Answer:
20m/s^2
Explanation:
Acceleration=Change in velocity/time taken for change
40-20/1
20m/s^2
Answer:
a) i₈ = 0.5 i₄, b) i₁₀ = 0.3 i₃, i₁₀ = 0.8 i₈
Explanation:
For this exercise we use ohm's law
V = i R
i = V / R
we assume that the applied voltage is the same in all cases
let's find the current for each resistance
R = 4 Ω
i₄ = V / 4
R = 8 Ω
i₈ = V / 8
we look for the relationship between these two currents
i₈ /i₄ = 4/8 = ½
i₈ = 0.5 i₄
R = 3 Ω
i₃ = V3
R = 10 Ω
i₁₀ = V / 10
we look for relationships
i₁₀ / 1₃ = 3/10
i₁₀ = 0.3 i₃
i₁₀ / 1₈ = 8/10
i₁₀ = 0.8 i₈
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:
Explanation:
The apparent brightness follows an inverse square law, therefore we can write:
where I is the apparent brightness and r is the distance from the Sun.
We can also rewrite the law as
(1)
where in this problem, we have:
apparent brightness at a distance 
, where
million km
We want to estimate the apparent brightness at 
, where is ten times , so
Re-arranging eq.(1), we find 
:
