A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cross section of the wire? (e = 1.60 × 10-19 C)

162500000.  

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

n=162500000 t

The number of electron passe per second will be 162500000.

weather forecasts are done by collecting data about the current state of the atmosphere mainly the temperature, humidity and wind and using meteorology, to determine the atmosphere that evolves over time. the h and l mean the high and low pressure. the red strip is a warm front. the blue strip is a cold front. the blue and red strip is a stationary front. the purple strip is a occluded front.

explanation:

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west