Answer:
the new diameter of the third ring = 0.607 mm
Explanation:
Consider the radius of 
bright ring when air is in between the lens and the plate ;
Using the expression: 
for the radius of the bright ring since the liquid (water ) of the refractive index 'n' is now in between the lens and the plate;
where;
= number of fringe
λ = wavelength
R = radius
n = refractive index of water
Now ;
the radius r of the third bright ring when the air is in between lens and plate = 
The new radius of the third bright fringe is 
Calculating the new diameter ; we have:
d(3) = 2(r(3))
d(3) = 2(0.3035)
d = 0.607 mm
Thus, the new diameter of the third ring = 0.607 mm
The gravitational potential energy of the brick is 25.6 J
Explanation:
The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.
Near the surface of a planet, the gravitational potential energy is given by
where
m is the mass of the object
g is the strength of the gravitational field
h is the height of the object relative to the ground
For the brick in this problem, we have:
m = 8 kg is its mass
g = 1.6 N/kg is the strenght of the gravitational field on the moon
h = 2 m is the height above the ground
Substituting, we find:
Learn more about potential energy:
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Answer:
5.5 × 10^14 Hz or s^-1
no orange light has less frequency so no photoelectric effect
Explanation:
hf = hf0 + K.E
HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s
f is frequency of incident photon and f0 is threshold frequency
hf0 = hf- k.E
6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20
6.63 × 10 ^-34 × f0 = 3.64158×10^-19
f0 = 3.64158×10^-19/ 6.63 × 10 ^-34
f0 = 5.4925 × 10^14
f0 =5.5 × 10^14 Hz or s^-1
frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light
kinetic energy is given as
KE = (0.5) m v²
given that : v = speed of the bottle in each case = 4 m/s
when m = 0.125 kg
KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J
when m = 0.250 kg
KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J
when m = 0.375 kg
KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J
when m = 0.0.500 kg
KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J
