Answer:
OT
It is important to wear protective equipments during games and practices.
Explanation:
Practice is a review of what is about to happen in some time to come and with practice ,the future can be trusted.
During practice it is very important to use protective gears or equipments because without them, injuries and accidents are liable to occur.
During the major game it is also important to prevent accidents or big injuries as the equipments protect each vital part of the body.
<h2>Answer:</h2>
<u>Turning a magnet very quickly would be BEST used to create an electric current</u>
<h2>Explanation:</h2>
In Electromagnetic waves electric field produces magnetic field and vice versa. A moving magnet can produce electric current. Dynamo is the best example for it. In dynamo armature is rotated between the magnets which results in the development of electric field and hence an electric current is produced in it.
MgCl2
Mg = magnesium
Cl = chlorine
Magnesium + chlorine = magnesium chloride.
This is because compounds are always written with the METAL FIRST and the NON METAL SECOND. the non metal ends in - ide when it reacts with a metal.
So ur answer would be magnesium chloride. :)
In a longitudinal wave the particle displacement is parallel to the direction of wave propagation. ... The particles do not move down the tube with the wave; they simply oscillate back and forth about their individual equilibrium positions.Answer:
Explanation:
The emf induced in the second coil is given by:
V = -M(di/dt)
V = emf, M = mutual indutance, di/dt = change of current in the first coil over time
The current in the first coil is given by:
i = i₀
i₀ = 5.0A, a = 2.0×10³s⁻¹
i = 5.0e^(-2.0×10³t)
Calculate di/dt by differentiating i with respect to t.
di/dt = -1.0×10⁴e^(-2.0×10³t)
Calculate a general formula for V. Givens:
M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)
Plug in and solve for V:
V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))
V = 320e^(-2.0×10³t)
We want to find the induced emf right after the current starts to decay. Plug in t = 0s:
V = 320e^(-2.0×10³(0))
V = 320e^0
V = 320 volts
We want to find the induced emf at t = 1.0×10⁻³s:
V = 320e^(-2.0×10³(1.0×10⁻³))
V = 43 volts
