Answer:
the maximum current is 500 A
Explanation:
Given the data in the question;
the B field magnitude on the surface of the wire is;
B = μ₀i / 2πr
we are to determine the maximum current so we rearrange to find i
B2πr = μ₀i
i = B2πr / μ₀
given that;
diameter d = 2 mm = 0.002 m
radius = 0.002 / 2 = 0.001 m
B = 0.100 T
we know that permeability; μ₀ = 4π × 10⁻⁷ Tm/A
so we substitute
i = (0.100)(2π×0.001 ) / 4π × 10⁻⁷
i = 500 A
Therefore, the maximum current is 500 A
<span>Radio waves are the type of wave that most likely transmit cell phones messages. Radio waves can be used to transmit data and information at various frequencies. Gamma rays are high frequency and are dangerous to our health. Ultraviolet rays come from the sun. Infrared waves are ones we experience every day, even though it is invisible to the naked eye. This is the heat we feel from the sun. </span>
To solve the exercise, the key concept to be addressed is the Mass Center.
The center of mass of an object is measured as,
Our values are given by,
Replacing the values in our previous equation we have,
Therefore the mass of the meter stick is 7.928g
Answer: The correct answer is vacuum.
Explanation:
Sound wave is a longitudinal wave in which the particles vibrate parallel to the direction of the motion of the wave. Sound wave consists of compression and rarefaction.
It needs a medium to travel. It can travel in solid, liquid and gas. It cannot travel in vacuum.
Denser the medium, more will be the speed of the wave.
Solid is more denser in comparison to liquid.
The speed of the sound in the solid medium is more in comparison to the liquid. The speed of the sound in liquid is more than the speed in the gas.
Therefore, vacuum is least effective medium for traveling sound waves.
