Explanation:
The total displacement is 3 miles
What is displacement?
Simply put it is the sum total of the changes in position of an object from it initial position to it final position, It is a vector quantity and has a direction and magnitude.
In our example the displacement the total points of movement can be annotated as
From A to B to C
From A to B= 1 mile
From B to C= 1 miles
Displacement is 1 + 2= 3 miles
A man starts from rest and accelerates at 4.00 m/s2. If he covers a distance of 525 m, how long does he accelerate?
Answer:
16.2 s
Explanation:
Given:
Δx = 525 m
v₀ = 0 m/s
a = 4.00 m/s²
Find: t
Δx = v₀ t + ½ at²
525 m = (0 m/s) t + ½ (4.00 m/s²) t²
t = 16.2 s
A 11.0kg bucket is lowered vertically by a rope in which there is 164N of tension at a given instant.
Determine the magnitude of the acceleration of the bucket.
Express your answer to two significant figures and include the appropriate units.
Answer:
The magnitude of the acceleration is 5.11 m/s²
Explanation:
Given;
Tension on the rope, T = 164 N
mass of the bucket, m = 11 kg
Weight of the rope is given by;
W = mg = 11 x 9.8 = 107.8 N
According to Newton's second law of motion, the tension on the rope is given by;
T = W - ma
ma = W - T
ma = 107.8N - 164N
ma = -56.2 N
a = -56.2 / m
a = -56.2 / 11
a = -5.11 m/s²
The magnitude of the acceleration is 5.11 m/s²
For the microscope to be in focus, how far should the objective lens be from the specimen?
Answer:
p ≈ f_ objective Therefore for the object to be in focus it must be close to the focal length
Explanation:
A microscope is an optical instrument that uses two lenses, or a long focal length objective lens that forms a real image of the object and an eyepiece that forms a virtual image of the object. Therefore for the object to be in focus it must be close to the focal length
p ≈ f_ objective
p distance objetive
A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stretched to a tension of 122 N. During what time interval will a transverse wave travel the entire length of the two wires
Answer:
The time taken is [tex]t = 0.356 \ s[/tex]
Explanation:
From the question we are told that
The length of steel the wire is [tex]l_1 = 31.0 \ m[/tex]
The length of the copper wire is [tex]l_2 = 17.0 \ m[/tex]
The diameter of the wire is [tex]d = 1.00 \ m = 1.0 *10^{-3} \ m[/tex]
The tension is [tex]T = 122 \ N[/tex]
The time taken by the transverse wave to travel the length of the two wire is mathematically represented as
[tex]t = t_s + t_c[/tex]
Where [tex]t_s[/tex] is the time taken to transverse the steel wire which is mathematically represented as
[tex]t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ][/tex]
here [tex]\rho_s[/tex] is the density of steel with a value [tex]\rho_s = 8920 \ kg/m^3[/tex]
So
[tex]t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]
[tex]t_s = 0.235 \ s[/tex]
And
[tex]t_c[/tex] is the time taken to transverse the copper wire which is mathematically represented as
[tex]t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ][/tex]
here [tex]\rho_c[/tex] is the density of steel with a value [tex]\rho_s = 7860 \ kg/m^3[/tex]
So
[tex]t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]
[tex]t_c =0.121[/tex]
So
[tex]t = t_c + t_s[/tex]
[tex]t = 0.121 + 0.235[/tex]
[tex]t = 0.356 \ s[/tex]
What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV. (in nm)
Answer: Wavelength of 424 nm can produce photoelectrons from silver
Explanation:
[tex]\phi=h\times \nu_o=\frac{hc}{\lambda}[/tex]
[tex]\phi[/tex] = work function = energy of photon
h = Planck's constant = [tex]6.63\times 10^{-34}Js[/tex]
[tex]\nu_0[/tex] = frequency of the metal
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] =longest wavelength of the radiation
Now put all the given values in the above formula, we get the wavelength of the photons.
