To solve this problem it is necessary to apply the concepts related to pressure as a unit that measures the force applied in a specific area as well as pressure as a measurement of the density of the liquid to which it is subjected, its depth and the respective gravity.
The two definitions of pressure can be enclosed under the following equations
[tex]P = \frac{F}{A}[/tex]
Where
F= Force
A = Area
[tex]P = \rho gh[/tex]
Where,
[tex]\rho =[/tex] Density
g = Gravity
h = Height
Our values are given as,
[tex]d = 0.8m \rightarrow r = 0.4m[/tex]
[tex]A = \pi r^2 = \pi * 0.4^2 = 0.503m^2[/tex]
If we make a comparison between the lid and the tube, the diameter of the tube becomes negligible.
Matching the two previous expressions we have to
[tex]\frac{F}{A} = \rho g h[/tex]
Re-arrange to find h
[tex]h = \frac{F}{A\rho g}[/tex]
[tex]h = \frac{390}{(0.503)(1000)(9.8)}[/tex]
[tex]h = 0.079m[/tex]
[tex]h = 7.9cm[/tex]
Therefore the height of water in the tube is 7.9cm
An audio engineer takes decibel readings at distances of r1 = 13 m and r2 = 24 m from a concert stage speaker during a sound check. When he is r1 from the speaker, the engineer registers a decibel level of β1 = 101 dB on his loudness meter.
Answer
given,
Distance for decibel reading
r₁ = 13 m
r₂ = 24 m
When the engineer is at r₁ reading is β₁ = 101 dB
now, Calculating the Intensity at r₁
Using formula
[tex]\beta = 10 log(\dfrac{I}{I_0})[/tex]
I₀ = 10⁻¹² W/m²
now inserting the given values
[tex]101= 10 log(\dfrac{I}{10^{-12}})[/tex]
[tex]10.1= log(\dfrac{I}{10^{-12}})[/tex]
[tex]\dfrac{I}{10^{-12}}= 10^{10.1}[/tex]
[tex]I =10^{10.1} \times 10^{-12}[/tex]
I = 0.01258 W/m²
now, calculating power at r₁
P₁ = I₁ A₁
P₁ = 0.01258 x 4 π r²
P₁ = 0.01258 x 4 π x 13²
P₁ = 26.72 W
Final answer:
The question deals with acoustics in physics, where an audio engineer uses the decibel scale to measure sound pressure levels and requires an understanding of the inverse square law and logarithmic measurements of sound intensity.
Explanation:
The student's question involves the concept of sound intensity levels and decibels, which are part of the field of acoustics, a topic in physics. Specifically, the student is dealing with the measurement of sound pressure levels at various distances from a sound source, which is an application of the inverse square law in acoustics. The decibel (dB) scale is a logarithmic unit used to measure sound intensity, with 0 dB representing the threshold of human hearing at an intensity of 10-12 W/m².
The loudness experienced by the audio engineer at various distances can be computed using the formula for sound intensity level (SIL), which is SIL = 10 log(I/I0), where I is the sound intensity and I0 is the reference intensity of 10-12 W/m².
9. A baseball pitcher hurls a 400 g baseball toward a batter with a speed of 30 m/s. The batter hits the pitch directly back toward the pitcher and it travels at 38 m/s. If the collision with the bat required 80 ms (milliseconds), find the force the bat applies to the ball. (Watch signs.)
Answer:
The force the bat applies to the ball is 340 N
Explanation:
given information:
mass of the ball, m = 400 g = 0.4 kg
initial velocity, [tex]v_{i}[/tex] = 30 m/s
final velocity, [tex]v_{f}[/tex] = 38 m/s
time, t = 80 ms = 0.08 s
the correlation between the change in momentum and the force is shown by the following equation
ΔP = FΔt
F = ΔP/Δt
where,
Δp = mΔv
= m([tex]v_{f} - v_{i}[/tex])
= m([tex]v_{f} + v_{i}[/tex])
[tex]v_{i}[/tex] = - because in opposite direction, thus
F = ΔP/Δt
= m([tex]v_{f} + v_{i}[/tex])/Δt
= (0.4)(38+30)/0.08
= 340 N
A compound die is used to blank and punch a large washer out of 6061ST aluminum alloy sheet stock 3.2 mm thick. The outside diameter of the washer is 25.0 mm, and the inside diameter is 12.0 mm. Determine the punch and die sizes for (a) the blanking operation, and (b) the punching operation.
