Answer:
Explanation:
Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .
Now the reflecting surface is twisted so that is becomes inclined at angle alpha .
The reflected light will be deviated from its original direction by angle
2 x alpha .
Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle
2 x beta
Hence angle between these two reflected beam
= 2 beta - 2 alpha
= 2 ( β - α )
So, angular separation between the rays reflected from the two surfaces
= 2 ( β - α ) .
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
A parallel-plate vacuum capacitor has 7.72 J of energy stored in it. The separation between the plates is 3.30 mm. If the separation is decreased to 1.45 mm, For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Stored energy. Part A what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed
Answer
3.340J
Explanation;
Using the relation. Energy stored in capacitor = U = 7.72 J
U =(1/2)CV^2
C =(eo)A/d
C*d=(eo)A=constant
C2d2=C1d1
C2=C1d1/d2
The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.
Initial separation between the plates =d1= 3.30mm .
Final separation = d2 = 1.45 mm
(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same
Energy=U =(1/2)q^2/C
U2C2 = U1C1
U2 =U1C1 /C2
U2 =U1d2/d1
Final energy = Uf = initial energy *d2/d1
Final energy = Uf =7.72*1.45/3.30
(A) Final energy = Uf = 3.340J
In a distant galaxy, whose light is just arriving from 10 billion light years away, our spectroscope should reveal that the most common element is
Answer:
In a distant galaxy, whose light is just arriving from 10 billion light years away, our spectroscope should reveal that the most common element is HELIUM
An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and inductor, or a capacitor. At time t = 0 the voltage is zero. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -Imax, where Imax is the current amplitude. What is the unknown element?
Answer:
Capacitor, is the right answer.
Explanation:
The unknown element is a Capacitor.
Below is the calculation that proves that it is a capacitor.
We know that for the Capacitor
i = Imax × sin(wt+(pi/2)).
i = Imax × sin ((2 × pi/T) × (T/4) + (pi/2))
i = Imax × sin(3.142) = 0 A
at, t = T/2
wt = (2 × pi/T) × (T/2) = pi
wt + (pi/2) = pi + (pi/2) = ( 3 × pi/2) =
i = Imax × sin(3 × pi/2) = -Imax
Which is in a correct agreement with capacitor therefore, the answer is a Capacitor.
A single-slit diffraction pattern is formed on a distant screen. Assume the angles involved are small. Part A By what factor will the width of the central bright spot on the screen change if the wavelength is doubled
Answer:
If the wavelength is doubled, the width of the central bright spot on the screen will increase by a factor of 2 (that is, it will also double).
Explanation:
For a single-slit diffraction, diffraction patterns are found at angles θ for which
w sinθ = mλ
where w is the width
λ is wavelength
m is an integer, m = 1,2,3, ....
From the equation, w sinθ = mλ
For the first case, where nothing was changed
w₁ = mλ₁ / sinθ
Now, If the wavelength is doubled, that is, λ₂ = 2λ₁
The equation becomes
w₂ = mλ₂ / sinθ
Then, w₂ = m(2λ₁) / sinθ
w₂ = 2(mλ₁) / sinθ
Recall that, w₁ = mλ₁ / sinθ
Therefore, w₂ = 2w₁
Hence, If the wavelength is doubled, the width of the central bright spot on the screen will increase by a factor of 2 (that is, it will also double).
Suppose that a sound source is emitting waves uniformly in all directions. If you move to a point twice as far away from the source, the frequency of the sound will be:________.
a. one-fourth as great.
b. half as great.
c. twice as great.
d. unchanged.
Answer:
d. unchanged.
Explanation:
The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.
In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from
v = fλ
that the frequency is tied to the wave, and does not change throughout the waveform.
where v is the speed of the sound wave
f is the frequency
λ is the wavelength of the sound wave.
By what angle should the second polarized sheet be rotated relative to the first to reduce the transmitted intensity to one-half the intensity that was transmitted through both polarizing sheets when aligned
Answer:
θ = 45º
Explanation:
The light that falls on the second polarized is polarized, therefore it is governed by the law of Maluz
I = I₀ cos² θ
in the problem they ask us
I = ½ I₀
let's look for the angles
½ I₀ = I₀ cos² θ
cos θ = √ ½ = 0.707
θ = cos 0.707
θ = 45º
Charge of uniform linear density (6.7 nC/m) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y
Thw question is not complete. The complete question is;
Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC
Answer:
Option C: E = 75 N/C
Explanation:
We are given;
Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m
Distance on the y-axis; d = 1.6 m
Now, the formula for electric field with uniform linear density is given as;
E = λ/(2•π•r•ε_o)
Where;
E is electric field
λ is uniform linear density = 6.7 × 10^(-9) C/m
r is distance = 1.6m
ε_o is a constant = 8.85 × 10^(-12) C²/N.m²
Thus;
E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))
E = 75.31 N/C ≈ 75 N/C
A car travels down the road for 535 m in 17.3 s. What is the velocity of the car in m/s and in km/h?
