In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
where D=5.00 m is the distance of the screen from the slits, and
is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
And from the relationship between frequency and wavelength,
, we can find the frequency of the light:
Answer:
- the fraction of the turbine work output used to drive the compressor is 50%
- the thermal efficiency is 29.6%
Explanation:
given information:
= 300 k
= 700 kPa
= 580 K
Qin = 950 kj/kg
efficiency, η = 86% = 0.86
Assume, Ideal gas specific heat capacities of air, = 1.005
k = 1.4
(a) the fraction of the turbine work output used to drive the compressor
Qin = ( - )
- = Qin /
= Qin / +
thus,
= = 1525 K
= \
=
so,
= \
= \
= 1525 K
= 875 K
= ( - )
= 1.005 kj/kgK (580 K - 300 K)
= 281.4 kJ/kg
= η ( - )
= 0.86 (1.005 kj/kgK) (1525 k - 875 K)
= 562.2 kJ/kg
therefore,
the fraction of turbine =
=
= 50%
(b) the thermal efficiency
η = ΔW/
= ( - )/
= /950
= 29.6%
Explanation:
Given parameters:
Distance = 15miles north = 24140.2m
Initial velocity = 0m/s
Final velocity = 4m/s
Unknown:
Speed, velocity and acceleration = ?
Solution:
The speed is the distance divide by time. It is a scalar quantity and has no directional attribute.
Speed =
The speed of the student is 4m/s
Velocity is the displacement divided by time. It is a vector quantity which specifies the direction and magnitude;
Velocity =
The velocity of the student is 4m/s due north
Acceleration is the change in velocity with time;
To find the acceleration, we use
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
4² = 0² + 2x a x 24140.2
a = = 0.00033m/s²
Answer:
The bond energy of F–F = 429 kJ/mol
Explanation:
Given:
The bond energy of H–H = 432 kJ/mol
The bond energy of H–F = 565 kJ/mol
The bond energy of F–F = ?
Given that the standard enthalpy of the reaction:
<u>H₂ (g) + F₂ (g) ⇒ 2HF (g)</u>
ΔH = –269 kJ/mol
So,
<u>ΔH = Bond energy of reactants - Bond energy of products.</u>
<u>–269 kJ/mol = [1. (H–H) + 1. (F–F)] - [2. (H–F)]</u>
Applying the values as:
–269 kJ/mol = [1. (432 kJ/mol) + 1. (F–F)] - [2. (565 kJ/mol)]
Solving for , The bond energy of F–F , we get:
<u>The bond energy of F–F = 429 kJ/mol</u>