Explanation:
work done is expressed as:
w=fd
where did force and d is distance
w=30*10
=300J
w is said to be done when a load is moved at a distance.
A high-voltage spark is delivered to the plugs via the primary winding. The principal magnetic field of the coil increases. Zero volts are reached by the secondary voltage.
The fundamental ideas are inductance and electromagnetism. Two sets of wire windings, or coils—get it—that make up an ignition coil can be found inside. An iron core is encircled by the two windings, referred to as a primary set and a secondary set. A magnetic field is created when the battery current enters the primary windings. The magnetic field collapses over the secondary windings when a switch—the distributor points opening or an electronic trigger—interrupts the flow of battery current.
To know more about ignition coil, visit;
brainly.com/question/28271692
#SPJ4
D
S T
Speed is 36m/12s
Speed is 3 m/s
Answer:
Electric field, E = 40608.75 N/C
Explanation:
It is given that,
Mass of electrons, ![m=9.1\times 10^{-31}\ kg](https://tex.z-dn.net/?f=m%3D9.1%5Ctimes%2010%5E%7B-31%7D%5C%20kg)
Initial speed of electron, u = 0
Final speed of electrons, ![v=3\times 10^7\ m/s](https://tex.z-dn.net/?f=v%3D3%5Ctimes%2010%5E7%5C%20m%2Fs)
Distance traveled, s = 6.3 cm = 0.063 m
Firstly, we will find the acceleration of the electron using third equation of motion as :
![a=\dfrac{v^2-u^2}{2s}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2s%7D)
![a=\dfrac{(3\times 10^7)^2}{2\times 0.063}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B%283%5Ctimes%2010%5E7%29%5E2%7D%7B2%5Ctimes%200.063%7D)
![a=7.14\times 10^{15}\ m/s^2](https://tex.z-dn.net/?f=a%3D7.14%5Ctimes%2010%5E%7B15%7D%5C%20m%2Fs%5E2)
Now we will find the electric field required in the tube as :
![ma=qE](https://tex.z-dn.net/?f=ma%3DqE)
![E=\dfrac{ma}{q}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bma%7D%7Bq%7D)
![E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7B9.1%5Ctimes%2010%5E%7B-31%7D%5Ctimes%207.14%5Ctimes%2010%5E%7B15%7D%7D%7B1.6%5Ctimes%2010%5E%7B-19%7D%7D)
E = 40608.75 N/C
So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.
Given:
u = 6.5 m/s, initial velocity
a = 1.5 m/s², acceleration
s = 100.0 m, displacement
Let v = the velocity attained after the 100 m displacement.
Use the formula
v² = u² + 2as
v² = (6.5 m/s)² + 2*(1.5 m/s²)*(100 m) = 342.25 (m/s)²
v = 18.5 m/s
Answer: 18.5 m/s