Answer:
96 m
Explanation:
Given:
v₀ = 0 m/s
a = 3 m/s²
t = 8 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (8 s) + ½ (3 m/s²) (8 s)²
Δx = 96 m
"A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.33. Wavelengths of 479 nm and 798 nm and no wavelengths between are intensified in the reflected beam. The thickness of the film is"
Answer:
t = 8.98 10⁻⁷ m
Explanation:
This is an exercise in interference by reflection, let's analyze what happens on each surface of the film.
* When the light ray shifts from a medium with a lower refractive index to a medium with a higher refractive index, the reflected ray has a reflection of 180
* The beam when passing to the middle its wavelength changes
λ = λ₀ / n
if we take this into account, the constructive interference equation for normal incidence is
2t = (m + ½) λ₀ / n
let's apply this equation to our case
for λ₀ = 479 nm = 479 10⁻⁹ m
t = (m + ½) 479 10⁻⁹ / 1.33
(m + ½) = 1.33 t / 479 10⁻⁹
for λ₀ = 798 nm = 798 10⁻⁹ m
t = (m' + ½) 798 10⁻⁹ /1.33
(m' + ½) = 1.33 t / 798 10⁻⁹
as they tell us that no other constructive interference occurs between the two wavelengths, the order of interference must be consecutive, let's write the two equat⁻ions
(m + ½) = 1.33 t / 479 10⁻⁹
((m-1) + ½) = 1.33 t / 798 10⁻⁹
(m + ½) = 1.33 t / 798 10⁻⁹ +1
resolve
1.33 t / 479 10⁻⁹ = 1.33 t / 798 10⁻⁹ +1
1.33 t / 479 10⁻⁹ = (1.33t + 798 10⁻⁹) / 798 10⁻⁹
1.33t = (1 .33t + 798 10⁻⁹) 479/798
1.33t = (1 .33t + 798 10⁻⁹) 0.6
1.33 t = 0.7983 t + 477.6 10⁻⁹
t (1.33 - 0.7983) = 477.6 10⁻⁹
t = 477.6 10⁻⁹ /0.5315
t = 8.98 10⁻⁷ m
A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted.
(a) What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.40 cm?
______Kn
(b) If a force of this magnitude is applied compressively, by how much (in mm) does the 26.0 cm long dowel shorten? (Enter the magnitude.)
mm
Answer:
a
[tex]F = 67867.2 \ N[/tex]
b
[tex]\Delta L = 2.6 \ mm[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]
The stress is [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]
The diameter is [tex]d = 2.40 \ cm = 0.024 \ m[/tex]
The radius is mathematically represented as
[tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]
The cross-sectional area is mathematically evaluated as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (0.012)^2[/tex]
[tex]A = 0.000452\ m^2[/tex]
Generally the stress is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
=> [tex]F = \sigma * A[/tex]
=> [tex]F = 1.50 *10^{8} * 0.000452[/tex]
=> [tex]F = 67867.2 \ N[/tex]
Considering part b
The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]
Generally Young modulus is mathematically represented as
[tex]E = \frac{ \sigma}{ strain }[/tex]
Here strain is mathematically represented as
[tex]strain = \frac{ \Delta L }{L}[/tex]
So
[tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]
[tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]
=> [tex]\Delta L = \frac{\sigma * L }{E}[/tex]
substituting values
[tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]
[tex]\Delta L = 0.0026[/tex]
Converting to mm
[tex]\Delta L = 0.0026 *1000[/tex]
[tex]\Delta L = 2.6 \ mm[/tex]
you check the weather and find that the winds are coming from the west at 15 milers per hour. this information describes the winds
Answer:
Velocity
Explanation:
We finds that the winds are coming from the west at 15 miles per hour. This information shows the velocity of the wind. Since, velocity is a vector quantity. It has both magnitude and direction. 15 miles per hour shows the speed of wind and west shows the direction of wind motion.
Hence, the given information describes wind velocity.
If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y-direction, the wave is traveling in the
Answer:
The wave is travelling in the ±z-axis direction.
Explanation:
An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the electric and magnetic field.