Answer:
24.616 mm
12.384 mm
Explanation:
[tex]D_b[/tex] = Blanking die diameter = 25 mm
[tex]D_h[/tex] = Punch diameter = 12 mm
a = Allowance = 0.06 for aluminium
t = Thickness = 3.2 mm
Clearance
[tex]c=at\\\Rightarrow c=0.06\times 3.2\\\Rightarrow c=0.192[/tex]
Diameter of punch is
[tex]d_p=D_b-2c\\\Rightarrow d_p=25-2\times 0.192\\\Rightarrow d_p=24.616\ mm[/tex]
Diameter of blanking punch is 24.616 m
For punching operation
[tex]d_d=D_h+2c\\\Rightarrow d_d=12+2\times 0.192\\\Rightarrow d_d=12.384\ mm[/tex]
The diameter of punching die is 12.384 mm
For the blanking operation, the punch diameter should be 12.0 mm and the die diameter should be 25.0 mm. For the punching operation, the punch diameter should be 12.0 mm and the die diameter should be slightly larger than 12.0 mm, typically by a clearance of 0.05 mm to 0.15 mm per side.
(a) Blanking Operation:
The blanking operation involves cutting the outside profile of the washer from the aluminum sheet stock. The punch and die sizes for this operation are determined by the outside and inside dimensions of the washer.
- The die diameter for the blanking operation should match the outside diameter of the washer, which is 25.0 mm. This is the size of the hole in the die through which the punch will pass to cut the washer's outer edge.
- The punch diameter for the blanking operation should match the inside diameter of the washer, which is 12.0 mm. This is the size of the punch that will push through the sheet stock to create the inner hole of the washer.
(b) Punching Operation:
The punching operation involves cutting the inside hole of the washer after the blanking operation has been completed. The punch and die sizes for this operation are determined by the inside diameter of the washer and the required clearance.
- The punch diameter for the punching operation should also be 12.0 mm, the same as the inside diameter of the washer. This ensures that the punch cuts the material exactly at the desired inner diameter.
- The die diameter for the punching operation should be slightly larger than the punch diameter to allow for clearance. Clearance is necessary to prevent the punch and die from binding together and to allow for the removal of the slug (the piece of material that is punched out). The typical clearance ranges from 0.05 mm to 0.15 mm per side. Therefore, the die diameter would be 12.0 mm plus two times the clearance. If we take the middle value of the clearance range, which is 0.10 mm per side, the die diameter would be 12.0 mm + 2 * 0.10 mm = 12.20 mm.
A compound microscope is made with an objective lens (f0 = 0.900 cm) and an eyepiece (fe = 1.10 cm). The lenses are separated by a distance of 10.0 cm. What is the angular magnification? (Assume the near point is 25.0 cm.)
Answer:
-252.52
Explanation:
L = Distance between lenses = 10 cm
D = Near point = 25 cm
[tex]f_o[/tex] = Focal length of objective = 0.9 cm
[tex]f_e[/tex] = Focal length of eyepiece = 1.1 cm
Magnification of a compound microscope is given by
[tex]m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52[/tex]
The angular magnification of the compound microscope is -252.52
A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave is 0.4. What is the largest ratio of tensile forces between the two ends of the rope before the rope starts to slide over the sheave?
Answer:3.51
Explanation:
Given
Coefficient of Friction [tex]\mu =0.4 [/tex]
Consider a small element at an angle \theta having an angle of [tex]d\theta [/tex]
Normal Force[tex]=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}[/tex]
[tex]N=T\cdot d\theta [/tex]
Friction [tex]f=\mu \times Normal\ Reaction[/tex]
[tex]f=\mu \cdot N[/tex]
and [tex]T+dT-T=f=\mu Td\theta [/tex]
[tex]dT=\mu Td\theta [/tex]
[tex]\frac{dT}{T}=\mu d\theta [/tex]
[tex]\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta [/tex]
[tex]\frac{T_2}{T_1}=e^{\mu \pi}[/tex]
[tex]\frac{T_2}{T_1}=e^{0.4\times \pi }[/tex]
[tex]\frac{T_2}{T_1}==e^{1.256}[/tex]
[tex]\frac{T_2}{T_1}=3.51[/tex]
How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C ? The specific heat of ice is 2090 J/(kg.K) and the latent heat of fusion of water is 33.5 x 10^4 J/kg .
Answer:209.98 kJ
Explanation:
mass of water [tex]m=456 gm[/tex]
Initial Temperature of Water [tex]T_i=25^{\circ}C[/tex]
Final Temperature of water [tex]T_f=-10^{\circ}C[/tex]
specific heat of ice [tex]c=2090 J/kg-K[/tex]
Latent heat [tex]L=33.5\times 10^4 J/kg[/tex]
specific heat of water [tex]c_{water}=4.184 KJ/kg-K[/tex]
Heat require to convert water at [tex]T=25^{\circ}C[/tex] to [tex]T=0^{\circ}C[/tex]
[tex]Q_1=0.456\times 4.184\times (25-0)=47.69 kJ[/tex]
Heat require to convert water at [tex]T=0^{\circ}[/tex] to ice at [tex]T=0^{\circ}[/tex]
[tex]Q_2=m\times L=0.456\times 33.5\times 10^4=152.76 kJ[/tex]
heat require to convert ice at [tex]T=0^{\circ} C\ to\ T=-10^{\circ} C[/tex]
[tex]Q_3=0.456\times 2090\times (0-(-10))=9.53 kJ[/tex]
Total heat [tex]Q=Q_1+Q_2+Q_3[/tex]
[tex]Q=47.69+152.76+9.53=209.98 kJ[/tex]
What is a particulate ? Name a couple of examples.
Answer:Particulates are small, distinct solids suspended in a liquid or gas and example are dust,soot,and salt particles
Explanation:
Please help I have marked the circles with numbers which one is it anybody knows?
Where is the natural light display called aurora borealis located?
Answer:
hi sandra!!
Explanation:
the number 3 is the correct!!
The Aurora is an incredible light show caused by collisions between electrically charged particles released from the sun that enter the earth’s atmosphere and collide with gases such as oxygen and nitrogen. The lights are seen around the magnetic poles of the northern and southern hemispheres.
Auroras that occur in the northern hemisphere are called ‘Aurora Borealis’ or ‘northern lights’ and auroras that occur in the southern hemisphere are called ‘Aurora Australis’ or ‘southern lights’.
Auroral displays can appear in many differents colours, but green is the most common. Colours such as red, yellow, green, blue and violet are also seen occasionally. The auroras can appear in many forms, from small patches of light that appear out of nowhere to streamers, arcs, rippling curtains or shooting rays that light up the sky with an incredible glow.
Answer:
Option-(3): The aurora borealis are the formation of the colored patterns on the pole of the planet, but other celestial bodies also experience such effects due to the heat flares or the sun flares when exposed to them.
Explanation:
Aurora borealis:
The natural phenomenon which is caused in most of the celestial bodies, but when our planet earth is exposed to such electromagnetic rays or flares released from the Sun, when the energy level on the surface increases and the flare is cut off from the environment of the sun. As, starts motion towards the different celestial bodies present in the solar system.
And when it reaches earth, the earth surface reflects the high in energy atoms coming towards its atmosphere. As the magnetic field present on the earth surface interacts with the Sun's flare and then it cut off's into two moving towards the two poles of the planet.
And when the molecules cools down, they are seen are different patterns on the poles.Which are also called as the aurora borealis in scientific terms.A 2.95 g sample of an unknown chlorofluorocarbon (an organic compound containing chlorine and fluorine) is decomposed and produces 581 mL of chlorine gas at a pressure of 752 mmHg and a temperature of 298 K. What is the percent by mass of chlorine in the chlorofluorocarbon?
Answer:
mass percent of chlorine is 56.5%
Explanation:
Given data:
Volume V = 581 mL = 0.581 L
Pressure P = 752 mm of Hg = 752/760 = 0.9894 atm
Temperature T = 298 K
Molar gas constant R = 0.08206 atm.L/mol.K
from Ideal gas equation we have following relation
PV = nRT solving for n
[tex]n = \frac{PV}{RT}[/tex]
[tex]n = \frac{0.9894 \times 0.581}{0.08206 \times 298}[/tex]
n = 0.0235 mole
mas of [tex]Cl_2[/tex] = moles × molar mass of [tex]Cl_2[/tex]
[tex]= 0.0235 \times 70.906 = 1.66 g[/tex]
Mass percent of chlorine [tex]= \frac{mass\ of\ Cl}{mass\ of\ sample}[/tex]
[tex]=\frac{1.66}{2.95} = 0.565 = 56.5 %[/tex]
During each cycle of operation a refrigerator absorbs 56 cal from the freezer compartment and expels 81 cal tothe room. If one cycle occurs every 10 s, how many minutes will it take to freeze 570 g of water, initially at 0°C?(Lv = 80 cal/g
Answer:
138.18 minutes
Explanation:
[tex]L_v[/tex] = Latent heat of water at 0°C = 80 cal/g
m = Mass of water = 570 g
Heat removed for freezing
[tex]Q=mL_v\\\Rightarrow Q=570\times 80\\\Rightarrow Q=45600\ cal[/tex]
Let N be the number of cycles and each cycle removes 56 cal from the freezer.
