
The two types of friction: Static friction and Kinetic friction. Static friction operates between two surfaces that aren't moving relative to each other, while kinetic friction acts between objects in motion.