[tex]\lambda=\frac{hc}{\phi}[/tex]
[tex]\lambda=\frac{6.63\times 10^{-34}\times 3\times 10^8m/s}{2.93\times 1.6\times 10^{-19}J}[/tex] ( as 1ev=[tex]1.6\times 10^{-19}J[/tex] )
[tex]\lambda=4.24\times 10^{-7}m[/tex]
[tex]1nm=10^{-9}m[/tex]
[tex]\lambda=424nm[/tex]
Here, wavelength of 424 nm can produce photoelectrons from silver
A 1500-kg car accelerates from 0 to 25 m/s in 7.0s with negligible friction and air resistance. What is the average power delivered by the engine? (1 hp 746 W)
Answer:
90 hp
Explanation:
Power = work / time
P = ½ (1500 kg) (25 m/s)² / 7.0 s
P = 67,000 W
P = 90 hp
2. A 15 kg mass fastened to the end of a steel wire of un-
stretched length 0.5 m is whirled in a vertical circle with an
angular velocity of 2 rev/s at the bottom of the circle. The cross
section of the wire is 0.02 cm2. Calculate the elongation of the
wire when the weight is at the lowest point of the path. Steel
has Y.M.= 2.0 x 1011 Pa. [1.66mm]
Explanation:
Elongation of the wire is:
ΔL = F L₀ / (E A)
where F is the force,
L₀ is the initial length,
E is Young's modulus,
and A is the cross sectional area.
ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)
ΔL = T (1.25×10⁻⁶ m/N)
T = (80,000 N/m) ΔL
Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.
Sum of forces in the centripetal direction:
∑F = ma
T − mg = mv²/r
T − mg = mω²r
T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)
T − 147 N = (2368.7 N/m) (0.5 m + ΔL)
Substitute:
(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)
(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL
(797631.3 N/m) ΔL = 1331.35 N
ΔL = 0.00167 m
ΔL = 1.67 mm
Define fluid flow. Write five difference between uniform and non uniform flow.
Answer:
Fluid Flow is a part of fluid mechanics and deals with fluid dynamics. Fluids such as gases and liquids in motion are called fluid flow. It involves the motion of a fluid subjected to unbalanced forces. This motion continues as long as unbalanced forces are applied.
Difference:
Whereas in real fluids velocity varies across the section. But when the size and shape of cross section are constant along the length of channels under consideration, the flow is said to be uniform. A non-uniform flow is one in which velocity is not constant at a given instant.
"Find the change in gravitational potential energy that the box undergoes as it rises to its final height."
Explanation:
Given that,
Mass of a box is 2.6 kg
(1) We need to find the work done on the box by this force as it is pushed up the 5.00-m ramp to a height of 3.00 m.
It means that the position of the object is 3 m i.e. h = 3 m
Work done = Fd
= mgh
So,
[tex]W=2.6\times 9.8\times 3\\\\W=76.44\ J[/tex]
(2) Now the gravitational potential energy that the box undergoes as it rises to its final height is equal to the work done by the box. So,
Change in potential energy = 76.44 J
An airplane is in level flight over Antarctica, where the magnetic field of the earth is mostly directed upward away from the ground. As viewed by a passenger facing toward the front of the plane, is the left or the right wingtip at higher potential? Does your answer depend on the direction the plane is flying?
Answer:
It does not depend on direction of plane and the left windtips more potential
Explanation:
Because if the Fleming right hand rule is applied the the right han is pointed in the direction of flight, and the fingers are curled in the direction of the magnetic lines. Thus , the lines are vertical and so are pointing down, thus by the right hand rule, the electrons move to the left hand side of the plane, although the potentials are equal and opposite in direction
A cyclist is standing still at the top of a hill and then begins to coast down the hill. The mass of the cyclist and bicycle is 64 kg total. The cyclist’s gravitational potential energy is converted into kinetic energy with an efficiency of 52%. What is her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started? Show all your work.
Her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started would be 10.1 m/s
What is mechanical energy?Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total potential energy stored energy in the system which is represented by total potential energy.
As given in the problem A cyclist is standing still at the top of a hill and then begins to coast down the hill. The mass of the cyclist and bicycle is 64 kg in total. The cyclist’s gravitational potential energy is converted into kinetic energy with an efficiency of 52%
The potential energy is getting converted into kinetic energy with an efficiency of 52 %
1/2mv² = 0.52 (mgh)
v²= 1.04gh
v = √(1.04gh)
v= √(1.04×9.81×10)
v = 10.1 m/s
Thus, her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started would be 10.1 m/s
Learn more about mechanical energy from here
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what single property was the most important in jesseca's material
Answer:
Jesseca wanted to create a material that reflected most of the light that fell on it.