So,
[tex]55\times N=45600\\\Rightarrow N=\frac{45600}{55}[/tex]
Each cycle takes 10 seconds so the total time would be
[tex]\frac{45600}{55}\times \frac{10}{60}=138.18\ minutes[/tex]
The total time taken to freeze 138.18 minutes
What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a final velocity of 7.3 m/s? Express the moment of inertia as a multiple of MR2, where M is the mass of the object and R is its radius. (Hint: Use total conservation of mechanical energy)
Answer:
I = 0.287 MR²
Explanation:
given,
height of the object = 3.5 m
initial velocity = 0 m/s
final velocity = 7.3 m/s
moment of inertia = ?
Using total conservation of mechanical energy
change in potential energy will be equal to change in KE (rotational) and KE(transnational)
PE = KE(transnational) + KE (rotational)
[tex]mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2[/tex]
v = r ω
[tex]mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}[/tex]
[tex]I = \dfrac{m(2gh - v^2)r^2}{v^2}[/tex]
[tex]I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}[/tex]
[tex]I =mr^2(0.287)[/tex]
I = 0.287 MR²
5.00 g of molten NaCl is at its melting temperature. The salt is added to 25.00 mL water at 25.0 °C in a coffee cup calorimeter. The salt becomes solid and the temperature of the water increases to 47.9 °C. a) Calculate the amount of heat transferred.
Answer:
heat absorbed by water = 2.39 KJ
Explanation:
given,
mass of molten NaCl = 5 g
mass of water = 25 g
Temperature of water = 25.0 °C
Temperature increased to = 47.9 °C
heat absorbed by water = m S ΔT
S is specific heat in J/gm °C
Specific heat of water = 4.184 J/gm °C
m is mass of water
heat absorbed by water = 25 x 4.184 x (47.9-25)
= 25 x 4.184 x 22.9
= 2395.34 J
heat absorbed by water = 2.39 KJ
The amount of heat transferred is 2.39 KJ.
What is Heat?
This is the form of energy which is transferred from one body to another as the result of a difference in temperature.
Parameters
mass of molten NaCl = 5 g
mass of water = 25 g
Temperature of water = 25.0 °C
Final Temperature of water = 47.9 °C
Specific heat of water = 4.184 J/gm °C
Heat = m S ΔT
Heat absorbed by water = 25 x 4.184 x (47.9 -25)
= 25 x 4.184 x 22.9 = 2395.34 J
= 2.39 KJ.
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A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 2.5 m away. He then wants to take a picture of an object that is 50 cm away.
Part A
In which direction must the lens move to focus on this second object?
Part B
How far must the lens move to focus on this second object?
Answer:
0.004 m away from the film
Explanation:
u = Object distance
v = Image distance
f = Focal length = 50 mm
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m[/tex]
The image distance is 0.051 m
When u = 50 cm
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m[/tex]
The image distance is 0.055 m
The lens has moved 0.055-0.051 = 0.004 m away from the film
To focus on the closer object, the lens must move away from the object. By applying the lens formula, we calculate that the lens must move exactly 7mm away from the initial position.
Explanation:This problem comes down to the concept of the lens formula in optical physics. The lens formula is 1/f = 1/v - 1/u where f is the focal length, v is the image distance, and u is the object distance.
Part A: To focus on the second object which is closer, the lens must move away from the object.
Part B: To calculate how far the lens must move, we need to apply the lens formula. Let's calculate the image distances for the two object distances.
For an object at 2.5 m, 1/v = 1/f + 1/u = 1/0.05 + 1/2.5 = 20 + 0.4 = 20.4. So, v = 0.049 m or 49 mm.
For an object at 0.5 m, 1/v = 1/f + 1/u = 1/0.05 - 1/0.5 = 20 - 2 = 18. So, v = 0.056 m or 56 mm.
Hence, the lens must move 56mm - 49mm = 7mm away from the initial position.
