Answer:
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
Explanation:
The decay of radioisotopes are represented by the following ordinary differential equation:
[tex]\frac{dm}{dt} = -\frac{t}{\tau}[/tex]
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]\tau[/tex] - Time constant, measured in seconds.
[tex]m[/tex] - Mass of the radioisotope, measured in grams.
The solution of this expression is:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]
Where [tex]m_{o}[/tex] is the initial mass of the radioisotope, measured in kilograms.
The ratio of current mass to initial mass is:
[tex]\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]
The time constant is now calculated in terms of half-life:
[tex]\tau = \frac{t_{1/2}}{\ln2}[/tex]
Where [tex]t_{1/2}[/tex] is the half-life of the radioisotope, measured in seconds.
Given that [tex]t_{1/2} = 88\,s[/tex], the time constant of the radioisotope is:
[tex]\tau = \frac{88\,s}{\ln 2}[/tex]
[tex]\tau \approx 126.957\,s[/tex]
Now, if [tex]\frac{m(t)}{m_{o}(t)} = \frac{1}{16}[/tex] and [tex]\tau \approx 126.957\,s[/tex], the time is:
[tex]t = -\tau \cdot \ln\frac{m(t)}{m_{o}}[/tex]
[tex]t = -(126.957\,s)\cdot \ln \frac{1}{16}[/tex]
[tex]t \approx 352\,s[/tex]
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
Suppose you want a telescope that would allow you to see distinguishing features as small as 3.5 km on the Moon some 384,000 km away. Assume an average wavelength of 550 nm for the light received.Required:What is the minimum diameter mirror on a telescope?
Explanation:
[tex]\theta=1.22 \frac{\lambda}{D}[/tex]
And, from equation ( 2 ), we get
[tex]\theta=\frac{x}{d}[/tex]
Thus,
[tex]\frac{x}{d} &=1.22 \frac{\lambda}{D}[/tex]
[tex]D &=1.22 \frac{\lambda d}{x}[/tex]
[tex]=1.22 \frac{550 \times 10^{-9} 3.84 \times 10^{8}}{5 \times 10^{3}}[/tex]
[tex]=0.0515 \mathrm{m}[/tex]
Thus, the diameter of the telescope's mirror that would allow us to see details as small as is
What is the force that attracts objects with mass toward each other?
Explanation:
gravitional force attracts objects with mass toward each other.
When light of wavelength 233 nm shines on a metal surface the maximum kinetic energy of the photoelectrons is 1.98 eV. What is the maximum wavelength (in nm) of light that will produce photoelectrons from this surface
Answer:
λmax = 372 nm
Explanation:
First we find the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
λ = Wavelength of Light = 233 nm = 2.33 x 10⁻⁷ m
c = speed of light = 3 x 10⁸ m/s
h = Planks Constant = 6.626 x 10⁻³⁴ J.s
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.33 x 10⁻⁷ m)
E = 8.5 x 10⁻¹⁹ J
Now, from Einstein's Photoelectric Equation:
E = Work Function + Kinetic Energy
8.5 x 10⁻¹⁹ J = Work Function + (1.98 eV)(1.6 x 10⁻¹⁹ J/1 eV)
Work Function = 8.5 x 10⁻¹⁹ J - 3.168 x 10⁻¹⁹ J
Work Function = 5.332 x 10⁻¹⁹ J
Since, work function is the minimum amount of energy required to emit electron. Therefore:
Work Function = hc/λmax
λmax = hc/Work Function
where,
λmax = maximum wavelength of light that will produce photoelectrons = ?
Therefore,
λmax = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5.332 x 10⁻¹⁹ J)
λmax = 3.72 x 10⁻⁷ m
λmax = 372 nm
An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9 000 m/s and an S-wave travels at 5 000 m/s. If P-waves are received at a seismic station 1.00 minute before an S-wave arrives, how far away is the earthquake center?
Assuming constant speeds, the P-wave covers a distance d in time t such that
9000 m/s = d/(60 t)
while the S-wave covers the same distance after 1 more minute so that
5000 m/s = d/(60(t + 1))
Now,
d = 540,000 t
d = 300,000(t + 1) = 300,000 t + 300,000
Solve for t in the first equation and substitute it into the second equation, then solve for d :
t = d/540,000
d = 300,000/540,000 d + 300,000
4/9 d = 300,000
d = 675,000
So the earthquake center is 675,000 m away from the seismic station.
A thick wire with a radius of 4.0 mm carries a uniform electric current of 1.0 A, distributed uniformly over its cross-section. At what distance from the axis of the wire, and greater than the radius of the wire, is the magnetic field strength equal to that at a distance 2.0 mm from the axis. distance
Answer:
8 mm
Explanation:
From the information given:
The Ampere circuital law can be used to estimate the magnetic field strength at two points when the distance is less than the radius and when the distance is greater than the radius.
when the distance is less than the radius ; we have:
[tex]B_1 = \dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2}[/tex]
when the distance is greater than the radius; we have:
[tex]B_2 = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
Equating both equations together ; we have :
[tex]\dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2} = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
[tex]\dfrac{1}{R}= \dfrac{r}{d^2}[/tex]
[tex]R= \dfrac{d^2}{r}[/tex]
where; d = radius of the wire and r = distance;
[tex]R =\dfrac{4^2}{2}[/tex]
[tex]R =\dfrac{16}{2}[/tex]
R = 8 mm
A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5 s. Determine 1. The acceleration of the car. 2. The distance it moves in the third second.
Answer:
Explanation:
Initial velocity , u = 30 m/s
final velocity , v = 10 m/s
time , t = 5 seconds
1. Acceleration = v - u / t
= 10 - 30 / 5
= -20 / 5
= - 4 m/s
Calculate the focal length (in m) of the mirror formed by the shiny bottom of a spoon that has a 3.40 cm radius of curvature. m (b) What is its power in diopters? D
Answer:
The power of the mirror in diopters is 58.8 D
Explanation:
Given;
radius of curvature of the spoon, R = 3.4 cm = 0.034 m
The focal length of a mirror is given by;
[tex]f = \frac{R}{2} \\\\f = \frac{0.034}{2} \\\\f = 0.017 \ m[/tex]
The focal length of the mirror is 0.017 m
(b) The power of the mirror is given by;
[tex]P = \frac{1}{f}[/tex]
where;
P is the power of the mirror
f is the focal length
[tex]P = \frac{1}{f}\\\\P= \frac{1}{0.017}\\\\P = 58.8 \ D[/tex]
Thus, the power of the mirror in diopters is 58.8 D
The positron has the same mass as an electron, with an electric charge of +e. A positron follows a uniform circular motion of radius 5.03 mm due to the force of a uniform magnetic field of 0.85 T. How many complete revolutions does the positron perform If it spends 2.30 s inside the field? (electron mass = 9.11 x 10-31 kg, electron charge = -1.6 x 10-19 C)
Answer:
5.465 × 10^10 revolutions
Explanation:
Formula for Magnetic Field = m. v/ q . r
M = mass of electron = mass of positron = 9.11 x 10^-31 kg,
radius of the positron = 5.03 mm
We convert to meters.
1000mm = 1m
5.03mm = xm
Cross multiply
x = 5.03/1000mm
x = 0.00503m
q = Electric charge = -1.6 x 10^-19 C
Magnetic field (B) = 0.85 T
Speed of the positron is unknown
0.85 = 9.11 x 10^-31 kg × v/ -1.6 x 10^-19 C × 0.00503
0.85 × 1.6 x 10^-19 C × 0.00503 = 9.11 x 10^-31 kg × v
v = 0.85 × -1.6 x 10^-19 C × 0.00503/9.11 x 10^-31 kg
v = 6.8408 ×10-22/ 9.11 x 10^-31 kg
v = 750911086.72m/s
Formula for complete revolutions =
Speed × time / Circumference
Time = 2.30s
Circumference of the circular path = 2πr
r =0.00503
Circumference = 2 × π × 0.00503
= 0.0316044221
Revolution = 750911086.72 × 2.30/0.0316044221
= 1727095499.5/0.0316044221
= 546541562294 revolutions
Approximately = 5.465 × 10^10 revolutions
A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish
Answer:
Apparent depth (Da) = 60.15 cm (Approx)
Explanation:
Given:
Distance from fish (D) = 80 cm
Find:
Apparent depth (Da)
Computation:
We know that,
Refractive index of water (n2) = 1.33
So,
Apparent depth (Da) = D(n1/n2)
Apparent depth (Da) = 80 (1/1.33)
Apparent depth (Da) = 60.15 cm (Approx)
The apparent depth of the fish is 60 cm.
To calculate the apparent depth of the fish, we use the formula below.
Formula:
R.F(water) = Real depth(D)/Apparent depth(D')R.F = D/D'.................... Equation 1Where:
R.F = Refractive index of waterMake D' The subject of the equation.
D' = D/R.F................... Equation 2From the question,
Given:
D = 80 cmR.F = 1.333Substitute these values into equation 2
D' = 80/1.33D' = 60.01D' = 60 cmHence, the apparent depth of the fish is 60 cm
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what effect does decreasing the field current below its nominal value have on the speed versus voltage characteristic of a separately excited dc motor
Answer
The effect is that it Decreases the field current IF and increases slope K1
An object is inside a room that has a constant temperature of 289 K. Via radiation, the object emits three times as much power as it absorbs from the room. What is the temperature (in kelvins) of the object
Answer:
T_object = 380.35 K
Explanation:
From Stefan–Boltzmann law, the power output is given by the formula:
P = σAT⁴
where;
σ is Stefan-Boltzmann constant
A is area of the radiating surface.
T is temperature of the body
Now, we are told that the power the object emitted is 3 times the power absorbed from the room.
Thus, we have;
P_e = 3P_a
Where P_e is power emitted and P_a is power absorbed.
So, we have;
σA(T_object)⁴ = 3σA (T_room)⁴
σA will cancel out to give;
(T_object)⁴ = 3(T_room)⁴
We are given T_room = 289 K
Thus;
(T_object)⁴ = 3 × 289⁴
(T_object) = ∜(3 × 289⁴)
T_object = 380.35 K
If a soap bubble is 130 nmnm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated normally by white light
Answer:
The question is not complete, here is the other part.
Assume that n = 1.36.
Express your answer to three significant figures and include the appropriate units.
λ = 707.2nm
Explanation:
n = 1.36
t = 130
2 n t= (m+1/2) λ
To solve for λ.
λ = 2 n t ÷ m+1/2
λ = 2 × 1.36 × 130 ÷ m +1/2
λ =
If m= 0
λ= 353.6 ÷ 0+ 1/2
λ = 353.6 × 2
= 707.2nm
A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field F(x, y, z).
Required:
Find the work done.
Answer:
the net work is zero
Explanation:
Work is defined by the expression
W = F. ds
Bold type indicates vectors
In this problem, the friction force does not decrease, therefore it will be zero.
Consequently for work on a closed path it is zero.
The work in going from the initial point (0, 0, 0) to the end of each segment is positive and when it returns from the point of origin the angle is 180º, therefore the work is negative, consequently the net work is zero
A car accelerates uniformly from rest and reaches a speed of 22.7 m/s in 9.02 s. Assume the diameter of a tire is 58.5 cm. (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. rev (b) What is the final angular speed of a tire in revolutions per second? rev/s
(a) The car is undergoing an acceleration of
[tex]a=\dfrac{22.7\frac{\rm m}{\rm s}-0}{9.02\,\mathrm s}\approx2.52\dfrac{\rm m}{\mathrm s^2}[/tex]
so that in 9.02 s, it will have covered a distance of
[tex]x=\dfrac a2(9.02\,\mathrm s)^2\approx102\,\mathrm m[/tex]
The car has tires with diameter d = 58.5 cm = 0.585 m, and hence circumference π d ≈ 1.84 m. Divide the distance traveled by the tire circumference to determine how many revolutions it makes:
[tex]\dfrac{102\,\mathrm m}{1.84\,\mathrm m}\approx55.7\,\mathrm{rev}[/tex]
(b) The wheels have average angular velocity
[tex]\omega=\dfrac{\omega_f+\omega_i}2=\dfrac{\theta_f-\theta_i}{\Delta t}[/tex]
where [tex]\omega[/tex] is the average angular velocity, [tex]\omega_i[/tex] and [tex]\omega_f[/tex] are the initial and final angular velocities (rev/s), [tex]\theta_i[/tex] and [tex]\theta_f[/tex] are the initial and final angular displacements (rev), respectively, and [tex]\Delta t[/tex] is the duration of the time between initial and final measurements. The second equality holds because acceleration is constant.
The wheels start at rest, so
[tex]\dfrac{\omega_f}2=\dfrac{55.7\,\rm rev}{9.02\,\rm s}\implies\omega_f\approx12.4\dfrac{\rm rev}{\rm s}[/tex]
A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the circle from which light escapes from the liquid into the air above the surface
Answer:
The radius is [tex]r = 3.1905 \ m[/tex]
Explanation:
From the question we are told that
The distance beneath the liquid is [tex]d = 2.70 \ m[/tex]
The refractive index of the liquid is [tex]n_i = 1.31[/tex]
Now the critical value is mathematically represented as
[tex]\theta = sin ^{-1} [\frac{1}{n_i} ][/tex]
substituting values
[tex]\theta = sin ^{-1} [\frac{1}{131} ][/tex]
[tex]\theta = 49.76^o[/tex]
Using SOHCAHTOA rule we have that
[tex]tan \theta = \frac{ r}{d}[/tex]
=> [tex]r = d * tan \theta[/tex]
substituting values
[tex]r = 2.7 * tan (49.76)[/tex]
[tex]r = 3.1905 \ m[/tex]
Based on The MOHS hardness Scale, which mineral could be scratched by a penny but not by a fingernail
A. Fluorite
B. Calcite
C. Gypsum
D. Talc
The correct answer is B. Calcite
Explanation:
Mohs hardness scale indicates the hardness of minerals using a scale from 1 to 10 as well as defining the objects or tools that can be used to scratch the minerals. These two features of minerals are shown in the table of the image. About this, it is shown gypsum and talc can be scratched by just a fingernail, considering minerals with a hardness of 2.5 or below can be scratched by a fingernail. In the case of calcite that has a hardness of 3, this cannot be scratched by a fingernail, but it can be scratched by a penny, which works for minerals with a hardness of 3.5 or below. Thus, the correct answer is Calcite.
Please help!
Much appreciated!
Answer:
F = 2.7×10¯⁶ N.
Explanation:
From the question given:
F = (9×10⁹ Nm/C²) (3.2×10¯⁹ C × 9.6×10¯⁹ C) /(0.32)²
Thus we can obtain the value value of F by carrying the operation as follow:
F = (9×10⁹) (3.2×10¯⁹ × 9.6×10¯⁹) /(0.32)²
F = 2.7648×10¯⁷ / 0.1024
F = 2.7×10¯⁶ N.
Therefore, the value of F is 2.7×10¯⁶ N.
For an object to move, a(n) _______ force must be applied. Question 1 options: Balanced Unbalanced
Answer:
Unbalenced
Explanation:
when balenced forces are applied to an object there is no motion. When you apply unbalenced force the object you are applying the force to will move in the opposite direction of the force.
Answer:
im pretty sure it unbalenced
Explanation:
i just am
A wire of 5.8m long, 2mm diameter carries 750ma current when 22mv potential difference is applied at its ends. if drift speed of electrons is found then:_________.
(a) The resistance R of the wire(b) The resistivity p, and(c) The number n of free electrons per unit volume.
Explanation:
According to Ohms Law :
V = I * R
(A) R (Resistance) = 0.022 / 0.75 = 0.03 Ohms
Also,
[tex]r = \alpha \frac{length}{area} = \alpha \frac{5.8}{3.14 \times 0.001 \times 0.001} [/tex]
(B)
[tex] \alpha(resistivity) = 1.62 \times {10}^{ - 8} [/tex]
Drift speed is missing. It is given as;
1.7 × 10^(-5) m/s
A) R = 0.0293 ohms
B) ρ = 1.589 × 10^(-8)
C) n = 8.8 × 10^(28) electrons
This is about finding, resistance and resistivity.