Explanation: The Graphite was the material in the passage that had reflected most of the light.
A scientist claims that a certain chemical will make fabric waterproof. Which
option describes a controlled experiment that will produce evidence that will
support or refute her claim?
Answer: She adds different amounts of the chemical to the material and then puts them in the water
Explanation:
Answer: One group of fabric is treated with the chemical, and the other group is not. Then each group is exposed to water.
Explanation:
A resistor is connected across an oscillating emf. The peak current through the resistor is 2.0 A. What is the peak current if:
a. The resistance R is doubled?
b. The peak emf εo is doubled?
c. The frequency ω is doubled?
Answer:
(a) When the resistance R is doubled, I = 1 A
(b) When the peak emf εo is doubled, I = 4 A
(c) When the frequency ω is doubled, I = 2 A
Explanation:
Given;
peak current through the resistor, I = 2.0 A
According to ohms law the peak current through the circuit is given by;
[tex]I = \frac{V}{R}[/tex]
(a) When the resistance R is doubled;
[tex]I = \frac{V_R}{R} \\\\I_1R_1 = I_2R_2\\\\I_2 = \frac{I_1R_1}{R_2} \\\\I_2 = \frac{2*R_1}{2R_1} \\\\I_2 = 1 \ A[/tex]
(b)When the peak emf εo is doubled
[tex]I = \frac{V}{R} = \frac{\epsilon_o}{R} \\\\R = \frac{\epsilon_ o}{I} \\\\\frac{\epsilon_ o_1}{I_1} = \frac{\epsilon_ o_2}{I_2} \\\\I_2 = \frac{\epsilon_ o_2 *I_1}{\epsilon _o_1} \\\\I_2 = \frac{2 \epsilon_ o_1 *2}{\epsilon _o_1} \\\\I_2 = 4 \ A[/tex]
(c) When the frequency ω is doubled
Peak current through resistor is independent of frequency
I₂ = 2.0 A
A 50g marble is moving at 2m/s when it strikes a 20g marble at rest. Immediately after the collision, the 50g ball is moving at 1m/s. Is this an elastic collision
Answer:
Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.
Explanation:
Given;
mass of first marble, m₁ = 50g = 0.05 kg
initial velocity of the first marble, u₁ = 2 m/s
mass of second marble, m₂ = 20 g = 0.02 kg
initial velocity of the second marble, u₂ = 0
final velocity of the first marble, v₁ = 1 m/s
Let the final velocity of the second marble, = v₂
Determine the final velocity of the second marble by applying principle of momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.05 x 2 + 0.02 x 0 = 0.05 x 1 + 0.02v₂
0.1 = 0.05 + 0.02v₂
0.02v₂ = 0.1 - 0.05
0.02v₂ = 0.05
v₂ = 0.05 / 0.02
v₂ = 2.5 m/s
During inelastic collision both objects will move at the same velocity after collision.
During elastic collision both objects will move at different velocities after collision.
Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.
If you are pushing on a crate on a frictionless surface in one direction, and your friend is pushing on the crate in the opposite direction with an equal amount of force. Which of the following statements is the most accurate? a. The crate will not move as the forces cancel each other out b. Because the surface is frictionless, the crate will always move regardless of who is pushing c. The crate can continue to move, but it will move at a constant velocity d. The net force is towards the direction that you are pushing, as you started the crate's motion
Answer:
Its not A..
Explanation:
I chose A - was incorrect
g Radiation of an unknown wavelength is used in a photoelectric effect experiment on a sodium surface. The maximum kinetic energy of the observed electrons is 0.7 eV. What is the wavelength of the light
Answer:
λ = 4.1638 10⁻⁷ m
Explanation:
The photoelectric effect was explained by Einstein assuming that the radiation acts like particles and the equation that describes the process is
K = h f -Ф
where K is the kinetic energy of the emitted electrons, hf the energy of the photons according to Planck's equation and Ф the work function of the material
In this case they give us the kinetic energy of the electrons
K = 0.7 eV
The sodium work function is tabulated Ф = 2.28 eV
Let's find the frequency of the photons
f = (K + Ф) / h
Planck's constant is
h = 6.626 10⁻³⁴ J s (1 eV / 1.6 10⁻¹⁹ J) = 4.136 10⁻¹⁵ eV s
f = (0.7 + 2.28) / 4.136 10⁻¹⁵
f = 7.2050 10¹⁴ Hz
let's find the wavelength using the relationship between speed and frequency and wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 7.205 10¹⁴
λ = 4.1638 10⁻⁷ m
A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm.
Required:
What is the gauge pressure at the bottom of the cylinder?
Answer:
Pressure, P = 3724 Pa
Explanation:
Given that,
Depth of water, [tex]h_w=20\ cm =0.2\ m[/tex]
Depth of oil, [tex]h_o=40-20=20\ cm=0.2\ m[/tex]
The density of water, [tex]d_w=1000\ kg/m^3[/tex]
The densinty of oil, [tex]d_o=900\ kg/m^3[/tex]
We need to find the gauge pressure at the bottom of the cylinder. So, total pressure is equal to :
[tex]P=d_wgh_w+d_ogh_o\\\\P=(d_wh_w+d_oh_o)g\\\\P=(1000\times 0.2+900\times 0.2)\times 9.8\\\\P=3724\ Pa[/tex]
So, the gauge pressure at the bottom of the cylinder is 3724 Pa.
The half-life of element X is 20 years. If there are 48 g initially a) How much is there after 80 years
Answer:
After 80 years there will be 3 g of element X remaining
Explanation:
Given;
the half life of element X = 20 year
initial mass of element X = 48 g
a) How much is there after 80 years
0 year --------------------------> 48 g
20 years -----------------------> (48g / 2) = 24 g
40 years ------------------------> 12 g
60 years ------------------------> 6 g
80 years --------------------------> 3 g
Therefore, after 80 years there will be 3 g of element X remaining.
A part of a circuit contains an inductor. The current through the inductor is changing uniformly from 2.40A to 0.30A over the course of 1.75s. If the EMF ( voltage change ) from one side of the inductor to the other is 5.70 Volts, what is the value of the inductance
Answer:
The value of the inductance is 4.75 H
Explanation:
Given;
initial current through the inductor, I₁ = 2.4 A
final current through the inductor, I₂ = 0.3 A
duration of change of current, dt = 1.75 s
voltage change of the inductor, V = 5.7 Volts
The voltage change of the inductor is given by;
[tex]V_L = -L\frac{di}{dt}\\\\ V_L = -L(\frac{I_2-I_1}{dt} )\\\\V_L = L(\frac{I_1-I_2}{dt} )\\\\[/tex]
Where;
L is the inductance of the coil;
[tex]5.7 = L(\frac{2.4-0.3}{1.75} )\\\\5.7 = 1.2 L\\\\L = \frac{5.7}{1.2}\\\\ L = 4.75 \ H[/tex]
Therefore, the value of the inductance is 4.75 H
Suppose a big chunk of gold is submerged in water and its volume is found to be 12.5 cm?
Compute the mass of the chunk of gold in grams if you know the density is 19.3 g/cm2. Round
appropriately.
Answer:
The mass of the chunk of gold is 241.25 g
Explanation:
Since density is defined as mass divided by volume, we can solve for the mass (m) via its equation:
[tex]density=\frac{mass}{volume} \\19.3=\frac{m}{12.5} \\m=19.3\,(12.5)\, grams\\m = 241.25 \,\,g[/tex]
A metal sphere of radius 19 cm has a net charge of 2.4 × 10–8 C. (a) What is the electric field at the sphere's surface? (b) If V = 0 at infinity, what is the electric potential at the sphere's surface? (c) At what distance from the sphere's surface has the electric potential decreased by 370 V?
