Answer:
Δx = 3.477 x 10⁻³ m = 3.477 mm
Explanation:
The distance between two consecutive dark fringes is given by the following formula, in Young's Double Slit experiment:
Δx = λL/d
where,
Δx = distance between two consecutive dark fringes = ?
λ = wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m
L = Distance between slits and screen = 2.2 m
d = slit separation = 0.31 mm = 0.31 x 10⁻³ m
Therefore,
Δx = (4.9 x 10⁻⁷ m)(2.2 m)/(0.31 x 10⁻³ m)
Δx = 3.477 x 10⁻³ m = 3.477 mm
The distance (in mm) between the first and second dark fringes of the
interference pattern is 3.477 mm
This is calculated by using the formula in Young's Double Slit experiment:
Δx = λL/d
where,
Δx = distance between two consecutive dark fringes which is unknown
λ which is wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m
L which is distance between slits and screen = 2.2 m
d which is slit separation = 0.31 mm = 0.31 x 10⁻³ m
We then substitute them into the equation
Δx = (4.9 x 10⁻⁷ m ×2.2 m)/(0.31 × 10⁻³ m)
Δx = 3.477 x 10⁻³ m = 3.477 mm
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The 10.2-ft steel beam is being hoisted from its horizontal position by the two cables attached at A and B. If the initial angular accelerations are α1 = 0.52 rad/sec2 and α2 = 0.35 rad/sec2 , determine the initial values of
(a) the angular acceleration α (positive if counterclockwise, negative if clockwise) of the beam,
(b) the acceleration of point C (positive if up, negative if down), and
(c) the distance d (positive if to the right, negative if to the left) from A to the point on the centerline of the beam which has zero acceleration.
Required information Problem 16.048 - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. A uniform slender rod AB rests on a frictionless horizontal surface, and a force P of magnitude 0.25 lb is applied at A in a direction perpendicular to the rod. Assume that the rod weighs 2.2 lb. Problem 16.048.a - Acceleration A of slender rod on frictionless surface Determine the acceleration of Point A. (You must provide an answer before moving on to the next part.) The acceleration of Point A is 14.6 14.6 Correct ft/s2 →.
NOTE: The diagram is attached to this solution
Answer:
The acceleration of point A = 14.64 ft/s²
Explanation:
By proper analysis of the diagram, acceleration of point A will be: (Check the free body diagram attached)
[tex]a_{A} = \bar{a} + \frac{\alpha L}{2}[/tex]
Weight, W = mg
g = 32.2 ft/s²
m = W/g
[tex]p = m \bar{a}\\p = w \bar{a} /g[/tex]
[tex]\bar{a} = pg/w\\\bar{a} = 0.25g/2.2\\\bar{a} =3.66[/tex]
[tex]\frac{pL}{2} = I \alpha[/tex]
but [tex]I = \frac{wL^{2} }{12g}[/tex]
[tex]\frac{pL}{2} = \frac{\alpha wL^{2} }{12g}\\\alpha = \frac{6 g p}{wL}\\\alpha = \frac{6*g*0.25}{2.2L} \\\alpha = 21.96/L[/tex]
[tex]a_{A} = 3.66 + \frac{(21.96/L ) * L}{2}\\a_{A} = 3.66 + 10.98\\a_{A} = 14. 64 ft/s^{2}[/tex]
in hooke's law is force directly proportional to extension?
Answer:
Explanation:
Hooke's Law is a principle of physics that states that the that the force needed to extend or compress a spring by some distance is proportional to that distance. ... In addition to governing the behavior of springs, Hooke's Law also applies in many other situations where an elastic body is deformed
Answer:
Yeah ,The extension of the spring is directly proportional to the force applied.
Problem: A barbell consists of two small balls, each with mass m at the ends of a very low mass rod of length d. The barbell is mounted on the end of a low-mass rigid rod of length b. This apparatus is started in such a way that while the rod rotates clockwise with angular speed ω1, the barbell rotates clockwise about its center with an angular speed ω2. What is the total angular momentum of this system about point B?
Answer:
mass of ball 1=m1
mass of ball 2=m2
velocity of ball=r1w1
velocity of ball 2=r2w2
Total angular momentum=m1*v1+m2*v2
but
v1=r1*w1
v2=r2*w2
Substitute values in above equation
Total angular momentum of the system=m1*r1*w1+m2*r2*w2
The total angular momentum of the system at point B will be [tex]\rm L=m_1 r_1\omega_1 +m_2 r_2 \omega_2[/tex].
What is angular momentum?The rotating counterpart of linear momentum is angular momentum also known as moment of momentum or rotational momentum.
The given data in the problem is;
m₁ is the mass of ball 1
m₂ mass of ball 2=m2
v₁ is the velocity of ball=r₁ω₁
v₂ is the velocity of ball 2=r₂ω₂
The total angular momentum of a system at point B is given as;
[tex]\rm V_{total}= r_1\omega_1 + r_2 \omega_2 \\\\ \rm L=m_1 r_1\omega_1 +m_2 r_2 \omega_2[/tex]
Hence the total angular momentum of the system at point B will be [tex]\rm L=m_1 r_1\omega_1 +m_2 r_2 \omega_2[/tex].
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A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experimenter then briefly touches the electroscope with a finger. The finger is removed, followed by the removal of the positively charged object. What happens to the leaves of the electroscope when a negative charge is now brought near but not in contact with the top of the electroscope?
Answer:
The leaves of the electroscope move further apart.
Explanation:
This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.
Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.
So, the leaves move further apart.
The electroscope initially becomes positively charged, but is neutralized when touched. Once a negatively charged object is brought near, the leaves become negatively charged and repel each other.
Explanation:When a positively charged object is brought near an uncharged gold leaf electroscope, the free electrons in the electroscope are attracted towards the positive charge, leaving the leaves positively charged and causing them to repel each other.
