Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
Ohms Law: V = IR
V is the voltage in volts
I is the current in amps
R is the resistance in Ohms
Rearrange: R = V/I
R = (110)/(0.050)
R = 2200
There are 2200 Ohms of resistance in the circuit.
Explanation:
In everyday use and in kinematics, the speed of an object is the magnitude of the rate of change of its position with time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity.
SI unit: m/s, m s−1
s=d/t
Answer:
796.18 Hz
Explanation:
Applying,
Maximum velocity = Amplitude×Angular velocity
Therefore,
V' = A(2πf)............... Equation 1
Where V' = maximum velocity of the eardrum, A = Amplitude of vibration of the eardrum, f = frequency of the eardrum vibration, π = pie
make f the subject of the equation
f = V'/2πA................ Equation 2
From the question,
Given: V' = 3.6×10⁻³ m/s, A' = 7.2×10⁻⁷ m,
Constant: 3.14.
Substitute these values into equation 2
f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)
f = 796.18 Hz
The formula that we will going to use in this question is simply P1V1/T1 = P2V2/T2 where T is the constant.
P1V1 = P2V2
P2 = P1V1/V2 = (V1 / V2) x P2 = (13.0 L / 6.5 L) x 0.76 atm = 1.5 atm
The answer in this question is 1.5 atm