In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m2. What is the maximum value of the magnetic field in the wave
Answer:
7.78x10^-8T
Explanation:
The Pointing Vector S is
S = (1/μ0) E × B
at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,
S = (1/μ0) E B
where S, E and B are magnitudes. The average value of the Pointing Vector is
<S> = [1/(2 μ0)] E0 B0
where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)
Also at any instant,
E = c B
where E and B are magnitudes, so it must also be true at the instant of peak values
E0 = c B0
Substituting for E0,
<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²
Solve for B0.
Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)
= 7.79 x10 ^-8 T
A sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x, t) = 1.99 cos(15.2x − 869t) where s is in micrometers, x is in meters, and t is in seconds. (a) Find the amplitude of this wave. µm (b) Find the wavelength of this wave. cm (c) Find the speed of this wave. m/s (d) Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x = 0.050 9 m at t = 2.94 ms. µm (e) Determine the maximum speed of a element's oscillatory motion. mm/s
Answer:
a) A = 1.99 μm , b) λ = 0.4134 m , c) v = 57.2 m / s , d) s = - 1,946 nm ,
e) v_max = 1,739 mm / s
Explanation:
A sound wave has the general expression
s = s₀ sin (kx - wt)
where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is
s = 1.99 sin (15.2 x - 869 t)
a) the amplitude of the wave is
A = s₀
A = 1.99 μm
b) wave spectrum is
k = 2π /λ
in the equation k = 15.2 m⁻¹
λ = 2π / k
λ = 2π / 15.2
λ = 0.4134 m
c) the speed of the wave is given by the relation
v = λ f
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 869 / 2π
f = 138.3 Hz
v = 0.4134 138.3
v = 57.2 m / s
d) To find the instantaneous velocity, we substitute the given distance and time into the equation
s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
s = 1.99 sin (0.77368 - 2.55486)
remember that trigonometry functions must be in radians
s = 1.99 (-0.98895)
s = - 1,946 nm
The negative sign indicates that it shifts to the left
e) the speed of the oscillating part is
v = ds / dt)
v = - s₀(-w) cos (kx -wt)
the maximum speed occurs when the cosines is 1
v_maximo = s₀w
v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
let's reduce to mm / s
v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
v_max = 1,739 mm / s
a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s
Calculation of Wavelength
When A sound wave has the general expression is:
Then, s = s₀ sin (kx - wt)
Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is
s is = 1.99 sin (15.2 x - 869 t)
a) When the amplitude of the wave is
A is = s₀
Thus, A = 1.99 μm
b) When the wave spectrum is
k is = 2π /λ
Now, in the equation k = 15.2 m⁻¹
Then, λ = 2π / k
After that, λ = 2π / 15.2
Thus, λ = 0.4134 m
c) When the speed of the wave is given by the relation is:
Then, v = λ f
Now, the angular velocity and frequency are related is:
w is = 2π f
Then, f = w / 2π
After that, f = 869 / 2π
Now, f = 138.3 Hz
Then, v = 0.4134 138.3
Thus, v = 57.2 m / s
d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation
Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
After that, s = 1.99 sin (0.77368 - 2.55486)
Then remember that trigonometry functions must be in radians
After that, s = 1.99 (-0.98895)
Thus, s = - 1,946 nm
When The negative sign indicates that it shifts to the left
e) When the speed of the oscillating part is
Then, v = ds / dt)
Now, v = - s₀(-w) cos (kx -wt)
When the maximum speed occurs when the cosines is 1
Then, v_maximo = s₀w
After that, v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
Now, let's reduce to mm / s
Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
Therefore, v_max = 1,739 mm / s
Finf more informmation about Wavelength here:
https://brainly.com/question/6352445
If the solenoid is 45.0 cm long and each winding has a radius of 8.0 cm , how many windings are in the solenoid
Answer:
The number of windings is 1.
Explanation:
The radius of the solenoid = 8.0 cm = 0.08 m
Length of the solenoid = 45.0 cm = 0.45 m
number of turn = ?
circumference of each winding = 2πr = 2 x 3.142 x 0.08 = 0.503 m
The number of windings = (Length of the solenoid)/(circumference of each winding)
==> 0.45/0.503 = 0.89 ≅ 1
A donkey is attached by a rope to a wooden cart at an angle of 23° to the horizontal. the tension in the rope is 210 n. if the cart is dragged horizontally along the floor with a constant speed of 7 km/h, calculate how much work the donkey does in 35 minutes.
Answer:
787528.7 J
Explanation:
Work done: This can be defined as the product of force and distance along the direction of force. The S.I unit of work is Joules (J).