When you touch the electroscope, you provided a grounding path which allows electrons to flow from the ground to the positively charged electroscope to neutralize it again, causing the leaves to fall back together.
After the positively charged object is removed, the electroscope remains neutral until a negatively charged object is brought near it. The electrons in the electroscope will be repelled towards the leaves, giving them a negative charge and causing them to repel again.
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Due to a technical malfunction a space explorer had to crash land on Planet-X. She manages to fix her space ship and now she is preparing for launch. However she needs to know the gravitational acceleration on the surface of the planet in order to take off successfully. She builds a mathematical pendulum out of a piece of string and a left over steel bolt. The bolt has a mass of 35.5 g, and the string is 133 cm long. She attaches the pendulum to a fixed point and she lets it swing. She counts 12 complete oscillations in a time period of 70.7 seconds. What is the gravitational acceleration on the surface of Planet-X
Answer:
The gravitational acceleration on the surface of Planet-X is 1.5 [tex]ms^{-2}[/tex]
Explanation:
Firstly, we are to list out the parameters given:
Mass (m) = 35.5 g, Length of string (L) = 133 cm = 1.33 m, t = 70.7 seconds for 12 oscillations, T = 70.7 ÷ 12 = 5.89 seconds
The formula for calculating the period of a simple pendulum (assuming the angle of deflection is lesser than 15º) is given by:
[tex]T=2\pi\sqrt{\frac{L}{g}}\\[/tex]
We want to calculate for the gravitational acceleration (g), hence, we have to make g the subject of the formula
[tex]g=4\pi^{2}\frac{L}{T^{2}}\\[/tex]
Substitute the parameters into the equation, we have:
g = [tex]4\pi^{2}[/tex] * 1.33 ÷ [tex]5.89^{2}[/tex] = 1.51
g = 1.5 [tex]ms^{-2}[/tex]
The gravitational acceleration on the surface of Planet-X is 1.5 [tex]ms^{-2}[/tex]
Answer:
the gravitational acceleration of the Xplanet is 1.344m/s^2
Explanation:
You can use the formula for the calculation of the frequency of a pendulum, in order to find an expression for the gravitational constant:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{l}}\\\\g=4\pi^2 f^2l[/tex]
Where you can notice that mass ob the object does not influence of the gravitatiolan acceleration. By the information of the question, you have the values of f and l. By replacing these values (with units of meter and seconds) you obtain:
[tex]f=\frac{12}{70.7}=0.16s^{-1}\\\\g=4\pi^2(0.16s^{-1})^2(1.33m)=1.344\frac{m}{s^2}[/tex]
Hence, the gravitational acceleration of the Xplante is 1.344m/s^2
Rotation of the lever OA is controlled by the motion of the contacting circular disk of radius r = 300 mm whose center is given a horizontal velocity v = 1.64 m/s. Determine the angular velocity ω (positive if counterclockwise, negative if clockwise) of the lever OA when x = 880 mm.
Answer:
The angular velocity is
5.64rad/s
Explanation:
This problem bothers on curvilinear motion
The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s
We know that the velocity v is given as
v= ωr
Where ω is the angular velocity
r is 300mm to meter = 0.3m
the radius of the circle
described by the level
v=1.64m/s
Making ω subject of the formula and solving we have
ω=v/r
ω=1.64/0.3
ω=5.46 rad/s
Puck 1 (1 kg) travels with velocity 20 m/s to the right when it collides with puck 2 (1 kg) which is initially at rest. After the collision, puck 1 moves with a velocity of 5 m/s. Assume that no external forces are present and therefore the momentum for the system of pucks is conserved. What is the final velocity (in m/s) of puck 2 after the collision
Answer:
Explanation:
Parameters given:
Mass of Puck 1, m = 1 kg
Mass of Puck 2, M = 1 kg
Initial velocity of Puck 1, u = 20 m/s
Initial velocity of Puck 2, U = 0 m/s
Final velocity of Puck 1, v = 5 m/s
Since we are told that momentum is conserved, we apply the principle of conservation of momentum:
Total initial momentum of the system = Total final momentum of the system
mu + MU = mv + MV
(1 * 20) + (1 * 0) = (1 * 5) + (1 * V)
20 = 5 + V
V = 20 - 5 = 15 m/s
Puck 2 moves with a velocity of 15 m/s
A pipe carrying water narrows to (2/3) times its initial diameter. If the absolute pressure in the narrow section is 0.50 atm , and if the speed of the water in the narrow section is 18.0 m/s , then which of the following is the closest to the pressure of the water in the wide section? Assume there is no change in height.
Group of answer choices
1.5 atm
1.8 atm
2.4 atm
2.1 atm
To solve this problem we will apply the concepts related to continuity and later to Bernoulli's principle. With the continuity equation we will relate the diameters of the two measurements to find the speed. Later with the speed we will proceed to replace it in the Bernoulli equations to find the pressure.
By continuity equation
[tex]A_1 V_1 = A_2 V_2[/tex]
[tex](\frac{\pi D^2_1}{4})V_1 = (\frac{\pi D_2^2}{4})V_2[/tex]
[tex]D_1^2V_1 = D_2^2V_2[/tex]
With the relation of diameters we have
[tex]D_1^2V_1 = (\frac{2D_1}{3})^2V_2[/tex]
[tex]D_1V_1 = (\frac{4}{9})D_1^2V_2[/tex]
[tex]V_1 = \frac{4}{9}V_2[/tex]
Replacing the value of the Volume
[tex]V_1 = \frac{4}{9} (18)[/tex]
[tex]V_1 = 8m/s[/tex]
By Bernoulli's equation
[tex]P_1 + \frac{\rho V_1^2}{2} = P_2 + \frac{\rho V_2^2}{2}[/tex]
[tex]P_1 + \frac{(1000)(8)^2}{2} = 50662.5+\frac{(1000)(18)^2}{2}[/tex]
[tex]P_1 = 180662.5Pa (\frac{1atm}{101325Pa})[/tex]
[tex]P_1 = 1.8atm[/tex]
Therefore the water pressure in the wide section is 1.8atm
Antonio wants to replace the lightbulbs in his house. How much money would he save in one year by replacing 10 incandescent bulbs with 10 CFL bulbs?
incandescent$1.47 per bulb CFL $3.99
annual operating cost incandescent $32.86 CFL $7.67
Answer:
226.70
Explanation:
Answer: $226.70 on edg
Explanation:
Two hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area 1.20 × 10−3 m2. The two ends of the wire are connected to a resistor. The total resistance in the circuit is 18.0 . If an externally applied uniform longitudinal magnetic filed in the core changes from 1.65 T in one direction to 1.65 T in the opposite direction, how much charge flows through a point in the circuit during the change?
Answer:
The charge flows through a point in the circuit during the change is 0.044 C.
Explanation:
Given that,
Number of turns in the copper wire, N = 200
Area of cross section, [tex]A=1.2\times 10^{-3}\ m^2[/tex]
Resistance of the circuit, R = 118 ohms
If an externally applied uniform longitudinal magnetic filed in the core changes from 1.65 T in one direction to 1.65 T in the opposite direction.
We need to find the charge flows through a point in the circuit during the change. Due to change in magnetic field an emf is induced in it. It is given by :
[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]
Using Ohm's law :
[tex]\epsilon=IR[/tex]
[tex]IR=-\dfrac{d\phi}{dt}\\\\I=-\dfrac{1}{R}\dfrac{d\phi}{dt}[/tex]
Electric current is equal to the rate of change of electric charge. So,
[tex]dq=\dfrac{NA(B(0)-B(t))}{R}\\\\dq=\dfrac{200\times 1.2\times 10^{-3}(1.65+1.65)}{18}\\\\dq=0.044\ C[/tex]
So, the charge flows through a point in the circuit during the change is 0.044 C.
A 2 kg sample of air contained in a 2 m3 rigid container undergoes a process in which its pressure reduces from 100.45 kPa to 86.10 kPa while the volume remains the same. Please calculate (a) how much work was done, (b) by how much did the internal energy of the air change, (c) how much heat was transferred (please specify the direction of heat transfer).
Answer:
Explanation:
The pictures attached below shows the solution to the problem and i hope its explanatory enough. Thank you
A cable passes over a pulley. Because the cable grips the pulley and the pulley has nonzero mass, the tension in the cable is not the same on opposite sides of the pulley. The force on one side is 137 N, and the force on the other side is 43 N. Assuming that the pulley is a uniform disk of mass 1.21 kg and radius 0.723 m, find the magnitude of its angular acceleration. [For a uniform disk, I = (1/2)mr2 .] Answer in units of rad/s 2
Answer:
[tex]\alpha=214.8 rad/s^2[/tex]
Explanation:
We are given that
[tex]F_1=137 N[/tex]
[tex]F_2=43 N[/tex]
Net force=F=[tex]F_1-F_2=137-43=94 N[/tex]
Mass,m=1.21 kg
Radius,r=0.723 m
We have to find the magnitude of its angular acceleration.
Moment of inertia ,[tex]I=\frac{1}{2}mr^2[/tex]
Substitute the values
Torque ,[tex]\tau=I\alpha[/tex]
[tex]F_{net}\times r=\frac{1}{2}mr^2\alpha[/tex]
[tex]\alpha=\frac{2F_{net}}{mr}[/tex]
[tex]\alpha=\frac{2\times 94}{1.21\times 0.723}[/tex]
[tex]\alpha=214.8 rad/s^2[/tex]
Is Fahrenheit an SI unit?
Answer:
No
Explanation:
Because the people put in charge of defining SI units said so
An incompressible fluid flows steadily through two pipes of diameter 0.15 m and 0.2 m which combine to discharge in a pipe of 0.3 diameter. If the average velocities in the 0.15 m and 0.2 m diameter pipes are 2 m/s and 3 m/s respectively, then find the average velocity in the 0.3 m diameter pipe.
Answer:
Average velocity = 1.835 m/s
Explanation:
Detailed explanation and calculation is shown in the image below
The average velocity in the 0.3 m diameter pipe. The result is 1.833 m/s.
To find the average velocity in the 0.3 m diameter pipe, we'll use the principle of continuity for incompressible fluid flow. The flow rate (Q) must be the same at all points in the system.
First, we calculate the flow rates in the two initial pipes:
For the 0.15 m diameter pipe: Q1 = A₁ v₁ = π (0.075)² 2 m/s = π × 0.005625 m² × 2 m/s = 0.01125π m³/s
For the 0.2 m diameter pipe: Q2 = A2 v2 = π (0.1)² 3 m/s = π × 0.01 m² × 3 m/s = 0.03π m³/s
The combined flow rate entering the 0.3 m diameter pipe is:
[tex]Q_t = Q_1 + Q_2[/tex] = 0.01125π m³/s + 0.03π m³/s = 0.04125π m³/s
Next, we find the area of the 0.3 m diameter pipe:
A₃ = π (0.15)² = π 0.0225 m²
The average velocity in the 0.3 m diameter pipe is then:
v₃ = Qtotal / A₃ = (0.04125π m³/s) / (π 0.0225 m²) ≈ 1.833 m/s
Thus, the average velocity in the 0.3 m diameter pipe is 1.833 m/s.
A solid block of copper, which is a good conductor, has a cavity in its interior. Within the cavity, insulated from the conductor, is a point charge of 3.00 nCnC. The block of copper itself has an excess charge of -8.00 nCnC. How much charge is on the outer surface of the block and the surface of the cavity
Answer:
The charge on the outer surface of the block = -5.00 nC
The charge on the surface of the cavity (on the inner surface of the block) = -3.00 nC
Explanation:
The point charge within the cavity will induce a charge equal in magnitude and opposite in sign on the inside cavities of the copper block.
Charge of the point charge = 3.00 nC
Charge induced on the inner surface of the Copper block's cavity = -3.00 nC
Since the charge on a conductor should usually be neutral, the charge on the inner surface causes a charge equal in magnitude and also opposite in sign on the outer surface of the block; that is, 3.00 nC.
But this block already has an excess charge of -8.00 nC (which resides on the surface because excess charge for conductors reside on the surface of the conductors)
So, net charge on the outer surface of the Copper block = -8.00 + 3.00 = -5.00 nC.
Hope this Helps!!!
5 points total) The latent heat of vaporization (boiling) of helium at a pressure of 1 atm and a temperature of 4.2 K is 21.8 kJ kg-1 . The densities at 4.2 K of the liquid and vapor are 125 kg m-3 and 19 kg m-3 respectively. a. (2 points) What proportion of latent heat is involved in work against the interatomic attraction? b. (2 points) Estimate the depth e of the potential well which results from the forces between two helium atoms. Give your answer in Joules, electron-volts (eV) and kelvin (K), where the answer in kelvin corresponds to an energy kT. (Take the average number of nearest neighbors to be z = 10 and note that the atomic number of helium is 4.) c. (1 point) Why does helium turn into a gas at such low temperatures?
Answer:
a) proportion of latent heat involved in work against the interatomic attraction = 0.794
b)Depth of the well in Joules = 23 * 10^-24 J
ii) In eV, E = 0.000144 eV
III) in Kelvin, E = 1.67 K
C) Check the explanation section for C
Explanation:
a) Latent heat, Q = 21.8 kJ/kg
Vapor density, Vd = 19 kg/m^3
Liquid density, Ld = 125 kg/ m^3
Pressure, P = 1 atm = 1 * 10^5 Pa
Volume change from liquid to vapor = (1/Vd) - (1/ Ld)
Volume change = (1/19) - (1/125)
Volume change = 0.045 m^3
Work done in converting from liquid to vapor, W = P * (Volume change)
W = 1 *0.045 * 10^5
W = 4.5 kJ
Let the proportion of latent heat involved in work against the interatomic attraction be Pr
Pr = (Q - W)/Q
Pr = (21.8 - 4.5)/21.8
Pr = 0.794
b) To calculate the depth of the potential well :
I) In joules
n(L-W) = 0.5 z Na E
z = 10
Where E = depth of the well
4(21.8-4.5) = 0.5 * 10 * 6.02 * 10^23 * E
E = 23 * 10^-24 J
ii) In eV
E = ( 23 * 10^-24)/(1.6 * 10^-19)
E = 0.000144 eV
III) In Kelvin
E = ( 23 * 10^-24)/(1.38 * 10^-23)
E = 1.67 K
C) Helium turns to a gas at that low temperature because of the large workdone (4.5 kJ) against the interatomic attraction.
Final answer:
a. About 83% of the latent heat is involved in work against the interatomic attraction. b. The depth of the potential well is approximately 4.62 x 10^-23 J, 2.88 x 10^14 eV, and 33.6 K. c. Helium turns into a gas at low temperatures due to its weak interatomic attraction.
Explanation:
a. To calculate the proportion of latent heat involved in work against interatomic attraction, we need to first calculate the work done using the formula W = P(V2 - V1), where P is the pressure and V2 and V1 are the volumes of the vapor and liquid respectively. The difference in volume is V2 - V1 = 1 - 19 = -18 m^3. Since the work done is negative (work is done against the interatomic attraction), the proportion of latent heat involved in work is given by the ratio |W|/Q, where Q is the latent heat of vaporization. Therefore, the proportion is |(-1 atm)(-18 m^3)/(21.8 kJ/kg)| = 0.8292, or about 83%.
b. The depth e of the potential well can be estimated using the formula e = kT/2, where k is Boltzmann's constant and T is the temperature. The average number of nearest neighbors z is 10, and the atomic number of helium is 4. Therefore, the depth e can be calculated as e = (4 * (1.38 x 10^-23 J/K) * 4.2 K)/(2 * 10) = 4.62 x 10^-23 J. This is equivalent to approximately 2.88 x 10^14 eV and 33.6 K.
c. Helium turns into a gas at such low temperatures because its interatomic attraction is weak. The low mass and low atomic number of helium result in weak intermolecular forces, making it easier for helium atoms to overcome the attractive forces and transition from a liquid to a gas state at low temperatures.
A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b and outer radius c. The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length λ.
1. Calculate the electric field (a) at any point between the cylinders a distance r from the axis and (b) at any point outside the outer cylinder.
Answer:
(a) E=λ/(2\pi e0 r)
(b) E = 0
Explanation:
(a) We can use the Gaussian's Law to calculate the electric field at any distance r from the axis. By using a cylindrical Gaussian surface we have:
[tex]\int \vec{E}\cdot d\vec{r}=\frac{\lambda}{\epsilon_o}[/tex]
where λ is the total charge per unit length inside the Gaussian surface. In this case we have that the Electric field vector is perpendicular to the r vector. Hence:
[tex]E\int dr=E2\pi r=\frac{\lambda}{\epsilon_o}\\\\E=\frac{\lambda}{2\pi r \epsilon_o}[/tex]
(b) outside of the outer cylinder there is no net charge inside the Gaussian surface, because charge of the inner radius cancel out with the inner surface of the cylindrical conductor.
Hence, we have that E is zero.
hope this helps!!