Answer:
30.92m/sExplanation:
[tex]Distance = 535m\\Time = 17.3s\\\\Velocity = \frac{Distane}{Time} \\\\V = \frac{535m}{17.3s} \\\\Velocity = 30.92m/s[/tex]
[tex]Distance = 535m\\\\535m \:to \: km=0.535km\\\\Time = 17.3s\\\\17.3s = 0.004805556hours\\\\Velocity = \frac{Distance}{Time}\\\\ V= \frac{0.535}{0.004805556} \\\\ V=111.329469472\\\\=111.33km/h[/tex]
An intergalactic rock star bangs his drum every 1.30 s. A person on earth measures that the time between beats is 2.50 s. How fast is the rock star moving relative to the earth
Answer:
v = 0.89 c = 2.67 x 10⁸ m/s
Explanation:
The time dilation consequence of the special theory of relativity shall be used here, From time dilation formula we have:
t = t₀/√[1 - v²/c²]
where,
t = time measured by the person on earth = 2.50 s
t₀ = rest time of the intergalactic rock star = 1.30 s
v = relative speed of the rock star = ?
Therefore,
2.5 s = (1.3 s)/√[1 - v²/c²]
√[1 - v²/c²] = 1.3/2.5
√[1 - v²/c²] = 0.52
[1 - v²/c²] = 0.52²
[1 - v²/c²] = 0.2074
v²/c² = 1 - 0.2074
v²/c² = 0.7926
v/c = √0.7926
v = 0.89 c
where,
c = speed of light = 3 x 10⁸ m/s
v = (0.89)(3 x 10⁸ m/s)
v = 0.89 c = 2.67 x 10⁸ m/s
A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses. But he loses them while travelling. Fortunately he has his old pair as a spare. (a) If the lenses of the old pair have a power of 2.25 diopters, what is his near point (measured from the eye) when wearing the old glasses, if they rest 2.0 cm in front of the eye
Answer:
30.93 cm
Explanation:
Given that:
A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses
The power of the old pair of lens p = 2.25 diopters
The focal point length = 1/p
The focal point length = 1/2.25
The focal point length = 0.444 m
The focal point length = 44.4 cm
The near point of the person from the glass = (85 -2)cm , This is because the glasses are usually 2 cm from the lens
The near point of the person from the glass = 83 cm
Let consider s' to be the image on the same sides of the lens,
∴ s' = -83 cm
We known that:
the focal length of a mirror image 1/f =1/u +1/v
Assume the near point is at an excellent distance s from the glass where the person wears the corrective glasses.
Then:
1/f = 1/s + 1/s'
1/s = 1/f - 1/s'
1/s = (s' -f)/fs'
s = fs'/(s'-f)
s =( 44.4× -83)/(-83 - 44.4)
s = - 3685.2 / - 127.4
s = 28.93 cm
Thus , the near distance point measured from the eye wearing the old glasses, if they rest 2.0 cm in front of the eye = (28.93 +2.0)cm
= 30.93 cm
A uniform bar has two small balls glued to its ends. The bar is 2.10 m long and with mass 3.70 kg , while the balls each have mass 0.700 kg and can be treated as point masses.
Required:
Find the moment of inertia of this combination about an axis
a. perpendicular to the bar through its center.
b. perpendicular to the bar through one of the balls.
c. parallel to the bar through both balls.
d. parallel to the bar and 0.500 m from it.
Answer:
Explanation:
a )
moment of inertia in the first case will be sum of moment of inertia of two balls + moment of inertia of bar
= 2 x .700 x (2.1 / 2 )² + 3.7 x 2.1² / 12
= 1.5435 + 1.35975
= 2.90325 kg m²
b )
moment of inertia required
= moment of inertia of bar + moment of inertia of the other ball
= 3.70 x (2.1² / 3 ) + .7 x 2.1²
= 5.439 + 3.087
= 8.526 kg m²
c )
In this case moment of inertia of the combination = 0 as distance of masses from given axis is zero .
d )
masses = 3.7 + .7 = 4.4 kg
distance from axis = .5 m
moment of inertia about given axis
= 4.4 x .5²
= 1.1 kg m².