From the question,
W = Tcos∅(d)............. Equation 1
Where W = work done, T = tension in the rope, ∅ = the angle of the rope to the horizontal, d = distance.
But,
d = v(t)..................... equation 2
Where v = velocity, t = time
Substitute equation 2 into equation 1
W = Tcos∅(vt)............. Equation 3
Given: T = 210 N, ∅ = 23°, v = 7 km/h = 1.94 m/s, t = 35 min = 2100 s
Substitute into equation 3
W = 210(cos23°)(1.94×2100)
W = 787528.7 J
Suppose that you release a small ball from rest at a depth of 0.700 m below the surface in a pool of water. If the density of the ball is 0.200 that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball.)
Number________
Units ________
Answer:
photo
Explanation:
A small wave pulse and a large wave pulse approach each other on a string; the large pulse is moving to the right.
Sometime after the pulses have met and passed each other, which of the following statements is correct? (More than one answer may be correct)
- the large pulse continues moving to the right
- the large pulse continues unchanged, moving to the right
- the small pulse is reflected and moves off to the right with a smaller amplitude
- the small pulse is reflected and moves off to the right with its original amplitude
- the two pulses combine into a single pulse moving to the right
Answer:
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
Explanation:
The waves fulfill the principle of superposition, which states that the value of the function at a point is the algebraic sum of the waves at a given instant.
The two waves in this exercise travel in the opposite direction, so when they are close, the resulting wave is the sum of the two waves, having a complicated shape. But when the waves follow their movement, they give in the same way as the initial a,
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
A major strike-slip earthquake on the San Andreas fault in California will cause a catastrophic tsunami affecting residents of San Francisco.
a) true
b) false
Answer:
a) true
Explanation:
The san andres is a transform fault that forms boundary between the Pacific and the North Atlantic plate and this slip strike is characterized by the latex motion the fault runs in the length of the California state. This plate is widely estimated for the high magnitude of earthquakes and varies from 7.7 to 8.3 magnitude. They are capable of producing a deadly tsunami that can devastate the pacific northwest.If a diode at 300°K with a constant bias current of 100μA has a forward voltage of 700mV across it, what will the voltage drop across this same diode be if the bias current is increased to 1mA? g
Answer:
the voltage drop across this same diode will be 760 mV
Explanation:
Given that:
Temperature T = 300°K
current [tex]I_1[/tex] = 100 μA
current [tex]I_2[/tex] = 1 mA
forward voltage [tex]V_r[/tex] = 700 mV = 0.7 V
To objective is to find the voltage drop across this same diode if the bias current is increased to 1mA.
Using the formula:
[tex]I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]
[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]
where;
[tex]V_r[/tex] = 0.7
[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}[/tex]
[tex]I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}[/tex]
[tex]\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]
[tex]\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]
Suppose n = 1
[tex]V_T = \dfrac{T}{11600} \\ \\ V_T = \dfrac{300}{11600} \\ \\ V_T = 25. 86 \ mV[/tex]
Then;
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { nV_T} -1} \end {pmatrix}[/tex]
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { 25.86} -1} \end {pmatrix}[/tex]
[tex]e^{\dfrac{V_r'}{nv_T}-1} = 5.699 \times 10^{12}[/tex]
[tex]{e^\dfrac{V_r'}{nv_T}} = 5.7 \times 10^{12}[/tex]
[tex]{\dfrac{V_r'}{nv_T}} =log_{e ^{5.7 \times 10^{12}}}[/tex]
[tex]{\dfrac{V_r'}{nv_T}} =29.37[/tex]
[tex]V_r'=29.37 \times nV_T[/tex]
[tex]V_r'=29.37 \times 25.86[/tex]
[tex]V_r'=759.5 \ mV[/tex]
[tex]Vr' \simeq[/tex] 760 mV
Thus, the voltage drop across this same diode will be 760 mV
An isolated system consists of two masses. The first, m1, has a mass of 1.90 kg, and is initially traveling to the east with a speed of 6.71 m/s. The second, m2, has a mass of 2.94 kg, and is initially traveling to the west with an unknown initial speed. The two masses collide head-on in a completely inelastic collision that stops them both. Calculate the initial kinetic energy of m2.
Answer:
m1v1=m2v2, v2=4.3m/s KE=(0.5)(2.94)(4.3)=6.2J
Can someone please help!!!
Answer:
W = F • ∆x
so for work to be done, a force and displacement has to be in the same direction. (Ex: a box is being pushed forward and it's also moving forward.)
Assume that Mars and Earth are in the same plane and that their orbits around the Sun are circles (Mars is ≈230×10^6km from the Sun and Earth is ≈150×10^6km from the Sun).
a) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is nearest Earth?
b) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is farthest from Earth?
Answer:
Explanation:
speed of light ≈ 300 000 000 m/s
a) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is nearest Earth?
(230e9 - 150e9) / 3e8 = 277 s or 4 minutes 27 seconds.
b) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is farthest from Earth?
(230e9 + 150e9) / 3e8 = 1267 s or 21 minutes 6 seconds.
g A projectile is fired from the ground at an angle of θ = π 4 toward a tower located 600 m away. If the projectile has an initial speed of 120 m/s, find the height at which it strikes the tower
Answer:
The projectile strikes the tower at a height of 354.824 meters.
Explanation:
The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:
Horizontal motion
[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]
Vertical motion
[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.
[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
The time spent for the projectile to strike the tower is obtained from first equation:
[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]
If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:
[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]
[tex]t \approx 7.071\,s[/tex]
Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])
[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]
[tex]y \approx 354.824\,m[/tex]
The projectile strikes the tower at a height of 354.824 meters.
A swimmer is treading water with their head above the surface of a pool and sees a penny at the bottom of the pool 5.0 mm below. How deep does the coin appear to be? (Index of refraction of water = 1.33) [Conceptual note: Does the coin appear to be shallower or deeper?]
Answer:
The apparent depth is [tex]D' = 0.00376 \ m[/tex]
Explanation:
From the question we are told that
The depth of the water is [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]
The refractive index of water is [tex]n = 1.33[/tex]
Generally the apparent depth of the coin is mathematically represented as
[tex]D' = D * [\frac{ n_a}{n} ][/tex]
Here [tex]n_a[/tex] is the refractive index of air the value is [tex]n_a = 1[/tex]
So
[tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]
[tex]D' = 0.00376 \ m[/tex]
The apparent depth will be 0.00376 m.
What is an index of refraction?
The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.
d is the depth of the water =5.0 mm =5.0 ×10⁻³
n is the refractive index of water =1.33
[tex]\rm n_a[/tex] is the refractive index of wire=1
The apparent depth of the coin is given as;
[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]
Hence the apparent depth will be 0.00376 m.
To learn more about the index of refraction refer to the link;
https://brainly.com/question/23750645
The capacitor is originally charged. How does the current I in the ammeter behave as a function of time after the switch is closed?
1. I = 0 always.
2. I = constant, not equal to 0.
3. I increases, then is constant.
4. I instantly jumps up, then slowly decreases.
5. None of the above.
Answer:
The current in the ammeter is zero.
(1) is correct option.
Explanation:
Given that,
The capacitor is charged.
We need find the current after closed switched
We know that,
When switch is closed then the capacitor behave as a short circuit, and the all current flows through it. the current is zero.
Then, the ammeter reads zero.
Hence, The current in the ammeter is zero.
(1) is correct option.
A small insect viewed through a convex lens is 1.5 cmcm from the lens and appears 2.5 times larger than its actual size. Part A What is the focal length of the lens
Answer:
The focal length of the lens is 2.5 cm
Explanation:
Use the two equations for thin lenses combined: the one for magnification (m), and the one that relates distances of object [tex]d_o[/tex], of image [tex]d_i[/tex], and focal length;
[tex]m=\frac{h_i}{h_o} =-\frac{d_i}{d_o} \\ \\\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}[/tex]
Since we know the value of the magnification (m), we can write the image distance in terms of the object distance, and then use it to replace the image distance in the second equation:
[tex]m=-\frac{d_i}{d_o} \\2.5=-\frac{d_i}{d_o}\\d_i=-2.5\,d_o[/tex]
then, solving for the focal distance knowing that the object distance is 1.5 cm:
[tex]\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}\\-\frac{1}{2.5\,d_o} +\frac{1}{d_o} =\frac{1}{f}\\(2.5\,d_o\,f)\,(-\frac{1}{2.5\,d_o} +\frac{1}{d_o}) =\frac{1}{f}\,(2.5\,d_o\,f)\\-f+2.5\,f=2.5\,d_o\\1.5\,f=2.5\,d_o\\f=\frac{2.5\,d_o}{1.5} \\f=\frac{2.5\,(1.5\,\,cm)}{1.5}\\f=2.5\,\,cm[/tex]
A thin film of soap with n = 1.37 hanging in the air reflects dominantly red light with λ = 696 nm. What is the minimum thickness of the film?