Answer:
1a. E(r) = lambda/ 2πrEo
1b. Electric field outside the outside cylinder = lambda/ 2πrEo
Explanation:
A beam of light is shined on a thin (sub-millimeter thick) single crystal wafer of material. The light source is special since it can be tuned to provide any wavelength of visible light on demand. The specimen is illuminated such that the wavelength of light is increased over time while the transmitted intensity of the light is measured. If the sample becomes transparent when the wavelength is greater than 650 nanometers, what is the band gap of the material, in eV
Answer:
The band gap of the material is 1.9113 eV
Explanation:
Given data:
λ = wavelength = 650 nm = 650x10⁻⁹m
Question: What is the band gap of the material, E = ?
[tex]E=\frac{hc}{\lambda }[/tex]
Here
h = Planck's constant = 6.626x10⁻³⁴J s
c = speed of light = 3x10⁸m/s
[tex]E=\frac{6.626x10^{-34}*3x10^{8} }{650x10^{-9} } =3.058x10^{-19} J=1.9113eV[/tex]
An LC circuit consists of a 3.400 capacitor and a coil with self-inductance 0.080 H and no appreciable resistance. At t = 0 the capacitor is fully charged so the potential between the plates is 1.588 V and the current in the inductor is zero. What is the charge on the plates? How long after t = 0 will the current in the circuit be maximum? What will be the maximum current? What is the total energy in the system?
Answer:
Explanation:
charge on the capacitor = capacitance x potential
= 1.588 x 3.4
= 5.4 C
Energy of capacitor = 1 / 2 C V ² , C is capacitance , V is potential
= .5 x 3.4 x 1.588²
= 4.29 J
If I be maximum current
energy of inductor = 1/2 L I² , L is inductance of inductor .
energy of inductance = Energy of capacitor
1/2 L I² = 4.29
I² = 107.25
I = 10.35 A
Time period of oscillation
T = 2π √ LC
=2π √ .08 X 3.4
= 3.275 s
current in the inductor will be maximum in T / 4 time
= 3.275 / 4
= .819 s.
Total energy of the system
= initial energy of the capacitor
= 4.29 J
In the 2016 Olympics in Rio, after the 50 m freestyle competition, a problem with the pool was found. In lane 1 there was a gentle 1.2 cm/s current flowing in the direction that the swimmers were going, while in lane 8 there was a current of the same speed but directed opposite to the swimmers' direction. Suppose a swimmer could swim the 50.0 m in 25.0 s in the absence of any current.
Part A: How would the time it took the swimmer to swim 50.0 m change in lane 1?
Part B: How would the time it took the swimmer to swim 50.0 m change in lane 8?
The time it would take for the swimmer to swim 50.0m would increase in lane 1, where there is a current flowing in the same direction of the swimmers. In lane 8, where there is a current flowing in the opposite direction, the time would decrease.
Explanation:In lane 1, where there is a 1.2 cm/s current flowing in the direction of the swimmers, the time it would take for the swimmer to swim 50.0 m would increase. This is because the current would push the swimmer backwards, making it harder for them to reach the finish line.
In lane 8, where there is a 1.2 cm/s current flowing in the opposite direction of the swimmers, the time it would take for the swimmer to swim 50.0 m would decrease. This is because the current would push the swimmer forward, helping them reach the finish line faster.
When light of wavelength 242 nm shines on a metal surface the maximum kinetic energy of the photoelectrons is 1.99 eV. What is the maximum wavelength (in nm) of light that will produce photoelectrons from this surface? (Use 1 eV = 1.602 ✕ 10^−19 J, e = 1.602 ✕ 10^−19 C, c = 2.998 ✕ 10^8 m/s, and h = 6.626 ✕ 10^−34 J · s = 4.136 ✕ 10^−15 eV · s as necessary.)
Answer:
Maximum wavelength will be [tex]3.96\times 10^{-7}m[/tex]
Explanation:
It is given wavelength [tex]\lambda =242nm=242\times 10^{-9}m[/tex]
Speed of light [tex]c=3\times 10^8m/sec[/tex]
Plank's constant [tex]h=6.6\times 10^{-4}Js[/tex]
So energy is equal to
[tex]E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{242\times 10^{-9}}=8.18\times 10^{-19}J[/tex]
Maximum kinetic energy is given
[tex]KE_{max}=1.99eV=1.99\times 1.6\times 10^{-19}=3.184\times 10^{-19}J[/tex]
Work function is equal to
[tex]w_0=E-KE_{max}=8.18\times 10^{-19}-3.184\times 10^{-19}=5\times 10^{-19}J[/tex]
[tex]\frac{hc}{\lambda _0}=5\times 10^{-19}[/tex]
[tex]\frac{6.6\times 10^{-34}\times 3\times 10^8}{\lambda _0}=5\times 10^{-19}[/tex]
[tex]\lambda _0=3.96\times 10^{-7}m[/tex]
(a) How long in seconds does it take a radio signal to travel 160 km from a transmitter to a receiving antenna? (b) We see a full Moon by reflected sunlight. How much earlier did the light that enters our eye leave the Sun? The Earth – Moon and Earth – Sun distances are 3.8x105 km and 1.5 × 108 km, respectively. (c) What is the round-trip travel time in seconds for light between Earth and a spaceship at a 3.6 × 108 km distance from Earth? (d) Suppose astronomers observe a supernova about 9300 light-years (ly) distant. How long ago in years did the explosion actually occur?