We are given;Length; L = 5.8 m
Diameter; d = 2mm = 0.002 m
Radius; r = d/2 = 0.001 m
Voltage; V = 22 mv = 0.022 V
Current; I = 750 mA = 0.75 A
Area; A = πr² = 0.001²π
Drift speed; v_d = 1.7 × 10^(-5) m/s
A) Formula for resistance is;R = V/I
R = 0.022/0.75
R = 0.0293 ohms
B) formula for resistivity is given by;ρ = RA/L
ρ = (0.0293 × 0.001²π)/5.8
ρ = 1.589 × 10^(-8)
C) Formula for current density is given by;J = n•e•v_d
Where;
J = I/A = 0.75/0.001²π A/m² = 238732.44 A/m²
e is charge on an electron = 1.6 × 10^(-19) C
v_d = 1.7 × 10^(-5) m/s
n is number of free electrons per unit volume
Thus;
238732.44 = n(1.6 × 10^(-19) × 1.7 × 10^(-5))
238732.44 = (2.72 × 10^(-24))n
n = 238732.44/(2.72 × 10^(-24))
n = 8.8 × 10^(28)
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Your favorite radio station broadcasts at a frequency of 91.5 MHz with a power of 11.5 kW. How many photons does the antenna of the station emit in each second?
Answer:
Number of photons emit per second = 1.9 × 10²⁹ (Approx)
Explanation:
Given:
Frequency = 91.5 MHz
Power = 11.5 Kw = 11,500 J/s
Find:
Number of photons emit per second
Computation:
Total energy with frequency (E) = hf
Total energy with frequency (E) = 6.626×10⁻³⁴ × 91.5×10⁶
Total energy with frequency (E) = 6.06×10⁻²⁶ J
Number of photons emit per second = 11,500 / 6.06×10⁻²⁶
Number of photons emit per second = 1897.689 × 10²⁶
Number of photons emit per second = 1.9 × 10²⁹ (Approx)
In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The centers of the first-order bright blue fringes lie at the outer edges of a screen that is located 0.497 m away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in what direction (toward or away from the slits) should the screen be moved, so that the centers of the first-order bright orange fringes just appear on the screen
Answer:
0.5639m
Explanation:
For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ
=sin⁻¹(mλ/d)
mλ /d =y/L
for the first order,
y= mλL/d
For ratio separation y₀/yD=1 and d= 1
y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]
1=λ ₀L₀/λD.LD.
λD.LD= λ ₀L₀
L₀= λD.LD/ λ ₀..............(1)
Then substitute the given values into (1) we have
L₀=471 *0.497/611
= 0.3831m
Distance by which the screen has to be moved towards the slit is
LD- Lo
0.947-0.3831= 0.5639m
Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r = 0.5 m and r = 1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?
The direction for each field vector is perpendicular to equipotential lines.
Take a snapshot of the simulation showing equipotential lines and paste to a word document.
....................
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
We have that for the Question, it can be said that the average induced emf in the coil is
E=0.028565V
From the question we are told
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?
Generally the equation for the Average emf induced is mathematically given as
[tex]Emf_a=-NA\frac{dB}{dt}\\\\Where\\\\Area\\\\a=\pir^2\\\\a=\pi(0.056)^2\\\\a=0.00985\\\\[/tex]
Hence
[tex]dB=0.24-0.53\\\\dB=-0.29T[/tex]
Therefore
[tex]E=-\frac{1*0.00985*-0.29 }{0.10}[/tex]
E=0.028565V
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A transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary
Answer:
12AExplanation:
Formula for calculating the relationship between the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:
[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex] where;
Vs and Vp are the emf in the secondary and primary coil respectively
Ns and Np are the number if turns in the secondary and primary coil respectively
Ip and Is are the currents in the secondary and primary coil respectively
Since the are all equal to each other, then we can equate any teo of the expression as shown;
[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
Given parameters
Np = 500-turns
Ns = 2000-turns
Is = 3.0Amp
Required
Current in the primary coil (Ip)
Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]
[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]
Hence the current in the primary coil is 12Amp
To a person swimming 0.80 m below the water surface of a swimming pool, the diving board directly overhead appears to be a height of 5.20 m above the swimmer. What is the actual height of the diving board above the water surface
Answer:
The actual height is 3.308 m.
Explanation:
The person is swimming below the water surface at distance = 0.80 m
The height of the diving board appears at a distance or height = 5.20 m
Now we have to find the actual distance of the diving board from the water surface.
We know the refractive index of water is 1.33.
Therefore, the actual height = (Distance that appears – distance below the water surface) / Refractive index.
The actual height = ( 5.20 - 0.80 ) / 1.33 = 3.308 m
A cart rolls 2 m to the right then rolls back 1 m to the left.
a. What is the total distance rolled by the cart?
