Answer:
spacing between the slits is 405.32043 ×
m
Explanation:
Given data
wavelength = 610 nm
angle = 2.95°
central bright fringe = 85%
to find out
spacing between the slits
solution
we know that spacing between slit is
I = 4
× cos²∅/2
so
I/4 = cos²∅/2
here I/4 is 85 % = 0.85
so
0.85 = cos²∅/2
cos∅/2 = √0.85
∅ = 2 ×
0.921954
∅ = 45.56°
∅ = 45.56° ×π/180 = 0.7949 rad
and we know that here
∅ = 2π d sinθ / wavelength
so
d = ∅× wavelength / ( 2π sinθ )
put all value
d = 0.795 × 610× / ( 2π sin2.95 )
d = 405.32043 × m
spacing between the slits is 405.32043 × m
Number 13 is D the heart has 4 chambers and pumps blood to the arteries and number 14 us C
Answer:
196000 N
Explanation:
The following data were obtained from the question:
Height (h) = 10 m
Area (A) = 2 m²
Force (F) =.?
Next, we shall determine the pressure in the tank.
This can be obtained as follow:
P = dgh
Where
P is the pressure.
d is the density of the liquid.
g is acceleration due to gravity
h is the height.
Height (h) = 10 m
Density (d) of water = 1000 kg/m³
Acceleration due to gravity (g) = 9.8 m/s²
Pressure (P) =...?
P = dgh
P = 1000 × 9.8 × 10
P = 98000 N/m²
Therefore, the pressure acting on the tank is 98000 N/m²
Finally, we shall determine the force of gravity acting on the column of water as follow:
Area (A) = 2 m²
Pressure (P) = 98000 N/m²
Force (F) =.?
Pressure (P) = Force (F) /Area (A)
P = F /A
98000 = F/ 2
Cross multiply
F = 98000 × 2
F = 196000 N
Therefore, the force of gravity acting on the column of water is 196000 N
The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
-----------------------------------------------------------------
20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.
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LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.
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LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.
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LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.
Explanation:
