Answer:
B
8.5 km/h east
Explanation:
Relative velocity= Va -Vb
=10-1.5
=8.5 km/h east
The concept relative speed is used when two or more bodies moving with some speed are considered. The relative speed of woman to the boat is 8.5 km/h east. The correct option is B.
What is relative speed?The relative speed of two bodies is defined as the sum of their speeds if they are moving in the opposite direction and it is the difference of their speeds if they are moving in the same direction.
The speed of the moving body with respect to the stationary body is known as the relative speed. The term relative means in comparison to. The relative speed is a scalar quantity.
Here both the boat and women are travelling in the same direction. So the relative speed is given as:
Relative speed = 10 - 1.5 = 8.5 km / h
Therefore the relative speed is 8.5 km/h east.
Thus the correct option is B.
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(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
pF
(b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
kV
Explanation:
(a) Given that,
Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]
The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]
The dielectric constant of, k = 2.1
When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :
[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]
Putting all the values we get :
[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]
(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]
The voltage difference between the plates at this critical voltage is given by :
[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]
or
V = 0.6 kV
We have that the Capacitance and potential difference is mathematically given as
[tex]Vmax=\frac{Q}{334.68pF}[/tex]C=334.68pF
Capacitance &potential differenceQuestion Parameters:
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
a)
Generally the equation for the Capacitance is mathematically given as
[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]
C=334.68pF
b)
Generally the equation for the Capacitance is mathematically given as
[tex]Vmax=\frac{Q}{C}[/tex]
Where
Q is the charge on the plates, and hence not given
Therefore, maximum potential difference is
[tex]Vmax=\frac{Q}{334.68pF}[/tex]
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A diffraction grating has 6000 lines per centimeter ruled on it. What is the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe when the grating is illuminated with a beam of light of wavelength 500 nm
Explanation:
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The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
What is diffraction?Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.
Given:
The number of lines = 6000 per cm,
The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m
Calculate the diffraction grating,
[tex]d = 1 / no\ of\ lines[/tex]
d = 10⁻² / 6000 m,
Calculate the second-order maxima angle and third-order maxima angle by the formula given below,
[tex]dsin\theta_1 = n_1 \lambda[/tex]
[tex]sin\theta_1 = n_1\lambda / d[/tex]
[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]
θ₁ = sin⁻¹(0.6)
θ₁ = 36.87°
Similarly, for θ₂,
θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)
θ₂ = sin⁻¹(0.9)
θ₂ = 64.16°
Calculate the separation as follows,
θ₂ - θ₁ = 64.16° - 36.87°
θ₂ - θ₁ = 27.29°
Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
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A clown 2 m tall looks at himself in a full-length mirror (floor-to-ceiling). Where in the mirror must he look to see his feet?
Answer:
Around the center of the mirror
During the first part of this lab, we want to determine how the object distance is related to what two quantities
Answer and Explanation:
The computation of the object distance related to two quantities is shown below:
It could find out by using the lens formula which is shown below:
[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]
where,
v = image distance
u = object distance
f = focal length
It could be found by applying the above formula i.e considering the image distance, object distance and the focal length
By what length will a slab of concrete that is originally 18 m long contract when the temperature drops from 24°C to -16°C? The coefficient of linear thermal expansion for this concrete is 1.0 × 10-5 C-1.
Explanation:
According to Thermal Expansion of solids:
[tex]dl = \alpha \times l \times dt[/tex]
[tex]dl = {10}^{ - 5} \times 18 \times 40 [/tex]
[tex]dl = 7.2 \times {10}^{ - 3} [/tex]
A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds. What is its rotation speed
Answer:
v = 6.28 m/s
Explanation:
It is given that,
A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds,
Number of revolution is half. It means angular velocity is 3.14 radians.
Let v is the angular speed. So,
[tex]v=\dfrac{\omega}{t}\\\\v=\dfrac{3.14}{0.5}\\\\v=6.28\ m/s[/tex]
So, the rotation speed is 6.28 m/s.
The angular velocity is the rotation speed, which is the angle of rotation
of the windmill per second, which is 2·π radians.
Response:
The rotation speed is 2·π rad/sHow can the rotational speed of the windmill be calculated?The given parameter are;
The angle of rotation the windmill rotates in 0.5 seconds = One-half a
rotation.
Required:
The rotational speed (angular velocity)
Solution:
The angle of one rotation = 2·π radians
Angle of one-half ration = [tex]\frac{1}{2}[/tex] × 2·π radians = π radians
[tex]Rotational \ speed = \mathbf{\dfrac{Angle \ of \ rotation}{Time}}[/tex]
Which gives;
[tex]Rotational \ speed, \omega = \dfrac{\pi}{0.5 \ s} = \mathbf{2 \cdot \pi \ rad/s}[/tex]
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P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism articulated to the ground at its vertex A, while vertex C is attached to the vertical cord fixed to the ground. If the coefficient of friction between the prism and the blocks is 0.4; determine the maximum angle that measures the inclination of the AC face with respect to the horizontal so that the system remains in equilibrium.
Answer:
21.8°
Explanation:
Let's call θ the angle between BC and the horizontal.
Draw a free body diagram for each block.
There are 4 forces acting on block D:
Weight force P pulling down,
Normal force N₁ pushing perpendicular to AB,
Friction force N₁μ pushing parallel up AB,
and tension force T pushing parallel up AB.
There are 4 forces acting on block E:
Weight force P pulling down,
Normal force N₂ pushing perpendicular to BC,
Friction force N₂μ pushing parallel to BC,
and tension force T pulling parallel to BC.
Sum of forces on D in the perpendicular direction:
∑F = ma
N₁ − P sin θ = 0
N₁ = P sin θ
Sum of forces on D in the parallel direction:
∑F = ma
T + N₁μ − P cos θ = 0
T = P cos θ − N₁μ
T = P cos θ − P sin θ μ
T = P (cos θ − sin θ μ)
Sum of forces on E in the perpendicular direction:
∑F = ma
N₂ − P cos θ = 0
N₂ = P cos θ
Sum of forces on E in the parallel direction:
∑F = ma
N₂μ + P sin θ − T = 0
T = N₂μ + P sin θ
T = P cos θ μ + P sin θ
T = P (cos θ μ + sin θ)
Set equal:
P (cos θ − sin θ μ) = P (cos θ μ + sin θ)
cos θ − sin θ μ = cos θ μ + sin θ
1 − tan θ μ = μ + tan θ
1 − μ = tan θ μ + tan θ
1 − μ = tan θ (μ + 1)
tan θ = (1 − μ) / (1 + μ)
Plug in values:
tan θ = (1 − 0.4) / (1 + 0.4)
θ = 23.2°
∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.
A semi-circular loop consisting of one turn of wire is place in the x-y plane. A constant magnetic field B=1.7T points along the negative z-axis(into the page), and a current I=0.7A flows counterclockwisefrom the positive z-axis. The net magnetic force on the circular section of the loop points in what direction? What is the net magnetice force on the circular section of the loop?
Answer:
The direction of net magnetic force on the circular section of the loop is in the positive y-axis
The net magnetic force on the circular section of the loop is 3.74 N
Explanation:
The magnetic field strength [tex]B[/tex] = 1.7 T
the current [tex]I[/tex] = 0.7 A
The diameter of the loop = 2 m
the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2
==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m
The force on the semi-circular is given as
F = [tex]BIl[/tex] sin ∅
but the loop is perpendicular to the field, therefore
sin ∅ = sin 90° = 1
F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N
The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".
According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.50 s apart. The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.
Required:
How far away did the impact occur?
Answer:
The distance is [tex]d = 193.6 \ m[/tex]
Explanation:
From the question we are told that
The time interval between the sounds is k[tex]t_1 = k + t_2[/tex] = 0.50 s
The speed of sound in air is [tex]v_s = 343 \ m/s[/tex]
The speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]
Generally the distance where the collision occurred is mathematically represented as
[tex]d = v * t[/tex]
Now from the question we see that d is the same for both sound waves
So
[tex]v_c t = v_s * t_1[/tex]
Now
So [tex]t_1 = k + t[/tex]
[tex]v_c t = v_s * (t+ k)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]t = 0.0645 \ s[/tex]
So
[tex]d = 3000 * 0.0645[/tex]
[tex]d = 193.6 \ m[/tex]
A projectile is launched with speed of 128 m/s, at an angle of 60° with the horizontal. After 2.0 s, what is the vertical component of the projectile's velocity?
After 2.0s, what is the speed of the projectile?
Answer:
a) 91 m/s
b) 111 m/s
Explanation:
v = u + at
v = 128sin60 + (-9.8)(2.0) = 91.25125... m/s
v = √(vx² + vy²) = √((128cos60)² + 91.25125²) = 111.4575... m/s
A 500 nm wavelength light illuminates a soap film with an index of refraction 1.33 to make it look bright. If the beam of light is incident normal on the film, what is the minimum thickness of the film
Answer:
t(min) = 94nm
Explanation:
The wavelength of the light incident is 500 nm.
The refractive index of the film is 1.33.
The minimum thickness of the soap film required for constructive interference.
The thickness of the film for the constructive interference is given by:
2*t= (m + 1/2) λ′
Now, λ′ = λ/μ = 500/1.33 = 376nm
The minimum thickness of the film
′t′ will be at m=0 :
2*t(min) = (0 + 1/2) 376
t(min) = 94nm
A long solenoid of radius 3 cm has 1100 turns per meter. If the solenoid carries a current of 1.5 A, then calculate the magnetic field at the center of the solenoid.a. 2.1E^-3T b. 1.0E^-3 T c. 1.7E^-4T d. 7.0E^-2 T
Answer:
The magnetic field at the center of the solenoid is 2.1 × 10⁻³ T
Explanation:
The magnetic field B at the center of the solenoid is given by
B = μ₀ni where μ₀ = permeability of free space = 4π × 10⁻⁷H/m, n = number of turns per unit length of the solenoid = 1100 turns per meter and i = current in the solenoid = 1.5 A.
So B = μ₀ni
= 4π × 10⁻⁷H/m × 1100 × 1.5 A
= 4π × 10⁻⁷H/m × 1650 A-turns/m
= 20734.5 × 10⁻⁷T
= 2.07345 × 10⁻³ T
≅ 2.1 × 10⁻³ T
So the magnetic field at the center of the solenoid is 2.1 × 10⁻³ T
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 wire is stretched 30% what is the percentage change in resistance
Answer:
The percentage change in resistance of the wire is 69%.
Explanation:
Resistance of a wire can be determined by,
R = (ρl) ÷ A
Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.
When the wire is stretched, its length and area changes but its volume and resistivity remains constant.
[tex]l_{o}[/tex] = 1.3l, and [tex]A_{o}[/tex] = [tex]\frac{A}{1.3}[/tex]
So that;
[tex]R_{o}[/tex] = (ρ[tex]l_{o}[/tex]) ÷ [tex]A_{o}[/tex] = (ρ × 1.3l) ÷ ([tex]\frac{A}{1.3}[/tex])
= (1.3lρ) ÷ ([tex]\frac{A}{1.3}[/tex])
= [tex](1.3)^{2}[/tex] × [(ρl) ÷ A]
= 1.69R (∵ R = (ρl) ÷ A)
[tex]R_{o}[/tex] = 1.69R
Where [tex]R_{o}[/tex] is the new resistance, [tex]l_{o}[/tex] is the new length, and [tex]A_{o}[/tex] is the new area after stretching the wire.
The change in resistance of the wire = [tex]R_{o}[/tex] - R
= 1.69R - 1R
= 0.69R
The percentage change in resistance = [tex]\frac{0.69R}{R}[/tex] × 100
= 0.69 × 100
= 69%
The percentage change in resistance of the wire is 69%.
A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.
Required:
What is the charge on the dust particle?
