This is modelling.
Option D.
From the model, the conditions of the now are put into the model in order to predict the weather for a future date.
<span />
Answer:
1287.5 N
Explanation:
We are given that
Mass of vehicle ,m=2100 kg
Weight of vehicle ,F=20,600 N
Area,A=![0.56 cm^2=0.56\times 10^{-4} m^2](https://tex.z-dn.net/?f=0.56%20cm%5E2%3D0.56%5Ctimes%2010%5E%7B-4%7D%20m%5E2)
![1 cm^2=10^{-4} m^2](https://tex.z-dn.net/?f=1%20cm%5E2%3D10%5E%7B-4%7D%20m%5E2)
We have to find the minimum force that must be applied to the attached piston of area
in order to lift the vehicle.
![A'=0.035cm^2=0.035\times 10^{-4} m^2](https://tex.z-dn.net/?f=A%27%3D0.035cm%5E2%3D0.035%5Ctimes%2010%5E%7B-4%7D%20m%5E2)
Apply the pascal's law
![\frac{F}{A}=\frac{F'}{A'}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BA%7D%3D%5Cfrac%7BF%27%7D%7BA%27%7D)
![\frac{20600}{0.56\times 10^{-4}}=\frac{F'}{0.035\times 10^{-4}}](https://tex.z-dn.net/?f=%5Cfrac%7B20600%7D%7B0.56%5Ctimes%2010%5E%7B-4%7D%7D%3D%5Cfrac%7BF%27%7D%7B0.035%5Ctimes%2010%5E%7B-4%7D%7D)
![F'=\frac{20600}{0.56\times 10^{-4}}\times 0.035\times 10^{-4}=1287.5 N](https://tex.z-dn.net/?f=F%27%3D%5Cfrac%7B20600%7D%7B0.56%5Ctimes%2010%5E%7B-4%7D%7D%5Ctimes%200.035%5Ctimes%2010%5E%7B-4%7D%3D1287.5%20N)
Answer:
Part a)
When rotated about the mid point
![\tau = 0.021Nm](https://tex.z-dn.net/?f=%5Ctau%20%3D%200.021Nm)
Part b)
When rotated about its one end
![\tau = 0.042 Nm](https://tex.z-dn.net/?f=%5Ctau%20%3D%200.042%20Nm)
Explanation:
As we know that the angular acceleration of the rod is rate of change in angular speed
so we will have
![\alpha = \frac{\Delta \omega}{\Delta t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5CDelta%20%5Comega%7D%7B%5CDelta%20t%7D)
![\alpha = \frac{2.4 - 0}{3.6}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B2.4%20-%200%7D%7B3.6%7D)
![\alpha = 0.67 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%200.67%20rad%2Fs%5E2)
Part a)
When rotated about the mid point
![I = 2mr^2](https://tex.z-dn.net/?f=I%20%3D%202mr%5E2)
![I = 2(0.48)(0.18)^2](https://tex.z-dn.net/?f=I%20%3D%202%280.48%29%280.18%29%5E2)
![I = 0.0311 kg m^2](https://tex.z-dn.net/?f=I%20%3D%200.0311%20kg%20m%5E2)
now torque is given as
![\tau = 0.0311 (0.67)](https://tex.z-dn.net/?f=%5Ctau%20%3D%200.0311%20%280.67%29)
![\tau = 0.021Nm](https://tex.z-dn.net/?f=%5Ctau%20%3D%200.021Nm)
Part b)
When rotated about its one end
![I = m(2r)^2](https://tex.z-dn.net/?f=I%20%3D%20m%282r%29%5E2)
![I = (0.48)(0.36)^2](https://tex.z-dn.net/?f=I%20%3D%20%280.48%29%280.36%29%5E2)
![I = 0.0622 kg m^2](https://tex.z-dn.net/?f=I%20%3D%200.0622%20kg%20m%5E2)
now torque is given as
![\tau = (0.0622)(0.67)](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%280.0622%29%280.67%29)
![\tau = 0.042 Nm](https://tex.z-dn.net/?f=%5Ctau%20%3D%200.042%20Nm)
A. an accelerating charged charged particle or changing magnetic fields
Answer:
A closed system.
Explanation:
The three major types of system are: open, closed and isolated. Open system interacts with its surroundings with respect to its particles and energy. A closed system interacts with its surroundings with respect to energy but not its particles. While an isolated system does not interact with its surroundings in any way.
Therefore, after the jar is sealed, it is an example of a closed system. This is because the emitted gas could not escape into the surroundings, but thermal energy was emitted into its surroundings after the chemical reaction has taken place.