Explanation:
It is given that,
Distance covered by the cart to the right is 2 m
Distance covered by the cart to the left is 1 m
We need to find the total distance rolled by the cart. Total distance is equal to the sum of the distances covered by an object. It does depend on the direction.
Total distance = 2 m + 1 m
D = 3 m
The cart rolled to a total distance of 3 m.
Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.)
(a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.97 cm, find the wavelength of the incident light.
(b) At what angle does the next set of bright fringes appear?
Answer:
a
[tex]\lambda = 1.667 nm[/tex]
b
[tex]\theta = 0.8681^o[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]
The is distance of the screen from the slit is [tex]D = 2.60 \ m[/tex]
The distance between the central bright fringe and either of the adjacent bright [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]
Generally the condition for constructive interference is
[tex]d sin \tha(\theta ) = n \lambda[/tex]
From the question we are told that small-angle approximation is valid here.
So [tex]sin (\theta ) = \theta[/tex]
=> [tex]d \theta = n \lambda[/tex]
=> [tex]\theta = \frac{n * \lambda }{d }[/tex]
Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes
Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as
[tex]y = D * sin (\theta )[/tex]
From the question we are told that small-angle approximation is valid here.
So
[tex]y = D * \theta[/tex]
=> [tex]\theta = \frac{ y}{D}[/tex]
So
[tex]\frac{n * \lambda }{d } = \frac{y}{D}[/tex]
[tex]\lambda =\frac{d * y }{n * D}[/tex]
substituting values
[tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]
[tex]\lambda = 1.667 *10^{-6}[/tex]
[tex]\lambda = 1.667 nm[/tex]
In the b part of the question we are considering the next set of bright fringe so n= 2
Hence
[tex]dsin (\theta ) = n \lambda[/tex]
[tex]\theta = sin^{-1}[\frac{ n * \lambda }{d} ][/tex]
[tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]
[tex]\theta = 0.8681^o[/tex]
The entropy of any substance at any temperature above absolute zero is called the: Select the correct answer below:
a. absolute entropy
b. Third Law entropy
c. standard entropy
d. free entropy
e. none of the above
Answer:
b. Third Law entropy
Explanation:
Third law entropy: In physics, the term "third law entropy" or "the third law of thermodynamics" states that the specific entropy of a particular system at "absolute zero" is considered as a "well-defined constant". It occurs because any system at "zero temperature" tends to exists or persists in its "ground state" in order for the entropy to be determined or described only by the "degeneracy" of the given ground state.
In the question above, the correct answer is option b.
A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 7.4 m/s2. For what value of the coefficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?
Answer:
The value is [tex]\mu = 0.76[/tex]
Explanation:
From the question we are told that
The acceleration is [tex]a = 7.4 \ m /s^2[/tex]
Generally the force by which the truck bed (truck) is moving with is mathematically represented as
[tex]F = ma[/tex]
Now for the truck cabinet to slip from the truck bed then the frictional force between the truck cabinet is equal the force by which the the truck bed is moving with that is
[tex]F_f = F[/tex]
Here [tex]F_f[/tex] is the frictional force which is mathematically represented as
[tex]F_f = \mu * m * g[/tex]
substituting into above equation
[tex]\mu * m * g = ma[/tex]
=> [tex]\mu = \frac{a}{g}[/tex]
substituting values
[tex]\mu = \frac{ 7.4 }{ 9.8}[/tex]
[tex]\mu = 0.76[/tex]
How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. the line density of the grating were doubled?
Answer:
a) the distance between the interference fringes is reduced by half
b) the distance between stripes is doubled
Explanation:
Interference experiments constructive interference is described by the expression
d sin θ = m λ
let's use trigonometry to find the distance between the interference fringes
tan θ= y / L
dodne y is the distance from the central maximum, L the distance from the slit to the observation screen. In general these experiments are carried out at very small angles
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ / d L
a) it asks us when the screen doubles its distance
L ’= 2 L
subtitute in the equation
y ’= m λ / (d 2L)
y ’=( m λ / d L) /2
y ’= y / 2
the distance between the interference fringes is reduced by half
b) the density of the network doubles
if the density doubles in the same distance there are twice as many slits, so the distance between them is reduced by half
d ’= d / 2
we substitute
y ’= m λ (L d / 2)
y ’= m λ / (L d) 2
y ’= y 2
the distance between stripes is doubled
