Answer:
The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Explanation:
Given that,
Distance between the slits = 0.04 mm
Width = 0.01 mm
Distance between the slits and screen = 1 m
Wavelength = 600 nm
We need to calculate the distance between the places where the intensity is zero due to the double slit effect
For constructive fringe
First minima from center
Second minima from center
The distance between the places where the intensity is zero due to the double slit effect
Put the value into the formula
Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Answer:
Explanation:
Impulse of a force is measured by force x time or F X t
Impulse also equals change in momentum or
F x t = m v₂ - m v₁
The given case is as follows
in the first case
F x t = mv - o = mv
F = mv / t
in the second case
F₁ x 4 t = mv
F₁ = 1/4 x mv /t
F₁ = F / 4
option a) is correct .
iii )
In the last case
F₂ X t = m v/2 -0
F₂ = 1/2 x mv / t
= 1/2 x F
F₂ = F/2
Option e ) is correct.
Liters
Grams
Degrees Celsius
The other answer choices are from the imperial system
I believe the acceleration would be 5m/s
All you would need to do is divide the final speed by the time it took to get there. I am only about 80 sure this answer is correct, so take my advise only if you feel comfortable.
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