Answer:
1.29*10^6 N/C
1135.6 V
9.18 cm
Explanation:
Given that
radius of the metal, r = 19 cm
charge of the metal, q = 2.4*10^-8 C
coulomb's constant, k = 8.99*10^9
to find the electric field, we use the formula E = kq/r², where
E = electric field
k = coulomb constant
q = charge on the metal and
r = radius of the metal
E = (8.99*10^9 * 2.4*10^-8) / 0.19²
E = 215.76 / 0.0361
E = 1.29*10^6 N/C
to find the electric potential, we use this relation
V = kq/r
V = (8.99*10^9 * 2.4*10^-8) / 0.19
V = 215.76 / 0.19
V = 1135.6 V
V = kq/r,
r = kq/V
r = (8.99*10^9 * 2.4*10^-8)/ (1135.6 - 370
r = 215.76 / 765.6
r = 0.281 = 28.1 cm
distance from the sphere
28.18 - 19 = 9.18 cm
Determine the inductance L of a 0.65-m-long air-filled solenoid 3.2 cm in diameter containing 8400 loops.
Answer:
The inductance is [tex]L = 0.1097 \ H[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 0.65 \ m[/tex]
The diameter is [tex]d = 3.2 cm = 0.032 \ m[/tex]
The number of loops is [tex]N = 8400[/tex]
Generally the radius is evaluated as
[tex]r = \frac{ 0.032 }{2} = 0.016 \ m[/tex]
The inductance is mathematically represented as
[tex]L = \frac{ \mu_o * N^2 * A }{ l }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
A is the cross-sectional area which is mathematically evaluated as
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.142 * (0.016)^2[/tex]
=> [tex]A = 0.000804 \ m^2[/tex]
=> [tex]L = \frac{ 4\pi * 10^{-7} * 8400^2 * 0.000804 }{ 0.65 }[/tex]
=> [tex]L = 0.1097 \ H[/tex]
Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.
What was his average speed in mph over the last 400 m?
Answer: His average speed in mph over the last 400 m is 7.7 m/s.
Explanation:
Given: Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.
We know that , speed = [tex]\dfrac{distance}{time}[/tex]
Here , distance = 400m and time = 51.9 s
Then, speed = [tex]\dfrac{400}{51.9}\approx7.7\ m/s[/tex]
Hence, his average speed in mph over the last 400 m is 7.7 m/s.
Answer: 17.2 MPH
Explanation:
Average speed = distance/time
400m÷ 51.9s= 7.7 m/s
Now convert m/s →MPH
m→km→mi and s→min→hr
[tex]\frac{7.7m}{s} x[/tex] [tex]\frac{1 km}{1000 m} x\frac{0.6214 mi}{1 km} x\frac{60 s}{1 min} x\frac{60 mins}{1 hr} = 17.2 mph[/tex]
As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:___________.
A) approaches zero.
B) approaches infinity.
C) approaches unity.
D) none of the given answers
Answer:
B) approaches infinity
Explanation:
The capacitive reactance of an AC capacitor is given by;
[tex]X_C = \frac{1}{\omega C } = \frac{1}{2\pi f C}[/tex]
Where;
C is the capacitance
f is the frequency of the ac voltage
[tex]X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi (0) C} \\\\X_C = \frac{1}{0} \\\\X_C = \ infinity[/tex]
Therefore, as the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor approaches infinity.
The correct option is (B) approaches infinity
a. What quantum number of the hydrogen atom comes closest to giving a 61-nm-diameter electron orbit?
b. What are the electron's speed and energy in this state?
Answer:
a
[tex]n = 23[/tex]
b
[tex]v = 87377.95 \ m/s[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 61\ nm = 61 *10^{-9} \ m[/tex]
Generally the radius electron orbit is mathematically represented as
[tex]r = \frac{61 *10^{-9}}{2}[/tex]
=> [tex]r = 3.05*10^{-8} \ m[/tex]
This radius can also be represented mathematically as
[tex]r = n^2 * a_o[/tex]
Here n is the quantum number and [tex]a_o[/tex] is the Bohr radius with a value
[tex]a_o = 0.0529 *10^{-9} \ m[/tex]
So
[tex]n = \sqrt{\frac{3.05*10^{-8}}{ 0.059*10^{-9}} }[/tex]
=> [tex]n = 23[/tex]
Generally the angular momentum of the electron is mathematically represented as
[tex]L = m * v * r = \frac{n * h }{2 \pi}[/tex]
Here h is the Planck constant and the value is [tex]h = 6.626*10^{-34} J \cdot s[/tex]
m is the mass of the electron with values [tex]m = 9.1*10^{-31} \ kg[/tex]
So
[tex]v = \frac{23 * 6.626*10^{-34} }{2\pi * 9.1 *10^{-31} * 3.05*10^{-8} }[/tex]
[tex]v = 87377.95 \ m/s[/tex]
While taking the stairs it takes you 10 seconds to reach the top. The next time you take the same stairs, it takes you 5 seconds to reach the top stair. During which of these trips up the stairs did you use more power to climb? Explain your answer in complete sentences with proper spelling, grammar, and other language mechanics.