A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire?
Answer:
The direction of the force will be towards the east
Explanation:
From the question we are told that
The direction of the downward
Generally according to Fleming's right-hand rule(
Thumb - direction of force
Middle finger - direction of current
Index finger - direction of the magnetic field
) and the fact that the earth magnetic field acts from south to north with respect to the four cardinal points then the direction of the force will be toward the east with respect to the four cardinal point on the earth
Light of wavelength 520 nm is used to illuminate normally two glass plates 21.1 cm in length that touch at one end and are separated at the other by a wire of radius 0.028 mm. How many bright fringes appear along the total length of the plates.
Answer:
The number is [tex]Z = 216 \ fringes[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 520 \ nm = 520 *10^{-9} \ m[/tex]
The length of the glass plates is [tex]y = 21.1cm = 0.211 \ m[/tex]
The distance between the plates (radius of wire ) = [tex]d = 0.028 mm = 2.8 *10^{-5} \ m[/tex]
Generally the condition for constructive interference in a film is mathematically represented as
[tex]2 * t = [m + \frac{1}{2} ]\lambda[/tex]
Where t is the thickness of the separation between the glass i.e
t = 0 at the edge where the glasses are touching each other and
t = 2d at the edge where the glasses are separated by the wire
m is the order of the fringe it starts from 0, 1 , 2 ...
So
[tex]2 * 2 * d = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=> [tex]2 * 2 * (2.8 *10^{-5}) = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=>
[tex]m = 215[/tex]
given that we start counting m from zero
it means that the number of bright fringes that would appear is
[tex]Z = m + 1[/tex]
=> [tex]Z = 215 +1[/tex]
=> [tex]Z = 216 \ fringes[/tex]
Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is
Answer:
The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Explanation:
Given;
intensity of light, I = 1 kW/m²
The radiation pressure of light is given as;
[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]
I kW = 1000 J/s
The energy flux density = 1000 J/m².s
The speed of light = 3 x 10⁸ m/s
Thus, the radiation pressure of the light is calculated as;
[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]
Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the other pipe?
Answer:
The length of the longer pipe is L = 2.30 m
Explanation:
Given that:
Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.
How long is the other pipe?
From above;
The formula for the frequency of open ended pipes can be expressed as:
[tex]f = \dfrac{nv}{2L}[/tex]
where n = 1 ( since half wavelength exist between those two pipes)
v = 343 m/s and L = 2.08 m
Thus, the shorter pipe produces a frequency of :
[tex]f = \dfrac{1*343}{2*2.08}[/tex]
[tex]f = \dfrac{343}{4.16}[/tex]
[tex]f =82.45 \ Hz[/tex]
Also; we know that the beat frequency was given as 8.0 Hz
Then,
The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz
The lower frequency of the longer pipe = 74.45 Hz
Finally;
From the above equation; make Length L the subject of the formula. Then,
The length of the longer pipe is L = [tex]\dfrac{nv}{2f}[/tex]
The length of the longer pipe is L = [tex]\dfrac{1*343}{2*74.45}[/tex]
The length of the longer pipe is L = [tex]\dfrac{343}{148.9}[/tex]
The length of the longer pipe is L = 2.30 m
A certain digital camera having a lens with focal length 7.50 cmcm focuses on an object 1.70 mm tall that is 4.70 mm from the lens.
Part A. How far must the lens be from the photocells?
s = cm
Part B. Is the image on the photocells erect or inverted? Real or virtual?
a. The image is erect and real.
b. The image is inverted and real.
c. The image is erect and virtual.
d. The image is inverted and virtual.