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Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a solid cylinder of mass 50 kg and diameter 25 cm. As this cylinder is turned with a crank, the rope raises the bucket. The mass of a bucket of water is 20 kg. Someone cranks the bucket up and then lets go of the crank, and the bucket of water falls down to the bottom of the well. Without friction or air resistance, what is the angular acceleration of the 50-kg cylinder?
To find the angular acceleration of the cylinder, calculate the moment of inertia of the cylinder, determine the torque due to the falling bucket, and then use Newton's second law for rotation to find the angular acceleration.
Explanation:The subject involves calculating the angular acceleration of a cylinder when the bucket of water tied to a rope around the cylinder is let go. We assume no friction or air resistance in this case.
First, we need to calculate the moment of inertia (I) of the solid cylinder, which can be calculated using the formula I = 0.5 * m * r^2. Here, m is the mass of the cylinder (50 kg) and r is the radius (half of the diameter, 0.125 m). After calculating I, use Newton's second law for rotation τ = I * α, where τ is torque and α is the angular acceleration.
The torque τ can be calculated as the product of the force exerted by the falling bucket (F) and the radius r of the cylinder. The force F is the weight of the falling bucket, m * g, where m is the mass of the bucket (20 kg) and g is acceleration due to gravity (about 9.8 m/s^2). Once you have calculated τ, you can solve for α by rearranging the equation to α = τ/I.
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The angular acceleration of the 50-kg cylinder is found to be approximately 62.7 rad/s².
To find the angular acceleration of the 50-kg cylinder when the bucket of water falls down to the bottom of the well, we will use the concept of rotational dynamics. The moment of inertia (I) of a solid cylinder is given by:
I = (1/2) [tex]\times[/tex] M[tex]\times[/tex] R²
where M is the mass and R is the radius. Given the mass M is 50 kg and the diameter is 25 cm, so the radius R is 0.125 m:
I = (1/2) * 50 kg[tex]\times[/tex] (0.125 m)² = 0.390625 kg m²
The torque (τ) caused by the falling bucket is:
τ = Force[tex]\times[/tex] RadiusThe force exerted by the bucket of water is its weight, F = m [tex]\times[/tex] g, where m is the mass of the bucket and water (20 kg) and g is the acceleration due to gravity (9.8 m/s²):
F = 20 kg [tex]\times[/tex] 9.8 m/s² = 196 NSo,
τ = 196 N [tex]\times[/tex] 0.125 m = 24.5 N mUsing Newton's second law for rotation:
τ = I[tex]\times[/tex] αwhere α is the angular acceleration:
24.5 N m = 0.390625 kg m²[tex]\times[/tex] αSolving for α:
α = 24.5 N m / 0.390625 kg m² ≈ 62.7 rad/s²Therefore, the angular acceleration of the 50-kg cylinder is approximately 62.7 rad/s².
A 20-cm-diameter vertical cylinder is sealed at the top by a frictionless 23 kg piston. The piston is 86 cm above the bottom when the gas temperature is 307°C. The air above the piston is at 1.00 atm pressure.
A) What is the gas pressure inside the cylinder?
B) What will the height of the piston be if the temperature is lowered to 13°C?
Answer:
108507.02596 Pa
42.42 cm
Explanation:
F = Force
m = Mass of piston = 23 kg
[tex]T_1[/tex] = Initial temperature = 307°C
[tex]T_2[/tex] = Final temperature = 13°C
[tex]P_o[/tex] = Outside pressure
[tex]P_i[/tex] = Pressure inside cylinder
A = Area of pistion
h = Height of piston
Change in pressure is given by
[tex]\Delta P=\frac{F}{A}\\\Rightarrow \Delta P=\frac{mg}{\pi r^2}\\\Rightarrow \Delta P=\frac{23\times 9.81}{\pi 0.1^2}\\\Rightarrow \Delta P=7182.02596\ Pa[/tex]
[tex]P_i-P_o=\Delta P\\\Rightarrow P_i=\Delta P+P_0\\\Rightarrow P_i=7182.02596+101325\\\Rightarrow P_i=108507.02596\ Pa[/tex]
Gas pressure inside the cylinder is 108507.02596 Pa
From ideal gas law we have the relation
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\Rightarrow \frac{Ah_1}{T_1}=\frac{Ah_2}{T_2}\\\Rightarrow \frac{h_1}{T_1}=\frac{h_2}{T_2}\\\Rightarrow h_2=\frac{h_1}{T_1}\times T_2\\\Rightarrow h_2=\frac{0.86}{307+273.15}\times (13+273.15)\\\Rightarrow h_2=0.42418\ m=42.42\ cm[/tex]
The height of the piston at 13°C will by 42.42 cm
(a) The gauge pressure inside the cylinder is 108,500 Pa.