Answer:
P₂ = 2 P₁
we conclude that in the second time the power used is double that in the first rise
Explanation:
In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.
Now we can analyze the required power,
P = W / t
From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,
therefore the first time the power is
P₁ = E / 10
P₁ = 0.1 E
for the second time the power is
P₂ = E / 5
P₂ = 0.2 E
we see that the power in the second case is
P₂ = 2 P₁
Therefore, we conclude that in the second time the power used is double that in the first rise.
A narrow beam of white light is incident on a sheet of quartz. Thebeam disperses in the quartz, with red light (λË700nm)traveling at an angle of 26.3o with respect to thenormal and violet light (λË400nm) traveling at25.7o. The index of refraction of quartz for red lightis 1.45. What is the index of refraction of quartz for violetlight?
Answer:
The index of refraction of quartz for violet light is 1.47.
Explanation:
It is given that, a narrow beam of white light is incident on a sheet of quartz.
The beam disperses in the quartz, with red light at an angle, [tex]\theta_r=26.3^{\circ}[/tex] wrt to the normal and violet light traveling at an angle of [tex]\theta_v=25.7^{\circ}[/tex]
The index of refraction of quartz for red light is 1.45.
We need to find the index of refraction of quartz for violet light.
Using Snell's law of red light as follows :
[tex]\mu_a\sin\theta_i=\mu_r\sin\theta_r[/tex]
Here,
[tex]\mu_a[/tex] is the refractive index of air
[tex]\theta_i[/tex] is the angle of incidence
We can find the value of angle of incidence as follows :
[tex]\sin\theta_i=\dfrac{\mu_r \sin\theta_r}{\mu_a}\\\\\sin\theta_i=\dfrac{1.45\times \sin(26.3)}{1}\\\\\theta_i=\sin^{-1}(0.642)\\\\\theta_i=39.79^{\circ}[/tex]
Now again using Snell's law for violet light as follows :
[tex]\mu_a\sin\theta_i=\mu_v\sin\theta_v\\\\\mu_v=\dfrac{\mu_a\sin\theta_i}{\sin\theta_v}\\\\\mu_v=\dfrac{1\times \sin(39.79)}{\sin(25.7)}\\\\\mu_v=1.47[/tex]
So, the index of refraction of quartz for violet light is 1.47.
What is the maximum distance allowed between the center of hole #2 and datum B as seen in the front view?
Answer:
4.003" (inches )
Explanation:
The maximum distance allowed between the center of hole #2 and datum B can be calculated by adding 4.000" + 0.003" ( perpendicularity of the of hole #2) as seen from the front view of the diagram .
Note :The hole 2 is sited below the workpiece when viewed from the front view while the Datum B is positioned on the left end of the workpiece also note that the diameter is
Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV. What is the maximum kinetic energy of the photoelectrons emitted from the metal
Answer:
1.41eV
Explanation:
Kinetic Energy of photoelectron(K. Emax)
E = Wo + K.Emax
E = hc/ λ
h = planck's constant = 6.63 * 10^-34
c = speed of light = 3×10^8 m/s
λ = 400nm
Work function (Wo) = 1.70eV
1 eV = 1×10^-19
E = [(6.63×10^-34) * (3×10^8)] / 400×10^-7
E = (19.89 × 10^(-26))/400×10^-7
E = 0.049725×10^-19
K.Emax = E - Wo
K.Emax = (0.049725×10^-19) - (1.7×10^-38)
0.049725×10^-19 interms of eV = (0.049725/1.6)×10^-19 =
K.Emax = 3.11eV - 1.70eV
K.Emax = 1.41eV