Answer:
The charge on the dust particle is [tex]q_d = 6.94 *10^{-13} \ C[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 2.0 \ m[/tex]
The width is [tex]w = 4.0 \ m[/tex]
The charge is [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]
The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]
Generally the electric field on the carpet is mathematically represented as
[tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]
Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]
[tex]E = -70621.5 \ N/C[/tex]
Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles
[tex]F__{E}} = F__{G}}[/tex]
=> [tex]q_d * E = m * g[/tex]
=> [tex]q_d = \frac{m * g}{E}[/tex]
=> [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]
=> [tex]q_d = 6.94 *10^{-13} \ C[/tex]
A single slit 1.5 mm wide is illuminated by 420- nm light. Part A What is the width of the central maximum (in cm ) in the diffraction pattern on a screen 4.5 m away
Answer:
The width is [tex]w_c = 0.00252 \ m[/tex]
Explanation:
From the question we are told that
The width of the single slit is [tex]a = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 420 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 4.5 \ m[/tex]
Generally the width of the central maximum is
[tex]w_c = 2 * y[/tex]
where y is the width of the first maxima which is mathematically represented as
[tex]y = \frac{\lambda * D}{a}[/tex]
=> [tex]y = \frac{ 420 *10^{-9} * 4.5}{ 1.5*10^{-3}}[/tex]
=> [tex]y = 0.00126 \ m[/tex]
So
[tex]w_c = 2 *0.00126[/tex]
[tex]w_c = 0.00252 \ m[/tex]
A laser beam takes 15 ms to travel from a rocket to the reflective surface of a planet and back to the rocket. How far is the rocket from this planet's surface
Explanation:
It took 7.5 ms or [tex]7.5×10^{-3}\:\text{s}[/tex] for the laser beam to reach the planet's surface. Since the speed of light is [tex]3×10^8\:\text{m/s}[/tex], the distance is
[tex]d = vt = (3×10^8\:\text{m/s})(7.5×10^{-3}\:\text{s})[/tex]
[tex]\:\:\:\:=2.25×10^{6}\:\text{m}[/tex]
An atom in the ground state has a collision with an electron, then emits a photon with a wavelength of 1240 nm. What conclusion can you draw about the initial kinetic energy of the electron
Answer:
attached below is the free body diagram of the missing illustration
Initial kinetic energy of the electron = 3 eV
Explanation:
The conclusion that can be drawn about the kinetic energy of the electron is
[tex]E_{e} = E_{3} - E_{1}[/tex]
E[tex]_{e}[/tex] = initial kinetic energy of the electron
E[tex]_{1}[/tex] = -4 eV
E[tex]_{3}[/tex] = -1 eV
insert the values into the equation above
[tex]E_{e}[/tex] = -1 -(-4) eV
= -1 + 4 = 3 eV
A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b) How long will it take for this person's temperature to return to 37°c if his mass is 90kg.
You slip a wrench over a bolt. Taking the origin at the bolt, the other end of the wrench is at x=18cm, y=5.5cm. You apply a force F? =88i^?23j^ to the end of the wrench. What is the torque on the bolt?
Answer:
The torque on the wrench is 4.188 Nm
Explanation:
Let r = xi + yj where is the distance of the applied force to the origin.
Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,
r = 0.18i + 0.055j
The applied force f = 88i - 23j
The torque τ = r × F
So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j
= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j
= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0 since i × i = 0, j × j = 0, i × j = k and j × i = -k
= 0 - 4.14k + 0.0484(-k) + 0
= -4.14k - 0.0484k
= -4.1884k Nm
≅ -4.188k Nm
So, the torque on the wrench is 4.188 Nm
21.-Una esquiadora olímpica que baja a 25m/s por una pendiente a 20o encuentra una región de nieve húmeda de coeficiente de fricción μr =0.55. ¿Cuánto desciende antes de detenerse?
Answer:
y = 12.82 m
Explanation:
We can solve this exercise using the energy work theorem
W = ΔEm
friction force work is
W = fr . s = fr s cos θ
the friction force opposes the movement, therefore the angle is 180º
W = - fr s
we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular
N -Wy = 0
N = mg cos θ
the friction force remains
fr = μ N
fr = μ mg cos θ
work gives
W = - μ mg s cos θ
initial energy
Em₀ = ½ m v²
the final energy is zero, because it stops
we substitute
- μ m g s cos θ = 0 - ½ m v²
s = ½ v² / (μ g cos θ)
let's calculate
s = ½ 20² / (0.55 9.8 cos 20)
s = 39.49 m
this is the distance it travels along the plane, to find the vertical distance let's use trigonometry
sin 20 = y / s
y = s sin 20
y = 37.49 sin 20
y = 12.82 m
A long, straight wire carries a 13.0-A current. An electron is fired parallel to this wire with a velocity of 250 km/s in the same direction as the current, 2.00 cm from the wire.