Explanation:
Given:
Distance between the transmitter and receiver = 160 km
Distance between Earth and Sun (D) = 150000000 km
Distance between Earth and Moon (D') = 380000 km
Distance between Earth and space ship (D") = 360000000 km
Distance between Earth and Supernova (d) = 9300 LY
We know that,
Speed of Light = 300000 km/sec
Lets us solve each problem now:
a) We know that the radio waves travle with speed of light. So the time(T) taken by the signal from transmitter to receiving antenna will be,
[tex]Time = \frac{Distance}{Speed}[/tex]
[tex]T = \frac{160}{300000}[/tex]
T = 0.53 millisecond
b) Time taken by sunlight to reach Earth is,
[tex]T_{1} = \frac{150000000}{300000} seconds[/tex]
T₁ = 500 seconds = 08 min 20 seconds
Time taken by light from Moon to reach Earth will be,
[tex]T_{2} = \frac{380000}{300000} seconds[/tex]
T₂ = 1.26 seconds
Let us assume that in the given situation Earth and Moon are at the same distance from the Sun. This means that Sun light will take the same time to reach Moon as it will take to reach Earth. So Sun light reaches the Moon in 500 seconds. After that it will be reflected towards Earth and will take 1.26 seconds more to reach Earth.
Thus the light left the Sun 501.26 seconds earlier.
c) Time taken by the light from space ship to reach Earth will be,
[tex]T = \frac{360000000}{300000} seconds[/tex]
T = 1200 seconds.
Thus round trip time will be = 2T = 2400 seconds
d) Lightyear is a unit of distance and is equal to the distance travelled by light in a year. If a supernova is at a distance of 9300 LY it means that light has taken 9300 years to reach Earth from that supernova. It means that the light we see today was actually generated from the supernova 9300 years ago.
It is evident that the explosion actually occurred 9300 years back.
We have calculated the times light and radio signals take to travel given distances at the speed of light; these times depend directly on the distances. For the observed supernova, we use the reverse calculation, determining the time light has traveled given the distance in light-years.
Explanation:(a) The radio signal is essentially light, so it travels at the speed of light, which is about 300,000 kilometers per second (3 × 10⁸ m/s). To find the time it takes to travel 160 km, we simply divide the distance by the speed: 160 km / 300000 km/s = 0.00053 seconds, or about 0.53 milliseconds.
(b) The light we see from the Moon has indeed traveled from the Sun, bounced off the Moon, and then hit our eyes. Therefore, it has traveled a distance of the Earth–Sun distance plus the Earth–Moon distance. Adding these distances, the light has traveled a total of 1.5 × 10⁸ km + 3.8 × 10⁵ km = 1.50038 × 10⁸ km. Dividing this distance by the speed of light gives the time this light has traveled: 1.50038 × 10⁸ km / 300000 km/s = 500.13 seconds, or about 8 minutes and 20 seconds.
(c) The round-trip travel time of light between Earth and a spaceship would be twice the time it would take light to make a one-way trip. The one-way trip time is calculated as the distance divided by the speed of light, which is 3.6 × 10⁸ km / 300000 km/s = 1200 seconds. Therefore, the round-trip travel time would be 2400 seconds or 40 minutes.
(d) A light-year is the distance that light travels in one year, so if a supernova is observed to be 9300 light-years away, this means the light from the supernova has been traveling for 9300 years before reaching Earth. Therefore, the supernova actually happened 9300 years ago.
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The constant pressure molar heat capacity, C_{p,m}C p,m , of nitrogen gas, N_2N 2 , is 29.125\text{ J K}^{-1}\text{ mol}^{-1}29.125 J K −1 mol −1 at 298\text{ K}298 K. Calculate the change in the internal energy when 20\text{ mol}20 mol of nitrogen gas at 298\text{ K}298 K is heated so that its temperature increases by 15.0^{\circ}\text{C}15.0 ∘ C. You may assume constant heat capacity over the temperature range.
Answer:
Explanation:
Constant pressure molar heat capacity Cp = 29.125 J /K.mol
If Cv be constant volume molar heat capacity
Cp - Cv = R
Cv = Cp - R
= 29.125 - 8.314 J
= 20.811 J
change in internal energy = n x Cv x Δ T
n is number of moles , Cv is molar heat capacity at constant volume , Δ T is change in temperature
Putting the values
= 20 x 20.811 x 15
= 6243.3 J.
are 1.8 mm apart is illuminated by a monochromatic coherent light source. A fringe pattern is observed on a screen 4.8 m from the slits. If there are 5.0 bright fringes/cm on the screen, what is the wavelength of the monochromatic light
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
A pair of narrow slits that are 1.8 mm apart is illuminated by a monochromatic coherent light source. A fringe pattern is observed on a screen 4.8 m from the slits. If there are 5.0 bright fringes/cm on the screen, what is the wavelength of the monochromatic light?
Given Information:
Distance from the double slits to the screen = D = 4.8 m
Double slit separation distance = d = 1.8 mm = 0.0018 m
Number of fringes = m = 5
Distance between fringes = y = 1 cm = 0.01 m
Required Information:
Wavelength of the monochromatic light = λ = ?
Answer:
Wavelength of the monochromatic light = λ = 7.5x10⁻⁷ m
Explanation:
The wavelength of the monochromatic light can be found using young's double slits formula,
λ = yd/mD
where d is the double slit separation distance, D is the distance from the double slits to the screen, y is the distance between bright fringes and m is number of fringes.
λ = 0.01*0.0018/5*4.8
λ = 0.00000075 m
λ = 7.5x10⁻⁷ m
Therefore, the wavelength of the monochromatic light is 7.5x10⁻⁷ m
Suppose a collection of quantum harmonic oscillators occupies the lowest 4 energy levels, and the spacing between levels is 0.04 eV. What is the complete emission spectrum for this system? That is, what photon energies will appear in the emissions? Include all energies, whether or not they fall in the visible region of the electromagnetic spectrum. Enter the photon energies in order of increasing energy.
a. Smallest photon energy = ____eV
b. Next larger photon energy = ____eV
c. Largest photon energy = ____eV
Answer:
a. Smallest photon energy = 0.04 eV
b. Next larger photon energy = 0.08 eV
c. Largest photon energy = 0.12 eV
Explanation:
Since the spacing between the levels is 0.04 even
The smallest photon energy, E= 0.04 eV
The next larger photon energy = 2E = 2×0.04 = 0.08 eV
Largest photon energy = 3E = 3×0.04 = 0.12
The relationships for the atomic transitions allowed to find all the transitions in the four-level system are:
The minimum energy is: E = 0.04 eV. Tthe following is: E = 0.08 eV. Tthe maximum energy is: E = 0.12 eV.