Part C. How tall is the image on the photocells?
|h?| = cm
Part D. A SLR digital camera often has pixels measuring 8.00?m
Answer:
a. 7.62cm
b. Real and inverted
c. 2.76 cm
d. 3450
Explanation:
We proceed as follows;
a. the lens equation that relates the object distance to the image distance with the focal length is given as follows;
1/f = 1/p + 1/q
making q the subject of the formula;
q = pf/p-f
From the question;
p = 4.70m
f = 7.5cm = 0.075m
Substituting these, we have ;
q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm
b. The image is real and inverted since the image distance is positive
c. We want to calculate how tall the image is
Mathematically;
h1 = (q/p)h0
h1 = (7.62/4.70)* 1.7
h1 = 2.76 cm
d. We want to calculate the number of pixels that fit into this image
Mathematically:
n = h1/8 micro meter
n = 2.76cm/8 micro meter = 2.76 * 10^-2/8 * 10^-6 = 3450
an electric device is plugged into a 110v wall socket. if the device consumes 500 w of power, what is the resistance of the device
Answer: R=24.2Ω
Explanation: Power is rate of work being done in an electric circuit. It relates to voltage, current and resistance through the following formulas:
P=V.i
P=R.i²
[tex]P=\frac{V^{2}}{R}[/tex]
The resistance of the system is:
[tex]P=\frac{V^{2}}{R}[/tex]
[tex]R=\frac{V^{2}}{P}[/tex]
[tex]R=\frac{110^{2}}{500}[/tex]
R = 24.2Ω
For the device, resistance is 24.2Ω.
The valid digits in a measurement are called _____ digits. Question 10 options: insignificant significant uncertain non-zero
Answer:
Significant
Explanation:
Valid digits in measurements are called significant digits, or also called significant figures.
These significant digits allow data and measurements to be more accurate and exact.
Answer:
Significant digits
Explanation
Took the test got it right
Give an example of a fad diet that is not healthy and one that is healthy. Explain how you know the difference.
Answer:
Good Diet: ! gallon of water a day, Fruits, Vegetables, White meats(Chicken), Don't eat past 3 PM.
Bad Diet: Pizza, Red meat, Baked goods, Eating at late hours.
Explanation: I know the difference because, When you drink water first thing in the morning it gets your metabolism running. Than means you can digest foods better, you want to feed your body good foods but you should not eat until you feel stuffed. You should eat until you are no longer starving. Than you should drink a cup of water in between meals. I know you should not eat past 3 pm because your body needs time to digest foods because you should never go to sleep with a full stomach. I know the difference between good food and bad food because when you eat healthy food and a balanced diet, your body will have more energy and you wont feel tired afterwards. Eating bad foods and food with artificial sugars will clump up in your kidneys, and your body will have small bursts of energy but you will feel lazy afterwards...Your body is supposed to stay energized from a healthy meal in order to give you the energy your body needs to exercise. If you feel droopy all the time and you don't want to do anything, than you are unhealthy.
Answer:
A vegetarian diet is an example of a good fad diet if you do it correctly. It can help you get lots of veggies and good nutrients from them while still following the non-meat diet you want. This can be effective and good for weight loss becasue you are still eating and getting all the good nutrients and calories from less fatty foods.
Vegan diet (some can be successful but many people fail and do not do good that is why I choose this) The problem with this fad diet is that it can cause nutritional deficiencies and lead to a host of additional health problems, including negatively impacting hormonal health and metabolism. Many people also struggle to find healthy vegan food and end up eating bad and fatty foods instead.
Explanation:
Got a 100
NASA is doing research on the concept of solar sailing. A solar sailing craft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion.
A) Should the sail be absorptive or reflective? Why?
B)The total power output of the sun is 3.90 × 1026 W . How large a sail is necessary to propel a 1.06 × 104 kg spacecraft against the gravitational force of the sun?
Answer:
A = 6.8 km²
Explanation:
A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.
B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;
F_rad = 2IA/c
I is given by the formula;
I = P/(4πr²)
Thus;
F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²
Where;
A is the area of the sail
r is the distance of the sail from the sun
c is the speed of light = 3 × 10^(8) m/s
P is total power output of the sun = 3.90 × 10^(26) W
Now,F_rad = F_g
Where F_g is gravitational force.
Thus;
PA/2cπr² = G•m•M_sun/r²
r² will cancel out to givw;
PA/2cπ = G•m•M_sun
Making A the subject, we have;
A = (2•c•π•G•m•M_sun)/P
Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg
G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²
Thus;
A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))
A = 6.8 × 10^(6) m² = 6.8 km²
Consider a hydraulic lift that uses an input piston with an area of 0.5m2. An input force of 15N is exerted on this piston. If the output piston has an area of 3.5m? What is the output force?
Answer:
The output force of the piston is 105 N.
Explanation:
Given;
the area of the input piston, A₁ = 0.5 m²
the input force of the piston, F₁ = 15 N
the area of the output piston, A₀ = 3.5 m²
the output force of the piston, F₀ = ?