(a) Find the magnitude and direction of the electron’s initial acceleration.
(b) What should be the magnitude and direction of a uniform electric field that will allow the electron to continue to travel parallel to the wire?
(c) Is it necessary to include the effects of gravity? Justify your answer.
Answer:
A.5.7 x 10^-12m/s²
B. 32.5V/m
C. Pls. See attached file for explanations
Explanation:
How does the direction of current flow in the coil affect the orientation of the magnetic field produced by the electromagnet
Answer:
The magnetic field produced by an electric current is always oriented perpendicular to the direction of the current flow. And.Direction of magnetic field is governed by the 'right hand thumb rule, The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of Force . Similar to the situation with electric field lines, the greater the number of lines (or the closer they are together) in an area the stronger the magnetic field.
A sphere of radius R has charge Q. The electric field strength at distance r > R is Ei.
What is the ratio Ef /Ei of the final to initial electric field strengths if (a) Q is halved, (b) R is halved, and (c) r is halved (but is still > R)? Each part changes only one quantity; the other quantities have their initial values.
Answer:
A. Ef/ Ei = 1/2
B. EF/ Ei = 1
C Ef / Ei = 4
Explanation:
To solve this we apply Coulomb's law which States that
E = Kq / r^2
Where
q = charge r = straight line distance from q to the point in question and
K = Coulomb's constant
Then
Ei = K Q / r^2
So
A) If Q is halved then
Ef = K Q / (2 r^2)
Ef/Ei = 1/2
B) If R is halved, the value of the E-f
at a distance r remains unchanged. So
Ef/Ei = 1
C) if r is now r/2 then
Ef = K Q / (r/2)^2 = K Q / r^2/4 = 4 K Q / r^2
Ef / Ei = 4
Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wavelength. It then falls on two slits separated by 0.490 mm . In the resulting interference pattern on a screen 2.12 m away, adjacent bright fringes are separated by 2.86 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Determining wavelength. Part A What is the wavelength of the light that falls on the slits
Answer:
λ = 6.61 x 10⁻⁷ m = 661 nm
Explanation:
From the Young's Double Slit experiment, the the spacing between adjacent bright or dark fringes is given by the following formula:
Δx = λL/d
where,
Δx = fringe spacing = 2.86 mm = 2.86 x ⁻³ m
L = Distance between slits and screen = 2.12 m
d = slit separation = 0.49 mm = 0.49 x 10⁻³ m
λ = wavelength of light = ?
Therefore,
2.86 x 10⁻³ m = λ(2.12 m)/(0.49 x 10⁻³ m)
(2.86 x 10⁻³ m)(0.49 x 10⁻³ m)/(2.12 m) = λ
λ = 6.61 x 10⁻⁷ m = 661 nm
Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00 µm.
Answer:
λ = 428.6 nm
Explanation:
Hello,
In this case, we must remember that the Young's double slit experiment is described by the expression :
d sin θ = m λ
For constructive interference , and:
d sin θ = (m + ½) λ
For destructive interference , whereas d accounts for the distance between the slits, λ for the wavelength and m for an integer that describes the order of interference . Thus, for the given angle 30º, the distance between the slits is 3.00 μm or 3.00 10⁻⁶ m and the order of interference is 3; we therefore use the destructive interference equation to compute the wavelength as shown below:
λ = 3x10⁻⁶ sin (30) / (3 +1/2)
λ = 4.286 10⁻⁷ m
Or in manometers:
λ = 428.6 nm
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A 0.100-kg metal rod carrying a current of 15.0 A glides on two horizontal rails 0.550 m apart and 2.0 m long,
(a) If the coefficient of kinetic friction between the rod and rails is 0.120, what vertical magnetic field is required to keep the rod moving at a constant speed?