Answer:
The thickness is [tex]t = 1.273 *10^{-7} \ m[/tex]
Explanation:
From the question we are told that
The refractive index of the film is [tex]n = 1.37[/tex]
The wavelength is [tex]\lambda = 696 \ nm = 696 *10^{-9 } \ m[/tex]
Generally the condition for constructive interference in a film is mathematically represented as
[tex]2 * t = [m + \frac{1}{2} ] \lambda_k[/tex]
Here t is the thickness of the film , m is the order number (0, 1, 2, 3 ... )
[tex]\lambda _k[/tex] is the wavelength of light that is inside the film , this is mathematically evaluated as
[tex]\lambda _k = \frac{ \lambda }{ n}[/tex]
[tex]\lambda _k = \frac{ 696 *10^{-9}}{ 1.37}[/tex]
[tex]\lambda _k = 5.095 *10^{-7 } \ m[/tex]
So for m = 0
[tex]t = [ 0 + \frac{1}{2} ] \lambda _k * \frac{1}{2}[/tex]
substituting values
[tex]t = [ 0 + \frac{1}{2} ] (5.095 *10^{-7}) * \frac{1}{2}[/tex]
[tex]t = 1.273 *10^{-7} \ m[/tex]
15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and direction of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current
Answer:
we see it is a linear relationship.
Explanation:
The magnetic flux is u solenoid is
B = μ₀ N/L I
where N is the number of loops, L the length and I the current
By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops
B = (μ₀ I / L) N
the amount between paracentesis constant, in the case of 4 loop the field is worth
B = cte 4
N B
4 4 cte
3 3 cte
2 2 cte
1 1 cte
as we see it is a linear relationship.
In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 998.0 rad/s while it is being read, and then is allowed to come to rest over 0.502 seconds , what is the magnitude of the average angular acceleration of the disk?
Answer:
1988.05 rad/s^2
Explanation:
The angular speed of the optical disk ω = 998.0 rad/s
the time taken to come to rest t = 0.502 s
The magnitude of the average angular acceleration ∝ = ω/t
∝ = 998.0/0.502 = 1988.05 rad/s^2
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.
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Two ice skaters push off against one another starting from a stationary position. The 45.0-kg skater acquires a speed of 0.375 m/s. What speed does the 60.0-kg skater acquire in m/s
Answer:
0.2812
Explanation:
Given that
mass of skater 1, m1 = 45 kg
mass of skater 2, m2 = 60 kg
speed of skater 1, v1 = 0.375 m/s
To attempt this question, we would be using the Law of conservation of momentum That says the momentum is constant, before and after the movement.
Thus, momentum p = mv
Law of conservation of momentum infers that,
m1v1 = m2v2
Now we proceed to substitute our values into the formula.
45 * 0.375 = 60 * v2
v2 = 16.875 / 60
v2 = 0.2812 m/s
Therefore the speed of the second skater has to be 0.2812 m/s
Two parallel metal plates, each of area A, are separatedby a distance 3d. Both are connected to ground and each plate carries no charge. A third plate carrying charge Qis inserted between the two plates, located a distance dfrom the upper plate. As a result, negative charge is induced on each of the two original plates. a) In terms of Q, find the amount of charge on the upper plate, Q1, and the lower plate, Q2. (Hint: it must be true that Q
Answer:
Upper plate Q/3
Lower plate 2Q/3
Explanation:
See attached file
B. CO
A wave has frequency of 2 Hz and a wave length of 30 cm. the velocity of the wave is
A. 60.0 ms
B. 6.0 ms
D. 0.6 ms
Answer:
0.6 m/s
Explanation:
2Hz = 2^-1 = 2 /s
30cm = .3m
Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.
