The interatomic spring stiffness for tungsten is determined from Young's modulus measurements to be 90 N/m. The mass of one mole of tungsten is 0.185 kg. If we model a block of tungsten as a collection of atomic "oscillators" (masses on springs), what is one quantum of energy for one of these atomic oscillators? Note that since each oscillator is attached to two "springs", and each "spring" is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above. Use these precise values for the constants: ℏ = 1.0546 10-34 J · s (Planck's constant divided by 2π) Avogadro's number = 6.0221 1023 molecules/mole kB = 1.3807 10-23 J/K (the Boltzmann constant)
Answers
Answer:
Explanation:
solution below
Final answer:
The quantum of energy for one atomic oscillator in tungsten, given the effective interatomic spring stiffness of 360 N/m, the mass of one tungsten atom as 3.074 x 10^-25 kg, and the reduced Planck's constant of 1.0546 x 10^-34 J · s, can be calculated to be approximately 1.33 x 10^-21 J.
Explanation:
To calculate the quantum of energy for one atomic oscillator in tungsten, we will consider the model of an atom being connected to two springs, both having an effective interatomic spring stiffness of four times the given value (90 N/m). This value thus becomes 360 N/m.
One mole of tungsten has a mass of 0.185 kg, thus the mass of one atom can be determined by dividing this value by Avogadro's number (6.0221 x 10^23 molecules/mole), which gives approximately 3.074 x 10^-25 kg.
The quantum of energy, or the energy of one quantum (the smallest possible energy increment), is given by the formula E = ħω, where ħ is the reduced Planck's constant (1.0546 x 10^-34 J · s) and ω is the angular frequency, given by sqrt(k/m), where k is the spring constant and m is the mass.
Substituting the known values into these equations gives ω= sqrt((360)/(3.074 x 10^-25)) and E= (1.0546 x 10^-34) x sqrt((360)/(3.074 x 10^-25)), which results in a quantum of energy of approximately 1.33 x 10^-21 J.
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Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 2 seconds.
Answers
Answer:
17.54N in -x direction.
Explanation:
Amplitude (A) = 3.54m
Force constant (k) = 5N/m
Mass (m) = 2.13kg
Angular frequency ω = √(k/m)
ω = √(5/2.13)
ω = 1.53 rad/s
The force acting on the object F(t) = ?
F(t) = -mAω²cos(ωt)
F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)
F(t) = -17.65 * cos (5.355)
F(t) = -17.57N
The force is 17.57 in -x direction
Answers
Answer:
The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.
Explanation:
(b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air?
(c) What was his total displacement relative to the air mass?
Answers
Answer:
a)
b)
c)
Explanation:
Given:
a)
duration of flight,
velocity of flight,
direction of flight, to the south of east
Now the total displacement:
b)
Velocity of air,
When the aircraft encounters a headwind in the opposite direction to the velocity of motion then the speed of the aircraft is lowered with respect to the ground.
But when the speed is observed with respect to the wind the reduced velocity of the aircraft is observed from an opposite moving wind having a magnitude equal to the difference in velocity of the aircraft. This results in no change in the apparent velocity of the aircraft.
Mathematically:
Velocity of the aircraft with respect to the ground:
Now the velocity of the aircraft with respect to the wind:
c)
Now the total displacement with respect to the wind:
Answers
Answer:
(a) 51428.59 J/C
(b) 25714.29 J/C
(c) 0 J/C
Explanation:
Parameters given:
Q1 = 2 * 10^-6 C
Q2 = 2 * 10^-6 C
Q3 = 2 * 10^-6 C
Q4 = 2 * 10^-6 C
=> Q1 = Q2 = Q3 = Q4 = Q
Side of the square = 2m
The center of the square is the midpoint of the diagonals, i.e. Using Pythagoras theorem:
BD² = 2² + 2²
BD² = 8
BD = √(8) = 2.8m
OD = 1.4m
(The attached diagram explains better)
Hence, the distance between the center and each point charge, r, is 1.4m.
Electric Potential, V = kQ/r
k = Coulombs constant
(a) If all charges are positive:
V(Total) = V1 + V2 + V3 + V4
V1 = Potential due to Q1
V2 = Potential due to Q2
V3 = Potential due to Q3
V4 =Potential due to Q4
Since Q1 = Q2 = Q3 = Q4 = Q
=> V1 = V2 = V3 = V4
=> V(Total) = 4V1
V = (4 * 9 * 10^9 * 2 * 10^-6)/1.4
V = 51428.59J/C
(b) If 3 charges are positive and 1 is negative:
Since Q1 = Q2 = Q3 = Q
and Q4 = -Q
The total potential becomes:
V(Total) = V1 + V2 + V3 - V4
Since V1, V2, V3 and V4 have the same value,
V(Total) = V1 + V2
V(Total) = 2V1
V(Total) = (2 * 9 * 10^9 * 2 * 10^-6)/1.4
V(Total) = 25174.29 J/C
(c) Two charges are positive and two are negative:
Since Q1 = Q2 = Q
and Q3 = Q4 = -Q
The total potential becomes:
V(Total) = V1 + V2 - V3 - V4
Since V1, V2, V3 and V4 have the same value,
V(Total) = 0 J/C
Answers
Answer:
Explanation:
If the electric field is uniform, the electric field between two points at potentials and
which are separated by a distance d will be given by the formula:
So in our case, we have
Answers
Answer: hello your question is incomplete below is the complete question
Water stands at a depth H in a large open tank whose side walls are vertical . A hole is made in one of the walls at a depth h below the water surface. Part B How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole
answer :
At Height ( h ) from the bottom of Tank
Explanation:
Determine how far above the bottom of the tank a second hole be cut
For the second hole to have the same range as the first hole
Range of first hole = Velocity of efflux of water * time of fall of water
= √ (2gh) * √( 2g (H - h) / g)
= √ ( 4(H-h) h)
Hence the Height at which the second hole should be placed to exercise same range of stream emerging = h from the bottom of the Tank
Final answer:
The second hole should be cut at the same height as the first hole to have the same range for the stream.
Explanation:
In order for the stream emerging from the second hole to have the same range as the first hole, the second hole should be cut at the same height as the first hole. This is because the range of the stream depends on the initial velocity and the vertical distance traveled. If the second hole is higher or lower than the first hole, the vertical distance traveled will be different and the range of the stream will be affected.
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