The pressure of the hydraulic lift is given by;
[tex]P = \frac{F}{A}[/tex]
where;
P is the hydraulic pressure
F is the piston force
A is the area of the piston
[tex]P = \frac{F}{A} \\\\\frac{F_o}{A_o} = \frac{F_i}{A_i} \\\\F_o = \frac{F_iA_o}{A_i} \\\\F_o = \frac{15*3.5}{0.5} \\\\F_o = 105 \ N[/tex]
Therefore, the output force of the piston is 105 N.
What is the mass of a rectangular block of
density 2.5 ×10³ k gm³that measures 10cm by 5 cm by 4 cm?
A. 0.002 kg
B. 0.080 kg
C. 0.200 kg
D. 0.500 kg
E. 1.000 kg
Answer:
Option (D) : 0.5 kg
Explanation:
[tex]mass = density \times volume[/tex]
[tex]mass = {2500} \times 0.1 \times 0.05 \times 0.04[/tex]
Mass of block = 0.5 kg
the mass of a rectangular block of density 2.5 ×10³ k gm³ that measures 10cm by 5 cm by 4 cm is 0.5 kg.
What is density ?Density is the ratio of mass to volume. it tells how much mass a body is having for its unit volume. for example egg yolk has 1027kg/m³ of density, means if we collect numbers of egg yolk and keep it in a container having volume 1 m³ then total amount of mass it is having will be 1027kg. Density is a scalar quantity. when we add egg yolk into the water, egg yolk has greater density than water( 997 kg/m³), because of higher density of egg yolk it contains higher mass in same volume as water. hence due to higher mass higher gravitational force is acting on the egg yolk therefore it goes down on the inside the water. water will float upon the egg yolk. same situation we have seen when we spread oil in the water. ( in that case water has higher density than oil. thats why oil floats on the water)
The Volume of the block is,
V = LBD, where L = length, B = breadth , D = depth of the block.
V = 10 × 5 × 4 = 200 cm³
Density of Block = 2.5 ×10³ kg/m³
Density = Mass / Volume
2.5 ×10³ kg/m³ = Mass / 200 cm³
2.5 ×10³ kg/m³ × 200 cm³ = Mass
2.5 ×10³ kg/m³ × 0.2 × 10⁻³ m³ = Mass
Mass = 0.5 kg
To know more about Mass :
https://brainly.com/question/19694949
#SPJ2.
Determine the magnitude and direction of the force between two parallel wires 30 m long and 6.0 cm apart, each carrying 30 A in the same direction.
Answer:
0.09N, attractive
Explanation:
It can be deducted from the question that the currents are arranged in parallel settings, then it is obvious that the force on each of the wire will be attractive toward the other wire.
the magnitude of force can be determined by using below formula;
F2 = (μ₀/2π)(I₁I₂/d)I₂
μ₀ = constant = 4π × 10^-7 H/m,
I₁, I₂ = currents= 30A
L = the length o the wire=30m
d = distance between these two wires= 0.06m
Since the current are arranged in the same direction, they exhibit attractive force on each other.
Then plugging the values Into the formula above we have
F₂ = (4π × 10^-7 T.m/A)/2π) × ((30A)²/ 0.06m)× 30 m
= 0.09 N, attractive
Therefore, the magnitude and direction of the force is 0.09 N, attractive
In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen is D=48.0 cm away from the slits. What is the linear distance Δx between the eighth order maximum and the fourth order maximum on the screen?
Answer:
Δx = 4.68 x 10⁻³ m = 4.68 mm
Explanation:
The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:
Δx = λD/d
So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:
Δx = 4λD/d
where,
Δx = distance between eighth order maximum and fourth order maximum=?
λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m
d = slit separation = 0.2 mm = 2 x 10⁻⁴ m
D = Distance between slits and screen = 48 cm = 0.48 m
Therefore,
Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)
Δx = 4.68 x 10⁻³ m = 4.68 mm
A belt is run over two drums. The larger drum has weight 4 lbs and a radius of gyration of 1.25 inches while the smaller drum has weight 2.7 lbs and a radius of gyration of 0.75 inches. The tension from the smaller drum is held constant at 6 lbs. If it is known that the speed of the belt is 11 ft/s after 0.16 s, what is the tension between the drums?