Atomic Transitions are the energy deference between two specific atomic levels.
[tex]\Delta E = E_f - E_i[/tex]
They indicate that we have 4 equally spaced states with an energy difference of 0.04 eV between each one.
In the attachment we can see a diagram of the system with an arrow indicating the possible transitions.
From level n = 2 to level n = 1 the energy is E₂₁ = 0.04 eV.
from level n = 3 to n = 1 the energy is E₃₁ = 0.08 eV.
all other transitions in the table.
Initial state final state Energy (eV)
4 1 0.12
3 1 0.08
2 1 0.04
4 2 0.08
3 2 0.04
4 3 0.04
These are all the possible transitions, we can see that some energies have several possible states, these currents that have several possibilities are called degenerate.
the energies are:
the minimum is: E = 0.04 eV.
the following is: E = 0.08 eV.
the maximum is E: = 0.12 eV.
In conclusion using the relationships for the atomic transitions we can find all the transitions in the four-level system are:
The minimum energy is: E = 0.04 eV The following is: E = 0.08 eV Tthe maximum energy is: E = 0.12 eV
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In a live fire exercise, an Army artillery team fires an artillery shell from a howitzer. The barrel of the howitzer makes a 60.0° angle above horizontal, and the speed of the shell upon exiting the barrel is 380 m/s. The shell hits a target on the side of a mountain 32.0 s after firing. Assuming the point where the shell exits the barrel to be the origin, and assuming as usual that the x-axis is horizontal and the y-axis is vertical, find the x and y coordinates, in meters, of the target.
Answer:
Explanation:
speed of shell =380 m /s
x - component = 380 cos 60
= 190 m /s
y- component = 380 sin 60
= 329 .1 m /s
time taken to hit target = 32 s
horizontal distance covered = horizontal velocity x time
= 190 x 32 = 6080 m
vertical distance covered
h = ut - 1/2 gt²
= 329.1 x 32 - .5 x 9.8 x 32²
= 10531.2 - 5017.6
= 5513.6 m .
x coordinate = 6080 m
y-coordinate = 5513.6 m
The current in a long solenoid of radius 5 cm and 17 turns/cm is varied with time at a rate of 5 A/s. A circular loop of wire of radius 7 cm and resistance 5 Ω surrounds the solenoid. Find the electrical current induced in the loop (in µA).
Answer:
The current is [tex]I =0.2mA[/tex]
Explanation:
From the question we are told that
The first radius is [tex]R_1 = 5cm = \frac{5}{100} = 0.05cm[/tex]
The number of turns is [tex]N = 17 \ turn/cm[/tex]
The current rate is [tex]\frac{dI}{dt} = 5 A/s[/tex]
The second radius is [tex]R_2 = 7cm = \frac{7}{100} = 0.07m[/tex]
The resistance is [tex]r = 5 \Omega[/tex]
Generally the magnetic flux induced in the solenoid is mathematically represented as
[tex]\O = B A[/tex]
Where is the magnetic field mathematically represented as
[tex]B = N \mu_o I[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a value of [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]
and A is the area mathematically represented as
[tex]A = \pi (R_2 - R_1)^2[/tex]
So
[tex]\O = N \mu I * \pi R^2[/tex]
Substituting values
[tex]\O = 17 * 4\pi *10^{-7} * \pi (7-5)^2I[/tex]
[tex]\O = 2.68*10^{-4}I[/tex]
The induced emf is mathematically represented as
[tex]\epsilon =- |\frac{d\O}{dt}|[/tex]
[tex]\epsilon = 2.68*10^{-4 } \frac{dI}{dt}[/tex]
substituting values
[tex]\epsilon =2.68 *10^{-4} * 5[/tex]
[tex]=1.3 *10^{-3} V[/tex]
From Ohm law
[tex]I = \frac{\epsilon }{r}[/tex]
Substituting values
[tex]I = \frac{1.3*0^{-3}}{5}[/tex]
[tex]I =0.2mA[/tex]
Which of these is a measure of force?
100 newtons
O 100 meters per second
100 watts
O 100 joules
Answer:
100 newtons
Explanation:
Recall force is the product of mass and acceleration from Newton's 2nd law given base units kgms^-2 which is equivalent to a newton( N)
A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225kg , and a 175-kg sack of gravel sits on the middle of it. The beam is oscillating in SHM, with an amplitude of 40.0cm and a frequency of 0.600cycles/s.
The sack of gravel falls off the beam when the beam has its maximum upward displacement. What is the frequency of the subsequent SHM of the beam?
What is the amplitude of the subsequent SHM of the beam?
If the gravel instead falls off when the beam has its maximum speed, what is the frequency of the subsequent SHM of the beam?
What is the amplitude of the subsequent SHM of the beam?
- **Frequency after gravel falls off at maximum displacement:**
[tex]\[ f' \approx 0.800 \text{ cycles/s} \][/tex]
- **Amplitude after gravel falls off at maximum displacement:**
[tex]\[ A' = 0.4 \text{ m} \][/tex]
- **Frequency after gravel falls off at maximum speed:**
[tex]\[ f' \approx 0.800 \text{ cycles/s} \][/tex]
- **Amplitude after gravel falls off at maximum speed:**
[tex]\[ A'' \approx 0.399 \text{ m} \][/tex]
Step 1
To analyze the subsequent simple harmonic motion (SHM) of the beam after the gravel falls off, we need to address the changes in the system's mass and how these changes affect the frequency and amplitude of the oscillations. Let's break down each part of the problem.