(.3m)*(2 /s) = 0.6 m/s
a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleration does it experience
Answer:
The angular acceleration is [tex]\alpha = 3.235 \ rad/s ^2[/tex]
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 0.034\ kg \cdot m^2[/tex]
The net torque is [tex]\tau = 0.11\ N \cdot m[/tex]
Generally the net torque is mathematically represented as
[tex]\tau = I * \alpha[/tex]
Where [tex]\alpha[/tex] is the angular acceleration so
[tex]\alpha = \frac{\tau }{I}[/tex]
substituting values
[tex]\alpha = \frac{0.1 1}{ 0.034}[/tex]
[tex]\alpha = 3.235 \ rad/s ^2[/tex]
An oil film (n = 1.48) of thickness 290 nm floating on water is illuminated with white light at normal incidence. What is the wavelength of the dominant color in the reflected light? A. Blue (470 nm) B. Green (541 nm) C. Violet (404 nm) D. Yellow (572 nm)
Answer:
The correct option is D
Explanation:
From the question we are told that
The refractive index of oil film is [tex]k = 1.48[/tex]
The thickness is [tex]t = 290 \ nm = 290*10^{-9} \ m[/tex]
Generally for constructive interference
[tex]2t = [m + \frac{1}{2} ]* \frac{\lambda}{k}[/tex]
For reflection of a bright fringe m = 1
=> [tex]2 * (290*10^{-9}) = [1 + \frac{1}{2} ]* \frac{\lambda}{1.48}[/tex]
=> [tex]\lambda = 5.723 *10^{-7} \ m[/tex]
This wavelength fall in the range of a yellow light
A long, horizontal hose of diameter 5.4 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.2 cm. Water squirts from the nozzle at velocity 20 m/sec. Assume that the water has no viscosity or other form of energy dissipation.
a) What is the velocity of the water in the hose?
b) What is the pressure differential between the water in the hose and water in the nozzle?
c) How long will it take to fill a tub of volume 120 liters with the hose?
Answer:
a) 0.988 m/s
b) 199512 Pa
c) 57.52 s
Explanation:
given that
A
A1 v1 = A2 v2
d1² v1 = d2² v2
v2 = [d1/d2]² v1
v2 = (1.2/5.4)² * 20
v2 = 0.049 * 20
v2 = 0.988 m/s
B
P + 1/2 ρ v² = K.
[p2 - p1] = 1/2 ρ [v1² - v2²]
[p2 - p1] = 1/2 * 1000 [20² - 0.988²]
[p2 - p1] = 500 * (400 - 0.976)
[p2 - p1] = 500 * 399
[p2 - p1] = 199512 Pa
C
Flow rate = AV = π [d²/ 4 ] * v
= π [0.012² / 4 ] * 20 = 0.00226 m³ /s
= π [0.054² / 4 ] * 0.988 = 0.00226 m³ /s
130 liters = 0.13 m³
t = 0.13/ 0.00226 = 57.52 s
Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tightly wound with a single layer of 1.50-mm-diameter superconducting wire.
What current is needed?
Answer:
The current needed is 1790.26 A
Explanation:
Given;
magnitude of magnetic field, B = 1.5 T
length of the solenoid, L = 1.8 m
diameter of the solenoid, d = 75 cm = 0.75 m
The magnetic field is given by;
[tex]B = \frac{\mu_o NI }{L}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
I is current in the solenoid
N is the number of turns, calculated as;
[tex]N = \frac{Length \ of\ solenoid}{diameter \ of \ wire} \\\\N = \frac{1.8}{1.5*10^{-3}} =1200 \ turns[/tex]
The current needed is calculated as;
[tex]I = \frac{BL}{\mu_o N} \\\\I = \frac{1.5 *1.8}{4\pi *10^{-7} *1200} \\\\I = 1790.26 \ A[/tex]
Therefore, the current needed is 1790.26 A.
Answer:
I = 1790.5 A
Explanation:
The magnetic field due to a solenoid is given by the following formula:
B = μ₀NI/L
where,
B = Magnetic Field Required = 1.5 T
μ₀ = 4π x 10⁻⁷ T/A.m
L = length of Solenoid = 1.8 m
I = Current needed = ?
N = No. of turns = L/diameter of wire = 1.8 m/1.5 x 10⁻³ m = 1200
Therefore,
1.5 T = (4π x 10⁻⁷ T/A.m)(1200)(I)/1.8 m
I = (1.5 T)(1.8 m)/(1200)(4π x 10⁻⁷ T/A.m)
I = 1790.5 A