Answer:
269 lb
Explanation:
We first find the tangential acceleration, a on the drums
a = Δv/Δt since the speed of the belt is 11 ft/s after 0.16 s, Δv = 11 ft/s and Δt = 0.16 s
a = Δv/Δt = 11 ft/s ÷ 0.16 s = 68.75 ft/s²
Since torque τ = Tk = Iα where I = moment of inertia of larger drum = Mk² where m = mass of larger drum = 4 lbs, k = radius of gyration of larger drum = 1.25 inches, T = tension due to larger drum and α = angular acceleration of larger drum.
So, T = Iα/k = Mk²α/k = Mαk = Ma (since a = αk )
T = 4 lbs × 68.75 ft/s² = 275 lb
The tension due to the smaller drum is T' = 6 lb .
So the net tension in the belt is T'' = T - T' = 275 lb - 6 lb = 269 lb
The intensity level 10 m from a point sound source is 85 dB. What is the intensity level 50 m away from the same source
Answer:
425dBExplanation:
Given the intensity level 10 m from a point sound source is 85 dB, then;
L1 = 10m, I1= 85dB ...1
The intensity level 50 m away from the same source cal be calculated using the equivalent expression;
when L2 = 50m, I2 = ? ... 2
Solving equation 1 nad 2;
10m = 85db
50m = x
Cross multiplying;
50 * 85 = 10 * x
10x = 50*85
10x = 4250
Divide both sides by 10
10x/10 = 4250/10
x = 425 dB
Hence, the intensity level 50 m away from the same source is 425dB
A 2-slit arrangement with 60.3 μm separation between the slits is illuminated with 482.0 nm light. Assuming that a viewing screen is located 2.14 m from the slits, find the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side. A. 24.1 mm B. 34.2 mm C. 68.4 mm D. 51.3 mm
Answer:
The distance is [tex]y = 0.03425 \ m[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 60.3 \mu m= 60.3 *10^{-6}\ m[/tex]
The wavelength is [tex]\lambda = 482.0 \ nm = 482.0 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 2.14 \ m[/tex]
Generally the distance of a fringe from the central maxima is mathematically represented as
[tex]y = [m + \frac{1}{2} ] * \frac{\lambda * D}{d}[/tex]
For the first dark fringe m = 0
[tex]y_1 = [0 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_1 = 0.00855 \ m[/tex]
For the second dark fringe m = 1
[tex]y_2 = [1 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_2 = 0.0257 \ m[/tex]
So the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side is
[tex]y = y_1 + y_2[/tex]
[tex]y = 0.00855 + 0.0257[/tex]
[tex]y = 0.03425 \ m[/tex]
The intensity at a certain distance from a bright light source is 7.20 W/m2 .
A. Find the radiation pressures (in pascals) on a totally absorbing surface and a totally reflecting surface.
B. Find the radiation pressures (in atmospheres) on a totally absorbing surface and a totally reflecting surface.
Answer:
A) P_rad.abs = 2.4 × 10^(-8) Pa and P_rad.ref = 4.8 × 10^(-8) Pa
B) P_rad.abs = 2.369 × 10^(-13) atm and P_rad.ref = 4.738 × 10^(-13) atm
Explanation:
A) The formula for radiation pressure for absorbed light is given as;
P_rad = I/c
Where I is the intensity = 7.20 W/m² and c is the speed of light = 3 × 10^(8) m/s
Thus;
P_rad = 7.2/(3 × 10^(8))
P_rad.abs = 2.4 × 10^(-8) Pa
Now formula for radiation pressure for reflected light is given as;
P_rad = 2I/c
Thus;
P_rad = (2 × 7.2)/(3 × 10^(8))
P_rad.ref = 4.8 × 10^(-8) Pa
B) Now, 1.013 × 10^(5) Pa = 1 atm
Thus, for the absorbed surface, we have;
P_rad.abs = (2.4 × 10^(-8))/(1.013 × 10^(5))
P_rad.abs = 2.369 × 10^(-13) atm
For the reflecting surface, we have;
P_rad_ref = (4.8 × 10^(-8))/(1.013 × 10^(5))
P_rad.ref = 4.738 × 10^(-13) atm
which objects would have a greater gravitational force between them, Objects A and B, or Objects B and C
Answer:
Objects that are closer together have a stronger force of gravity between them.
Explanation:
For example, the moon is closer to Earth than it is to the more massive sun, so the force of gravity is greater between the moon and Earth than between the moon and the sun.