Initial System Details
- **Mass of the beam, [tex]\( m_{\text{beam}} \)[/tex]:** 225 kg
- **Mass of the sack of gravel, [tex]\( m_{\text{gravel}} \)[/tex]:** 175 kg
- **Total initial mass, [tex]\( m_{\text{total}} \):** \( 225 \text{ kg} + 175 \text{ kg} = 400 \text{ kg} \)[/tex]
- **Amplitude, A:** 40.0 cm = 0.4 m
- **Frequency, f:** 0.600 cycles/s
Step 2
The angular frequency [tex](\(\omega\))[/tex] is related to the frequency by:
[tex]\[ \omega = 2 \pi f \]\[ \omega = 2 \pi \times 0.600 \text{ s}^{-1} = 1.2 \pi \text{ s}^{-1} \][/tex]
Step 3
Spring Constant and Natural Frequency
For a system undergoing SHM, the angular frequency is given by:
[tex]\[ \omega = \sqrt{\frac{k_{\text{eff}}}{m}} \][/tex]
Where [tex]\( k_{\text{eff}} \)[/tex] is the effective spring constant of the system, and m is the mass. Rearranging for [tex]\( k_{\text{eff}} \)[/tex]:
[tex]\[ k_{\text{eff}} = \omega^2 m_{\text{total}} \]\[ k_{\text{eff}} = (1.2 \pi)^2 \times 400 \text{ kg} \]\[ k_{\text{eff}} = (1.44 \pi^2) \times 400 \text{ kg} \]\[ k_{\text{eff}} = 1.44 \times 9.8696 \times 400 \text{ kg} \]\[ k_{\text{eff}} \approx 5680 \text{ N/m} \][/tex]
Step 4
1. Frequency of Subsequent SHM after Gravel Falls Off at Maximum Upward Displacement
When the gravel falls off, the total mass reduces to the mass of the beam:
[tex]\[ m_{\text{beam}} = 225 \text{ kg} \][/tex]
The new angular frequency [tex](\(\omega'\))[/tex] is given by:
[tex]\[ \omega' = \sqrt{\frac{k_{\text{eff}}}{m_{\text{beam}}}} \]\[ \omega' = \sqrt{\frac{5680 \text{ N/m}}{225 \text{ kg}}} \]\[ \omega' \approx \sqrt{25.2444 \text{ s}^{-2}} \]\[ \omega' \approx 5.024 \text{ s}^{-1} \][/tex]
The new frequency [tex]\( f' \)[/tex] is:
[tex]\[ f' = \frac{\omega'}{2 \pi} \]\[ f' \approx \frac{5.024}{2 \pi} \]\[ f' \approx 0.800 \text{ cycles/s} \][/tex]
Step 5
2. Amplitude of Subsequent SHM after Gravel Falls Off at Maximum Upward Displacement
When the gravel falls off at the maximum upward displacement, the energy is conserved. The amplitude remains the same because the position where the gravel falls off doesn't change the energy stored in the system at that instant.
So, the amplitude of the subsequent SHM is still:
[tex]\[ A' = 0.4 \text{ m} \][/tex]
Step 6
3. Frequency of Subsequent SHM after Gravel Falls Off at Maximum Speed
The frequency of SHM depends on the system's mass and the effective spring constant. When the gravel falls off at maximum speed, the frequency calculation remains the same as the first part since the only factor affecting frequency is the mass change.
Thus, the frequency remains:
[tex]\[ f' \approx 0.800 \text{ cycles/s} \][/tex]
Step 7
4. Amplitude of Subsequent SHM after Gravel Falls Off at Maximum Speed
At maximum speed, the kinetic energy is at its maximum. When the gravel falls off, the kinetic energy initially stored in the combined system transfers entirely to the beam. The initial total kinetic energy is:
[tex]\[ KE = \frac{1}{2} m_{\text{total}} \omega^2 A^2 \][/tex]
After the gravel falls off, the new amplitude [tex]\( A'' \)[/tex] can be found using:
[tex]\[ KE = \frac{1}{2} m_{\text{beam}} \omega'^2 A''^2 \][/tex]
Since the kinetic energy is conserved:
[tex]\[ \frac{1}{2} m_{\text{total}} \omega^2 A^2 = \frac{1}{2} m_{\text{beam}} \omega'^2 A''^2 \]\[ m_{\text{total}} \omega^2 A^2 = m_{\text{beam}} \omega'^2 A''^2 \][/tex]
Solve for [tex]\( A'' \)[/tex]:
[tex]\[ A''^2 = \frac{m_{\text{total}} \omega^2 A^2}{m_{\text{beam}} \omega'^2} \]\[ A'' = \sqrt{\frac{m_{\text{total}} \omega^2 A^2}{m_{\text{beam}} \omega'^2}} \]\[ A'' = \sqrt{\frac{400 \times (1.2 \pi)^2 \times 0.4^2}{225 \times (5.024)^2}} \][/tex]
Substitute the values:
[tex]\[ A'' = \sqrt{\frac{400 \times 1.44 \pi^2 \times 0.16}{225 \times 25.2444}} \]\[ A'' = \sqrt{\frac{400 \times 1.44 \times 9.8696 \times 0.16}{225 \times 25.2444}} \]\[ A'' = \sqrt{\frac{900.9792}{5670.99}} \]\[ A'' \approx \sqrt{0.1589} \]\[ A'' \approx 0.399 \text{ m} \][/tex]
Thus, the new amplitude is approximately [tex]\( 0.399 \text{ m} \).[/tex]