1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec
Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s
In 10 second, number of electrons are calculated as;
Number of electrons through the device = 3.125*10^19*10 = 3.125*10^20 electrons
When the object is at the focal point the angular magnification is 2.94.
Angular magnification:
The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.
Here we have to find the angular magnification when the object is at the focal point.
Focal length = 6.00 cm
Formula to calculate angular magnification:
Angular magnification = 25/f
= 25/ 8.5
= 2.94
Therefore the angular magnification of this thin lens is 2.94
To know more about angular magnification refer:: brainly.com/question/28325488
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Answer:
Wavelength of the sound wave that reaches your ear is 1.15 m
Explanation:
The speed of the wave in string is
where T= 200 N is tension in the string , 
=1.0 g/m is the linear mass density
Wavelength of the wave in the string is
The frequency is
The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)
