<u>Answer:</u>
The correct answer option is: they are equal and opposite.
<u>Explanation:</u>
A force is a push or a pull which acts upon any object as a result of its interaction with another object where the forces are called action and reaction.
And according to the Newton's third law: for every action, there is an equal and opposite reaction.
For example, if a bat hits a ball with a 1000-N force, the force exerted on the bat will also be equal to 1000 N but in opposite direction.
Answer:
![N = 1.2 \times 10^{23}](https://tex.z-dn.net/?f=N%20%3D%201.2%20%5Ctimes%2010%5E%7B23%7D)
Explanation:
As we know that the current through the battery is given as
![i = (0.88 A) e^{-t/6}](https://tex.z-dn.net/?f=i%20%3D%20%280.88%20A%29%20e%5E%7B-t%2F6%7D)
here from above equation we know that current will become zero when time elapsed is very large
so here we can say that charge will flow through the battery from t = 0 to t = infinite
now we have
![q = \int i dt](https://tex.z-dn.net/?f=q%20%3D%20%5Cint%20i%20dt)
![q = \int (0.88 A) e^{-t/6} dt](https://tex.z-dn.net/?f=q%20%3D%20%5Cint%20%280.88%20A%29%20e%5E%7B-t%2F6%7D%20dt)
![q = 0.88\times -6(3600)\times (e^{-t/6})](https://tex.z-dn.net/?f=q%20%3D%200.88%5Ctimes%20-6%283600%29%5Ctimes%20%28e%5E%7B-t%2F6%7D%29)
![q = -1.90 \times 10^{4} (0 - 1)](https://tex.z-dn.net/?f=q%20%3D%20-1.90%20%5Ctimes%2010%5E%7B4%7D%20%280%20-%201%29)
![q = 1.90 \times 10^{4} C](https://tex.z-dn.net/?f=q%20%3D%201.90%20%5Ctimes%2010%5E%7B4%7D%20C)
As we know that
![q = Ne](https://tex.z-dn.net/?f=q%20%3D%20Ne)
![N = \frac{q}{e}](https://tex.z-dn.net/?f=N%20%3D%20%5Cfrac%7Bq%7D%7Be%7D)
![N = \frac{1.90 \times 10^4}{1.6 \times 10^{-19}}](https://tex.z-dn.net/?f=N%20%3D%20%5Cfrac%7B1.90%20%5Ctimes%2010%5E4%7D%7B1.6%20%5Ctimes%2010%5E%7B-19%7D%7D)
![N = 1.2 \times 10^{23}](https://tex.z-dn.net/?f=N%20%3D%201.2%20%5Ctimes%2010%5E%7B23%7D)
Answer:
a
The magnetic field strength is ![B = 2 \mu T](https://tex.z-dn.net/?f=B%20%3D%202%20%5Cmu%20T)
Explanation:
From the question we are told that
The length line above the ground is ![R = 20m](https://tex.z-dn.net/?f=R%20%3D%2020m)
The current of the line is ![I = 200A](https://tex.z-dn.net/?f=I%20%20%3D%20200A)
The voltage of the line is ![V = 110kV](https://tex.z-dn.net/?f=V%20%3D%20110kV)
Generally magnetic field strength is mathematically represented as
![B = \frac{\mu_o I}{2 \pi R}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu_o%20I%7D%7B2%20%5Cpi%20R%7D)
Where
is the permeability of free space ![= 4\pi * 10^{-7} N/A^2](https://tex.z-dn.net/?f=%3D%204%5Cpi%20%2A%2010%5E%7B-7%7D%20N%2FA%5E2)
![B = \frac{(4\pi * 10^{-7} N/A^2) *200}{2 \pi *20}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%284%5Cpi%20%2A%2010%5E%7B-7%7D%20N%2FA%5E2%29%20%2A200%7D%7B2%20%5Cpi%20%2A20%7D)
![= (2.0*10^{-7})[\frac{200}{20} ]](https://tex.z-dn.net/?f=%3D%20%282.0%2A10%5E%7B-7%7D%29%5B%5Cfrac%7B200%7D%7B20%7D%20%5D)
![= 2*10^{-6}T](https://tex.z-dn.net/?f=%3D%202%2A10%5E%7B-6%7DT)
![= 2 \mu T](https://tex.z-dn.net/?f=%3D%202%20%5Cmu%20T)
Earths magnetic field is approximately given as ![50 \mu T](https://tex.z-dn.net/?f=50%20%5Cmu%20T)
So the percentage would be
![= \frac{Magnetic \ Field \ Intensity \ Of \ Line}{Earth's \ Magnetic \ Field} * 100](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7BMagnetic%20%5C%20Field%20%5C%20Intensity%20%5C%20Of%20%20%5C%20Line%7D%7BEarth%27s%20%5C%20Magnetic%20%5C%20Field%7D%20%2A%20100)
![= \frac{2 \mu T }{50 \mu T } * 100](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B2%20%5Cmu%20T%20%7D%7B50%20%5Cmu%20T%20%7D%20%2A%20100)
%
Ok, assuming "mj" in the question is Megajoules MJ) you need a total amount of rotational kinetic energy in the fly wheel at the beginning of the trip that equals
(2.4e6 J/km)x(300 km)=7.2e8 J
The expression for rotational kinetic energy is
E = (1/2)Iω²
where I is the moment of inertia of the fly wheel and ω is the angular velocity.
So this comes down to finding the value of I that gives the required energy. We know the mass is 101kg. The formula for a solid cylinder's moment of inertia is
I = (1/2)mR²
We want (1/2)Iω² = 7.2e8 J and we know ω is limited to 470 revs/sec. However, ω must be in radians per second so multiply it by 2π to get
ω = 2953.1 rad/s
Now let's use this to solve the energy equation, E = (1/2)Iω², for I:
I = 2(7.2e8 J)/(2953.1 rad/s)² = 165.12 kg·m²
Now find the radius R,
165.12 kg·m² = (1/2)(101)R²,
√(2·165/101) = 1.807m
R = 1.807m
Force is (mass × acceleration) measured in Newton
Pressure is the 'force' per unit area measured in Newton/m^2 (pascal)