(b) If the friction between the rod and rail is reduced zero, the rod will accelerate. If the rod starts from rest at the one end of the rails, what is the speed of the rod at the other end of the rails for this frictionless situation? Use the same field value you calculated in part (a).
Answer:
The speed of the rod is 2.169 m/s.
Explanation:
Given that,
Mass = 0.100 kg
Current = 15.0 A
Distance = 2 m
Length = 0.550 m
Kinetic friction = 0.120
(a). We need to calculate the magnetic field
Using relation of frictional force and magnetic force
[tex]F_{f}=F_{B}[/tex]
[tex]\mu mg=Bli[/tex]
[tex]B=\dfrac{\mu mg}{li}[/tex]
Where, l = length
i = current
m = mass
Put the value into the formula
[tex]B=\dfrac{0.120\times0.1\times9.8}{0.550\times15.0}[/tex]
[tex]B=0.01425\ T[/tex]
[tex]B=1.425\times10^{-2}\ T[/tex]
(b). If the friction between the rod and rail is reduced zero.
So, [tex]f_{f}=0[/tex]
We need to calculate the acceleration
Using formula of force
[tex]F_{net}=f_{f}+F_{B}[/tex]
[tex]F_{net}=0+Bil[/tex]
[tex]ma=Bil[/tex]
[tex]a=\dfrac{Bil}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{1.425\times10^{-2}\times15\times0.55}{0.1}[/tex]
[tex]a=1.176\ m/s^2[/tex]
We need to calculate the speed of the rod
Using equation of motion
[tex]v^2=u^2+2as[/tex]
Put the value into the formula
[tex]v^2=0+2\times1.176\times2[/tex]
[tex]v^2=\sqrt{4.704}\ m/s[/tex]
[tex]v=2.169\ m/s[/tex]
Hence, The speed of the rod is 2.169 m/s.
Red and orange stars are found evenly spread throughout the galactic disk, but blue stars are typically found
Answer:
only in or near star-forming clouds
Explanation:
When in the galactic disk, Red and orange stars are found evenly spread so here Blue stars are hot and therefore massive and therefore short-lived, that is means they never have time to venture far from the places, where they were born. so correct answer is blue stars are typically found only in or near star-forming clouds
A lab technician uses laser light with a wavelength of 650 nmnm to test a diffraction grating. When the grating is 42.0 cmcm from the screen, the first-order maxima appear 6.09 cmcm from the center of the pattern. How many lines per millimeter does this grating have?
Answer:
221 lines per millimetre
Explanation:
We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.
Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,
tanθ = 6.09 cm/42.0 cm = 0.145
From trig ratios, cot²θ + 1 = cosec²θ
cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969
sinθ = 1/cosecθ = 1/6.969 = 0.1435
Also, sinθ = mλ/d at the first-order maximum, m = 1. So
sinθ = (1)λ/d = λ/d
Equating both expressions we have
0.1435 = λ/d
d = λ/0.1435
Now, λ = 650 nm = 650 × 10⁻⁹ m
d = 650 × 10⁻⁹ m/0.1435
d = 4529.62 × 10⁻⁹ m per line
d = 4.52962 × 10⁻⁶ m per line
d = 0.00452962 × 10⁻³ m per line
d = 0.00452962 mm per line
Since d = width of grating/number of lines of grating
Then number of lines per millimetre = 1/grating spacing
= 1/0.00452962
= 220.77 lines per millimetre
≅ 221 lines per millimetre since we can only have a whole number of lines.
If a negatively charged rod is held near a neutral metal ball, the ball is attracted to the rod. This happens:_______
a. because of magnetic effects
b. because the ball tries to pull the rod's electrons over to it
c. because the rod polarizes the metal
d. because the rod and the ball have opposite charges
Answer:
c. because the rod polarizes the metal.
Explanation:
Bringing the negatively charged rod close to the neutral metal ball causes the neutral metal ball to be polarized with induced positive charge on it. The polarizing of the formally neutral metal ball is due to the negative charge on the metal rod (bodies induce a charge opposite of their own charge on a nearby neutral body). The ball and rod then attract themselves because bodies with opposite charges attract each other, unlike bodies with same charges that repel each other.