Answer:
Given the area A of a flat surface and the magnetic flux through the surface
it is possible to calculate the magnitude
.
Explanation:
The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux
is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (
). So 1 Wb=1 T.m².
For a flat surface S of area A in a uniform magnetic field B, with
being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:
![\Phi=B\ A\ cos\theta](https://tex.z-dn.net/?f=%5CPhi%3DB%5C%20A%5C%20cos%5Ctheta)
We are told the values of
and B, then we can calculate the magnitude
![\frac{\Phi}{A}=B\ cos\theta](https://tex.z-dn.net/?f=%5Cfrac%7B%5CPhi%7D%7BA%7D%3DB%5C%20cos%5Ctheta)
Answer:
28716.4740661 N
1.2131147541 m/s
51.2474965841%
Explanation:
m = Mass of plane = 74000 kg
s = Displacement = 3.7 m
f = Frictional force = 14000 N
t = Time taken = 6.1 s
u = Initial velocity = 0
v = Final velocity
![s=ut+\frac{1}{2}at^2\\\Rightarrow 3.7=0\times 6.1+\frac{1}{2}\times a\times 6.1^2\\\Rightarrow a=\frac{3.7\times 2}{6.1^2}\\\Rightarrow a=0.198871271164\ m/s^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%203.7%3D0%5Ctimes%206.1%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20a%5Ctimes%206.1%5E2%5C%5C%5CRightarrow%20a%3D%5Cfrac%7B3.7%5Ctimes%202%7D%7B6.1%5E2%7D%5C%5C%5CRightarrow%20a%3D0.198871271164%5C%20m%2Fs%5E2)
Force is given by
![F=ma+f\\\Rightarrow F=74000\times 0.198871271164+14000\\\Rightarrow F=28716.4740661\ N](https://tex.z-dn.net/?f=F%3Dma%2Bf%5C%5C%5CRightarrow%20F%3D74000%5Ctimes%200.198871271164%2B14000%5C%5C%5CRightarrow%20F%3D28716.4740661%5C%20N)
The force with which the team pulls the plane is 28716.4740661 N
![v=u+at\\\Rightarrow v=0+0.198871271164\times 6.1\\\Rightarrow v=1.2131147541\ m/s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20v%3D0%2B0.198871271164%5Ctimes%206.1%5C%5C%5CRightarrow%20v%3D1.2131147541%5C%20m%2Fs)
The speed of the plane is 1.2131147541 m/s
Kinetic energy is given by
![K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}\times 74000\times 1.2131147541^2\\\Rightarrow K=54450.9540448\ J](https://tex.z-dn.net/?f=K%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C%5CRightarrow%20K%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%2074000%5Ctimes%201.2131147541%5E2%5C%5C%5CRightarrow%20K%3D54450.9540448%5C%20J)
Work done is given by
![W=Fs\\\Rightarrow W=28716.4740661\times 3.7\\\Rightarrow W=106250.954045\ J](https://tex.z-dn.net/?f=W%3DFs%5C%5C%5CRightarrow%20W%3D28716.4740661%5Ctimes%203.7%5C%5C%5CRightarrow%20W%3D106250.954045%5C%20J)
The fraction is given by
![\dfrac{54450.9540448}{106250.954045}=0.512474965841](https://tex.z-dn.net/?f=%5Cdfrac%7B54450.9540448%7D%7B106250.954045%7D%3D0.512474965841)
The teams 51.2474965841% of the work goes to kinetic energy of the plane.
In physics, "work<span>" is when a force applied to an object moves the object in the same direction as the force. If someone pushes against a wall, no </span>work<span> is done on the system. It is calculated as follows:
Work = Force x distance
Work = 25 N x 4 meters
Work = 100 N.m</span>
Answer:
Explanation:
Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.
Answer:
mass of box 1 = 2.20 kg
mass of box 2 = 5.93 kg
Explanation:
Let the mass of box 1 and box 2 is respectively
and ![m_2](https://tex.z-dn.net/?f=m_2)
so we will have
Force applied on box 1 then acceleration
![a = \frac{F}{m_1 + m_2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BF%7D%7Bm_1%20%2B%20m_2%7D)
![1.50 = \frac{12.2}{m_1 + m_2}](https://tex.z-dn.net/?f=1.50%20%3D%20%5Cfrac%7B12.2%7D%7Bm_1%20%2B%20m_2%7D)
![m_1 + m_2 = 8.13](https://tex.z-dn.net/?f=m_1%20%2B%20m_2%20%3D%208.13)
Now we know that contact force between them in above case is given as
![F_n = m_2a](https://tex.z-dn.net/?f=F_n%20%3D%20m_2a)
![8.90 = m_2(1.5)](https://tex.z-dn.net/?f=8.90%20%3D%20m_2%281.5%29)
![m_2 = 5.93 kg](https://tex.z-dn.net/?f=m_2%20%3D%205.93%20kg)
now we have
![m_1 = 2.20 kg](https://tex.z-dn.net/?f=m_1%20%3D%202.20%20kg)