Answer:
The answer to the questions is;
In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)
The distance 4.1 cm is equivalent to λ/4
Explanation:
For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point
When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound
The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ
Therefore 4.1 cm is λ/4
Answer:
See below
Explanation:
<u>I will use 3 x 10^8 m/s for speed or wave</u>
speed = wavelength * frequency
3 x 10^8 = w * 7.34 x 10^2 <====== are you sure this isn't KILO Hz ?
w = <u>408719. 3 meters </u>
Answer: a) for 150 Angstroms 6.63 *10^-3 eV; b) for 5 Angstroms 6.02 eV
Explanation: To solve this problem we have to use the relationship given by De Broglie as:
λ =p/h where p is the momentum and h the Planck constant
if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m
Finally we have:
eΔV=p^2/2m= h^2/(2*m*λ^2)
replacing we obtained the above values.
Answer:
U₂ = 400 KJ
Explanation:
Given that
Initial energy of the tank ,U₁= 800 KJ
Heat loses by fluid ,Q= - 500 KJ
Work done on the fluid ,W= - 100 KJ
Sign -
1.Heat rejected by system - negative
2.Heat gain by system - Positive
3.Work done by system = Positive
4.Work done on the system-Negative
Lets take final internal energy =U₂
We know that
Q= U₂ - U₁ + W
-500 = U₂ - 800 - 100
U₂ = -500 +900 KJ
U₂ = 400 KJ
Therefore the final internal energy = 400 KJ
Answer: 580 N
Refer to attached figure.
The angle of inclination is 22 degrees
weight (gravitational force) acts downwards.
Normal force is a contact force which acts perpendicular to the point of contact.
The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.
Gravitational force on an object = mg
The normal force 
