Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
Answer:
applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²
Answer:
Her speed is 1.1 m/s, and her velocity is 0 m/s
Explanation:
Speed = Distance covered/Time
Given
Distance = 400m
Time = 6minutes = 6*60 = 360 secs
Substitute the given parameter into the formula;
Speed = 400/360
Speed = 1.1m/s
Since the track is a circular track, the displacement will be zero. She is only moving in a circular path (no direction)
Velocity = Displacement/Time
Velocity = 0/3600
Velocity = 0m/s
Hence her speed is 1.1 m/s, and her velocity is 0 m/s