(b) The height of the piston when the temperature is lowered to 13°C is 42.4 cm.
Change in gas pressure inside the cylinderThe change in the pressure of the gas inside the cylinder is calculated as follows;
ΔP = F/A
ΔP = mg/πr²
ΔP = (23 x 9.8)/(π x0.1²)
ΔP = 7,174.7 Pa
Gauge pressure inside the cylinderThe gauge pressure inside the cylinder is calculated as follows;
Pi = ΔP + Po
Pi = 7,174.7 + 101325
Pi = 108,500 Pa
Height of the pistonThe height of the piston is calculated as follows;
[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2} \\\\\frac{Ah_1}{T_1} = \frac{Ah_2}{T_2} \\\\h_2 = \frac{T_2h_1}{T_1} \\\\h_2 = \frac{(13 + 273) \times 0.86}{(307 + 273)} \\\\h_2 = 0.424 \ m[/tex]
h₂ = 42.4 cm
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A boy exerts a force of 11.8 N at 28.0° above the horizontal on a 6.15 kg sled. Find the work done by the boy and the final speed of the sled after it moves 2.75 m, assuming the sled starts with an initial speed of 0.370 m/s and slides horizontally without friction.
Explanation:
It is given that,
Force exerted by a boy, F = 11.8 N
Angle above the horizontal, [tex]\theta=28^{\circ}[/tex]
Mass of the sled, m = 6.15 kg
Distance moved, d = 2.75 m
Initial speed, u = 0.37 m/s
Let W is the work done by the boy. Using the expression for the work done to find it as :
[tex]W=Fd\ cos\theta[/tex]
[tex]W=11.8\times 2.75\ cos(28)[/tex]
W = 28.65 joules
Let v is the final speed of the sled. Using the work energy theorem to find it. It states that the work done is equal to the change in kinetic energy of an object. It is given by :
[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]
[tex]v=\sqrt{\dfrac{2W}{m}+u^2}[/tex]
[tex]v=\sqrt{\dfrac{2\times 28.65}{6.15}+(0.37)^2}[/tex]
v = 3.07 m/s
So, the final speed of the sled after it moves 2.75 m is 3.07 m/s. Hence, this is the required solution.
The final speed of the sled is approximately 3.073 m/s and the the work done by the boy on the sled is 28.64 J.
To solve this problem, we need to calculate the work done by the boy on the sled and then use that to find the final speed of the sled.
First, let's calculate the work done by the boy. Work (W) is defined as the product of the force (F) applied in the direction of motion and the distance (d) over which the force is applied. The force applied in the direction of motion is the horizontal component of the force exerted by the boy. We can calculate this using trigonometry:
[tex]\[ F_{\text{horizontal}} = F \cdot \cos(\theta) \][/tex]
[tex]\[ F_{\text{horizontal}} = 11.8 \, \text{N} \cdot \cos(28.0^\circ) \][/tex]
[tex]\[ F_{\text{horizontal}} \approx 11.8 \, \text{N} \cdot 0.8829 \][/tex]
[tex]\[ F_{\text{horizontal}} \approx 10.42 \, \text{N} \][/tex]
Now, we can calculate the work done:
[tex]\[ W = F_{\text{horizontal}} \cdot d \][/tex]
[tex]\[ W = 10.42 \, \text{N} \cdot 2.75 \, \text{m} \][/tex]
\[ W \approx 28.64 \, \text{J} \]
Next, we need to find the final speed of the sled. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy (KE):
[tex]\[ W = \Delta KE \][/tex]
[tex]\[ W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \][/tex]
Given that the initial speed [tex]\( v_i \)[/tex] is 0.370 m/s and the mass [tex]\( m \)[/tex] of the sled is 6.15 kg, we can solve for the final speed [tex]\( v_f \)[/tex]:
[tex]\[ 28.64 \, \text{J} = \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot v_f^2 - \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot (0.370 \, \text{m/s})^2 \][/tex]
[tex]\[ 28.64 \, \text{J} = 3.075 \, \text{kg} \cdot v_f^2 - 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ 28.64 \, \text{J} + 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]
[tex]\[ 29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]
[tex]\[ v_f^2 = \frac{29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{3.075 \, \text{kg}} \][/tex]
[tex]\[ v_f^2 \approx 9.444 \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ v_f \approx \sqrt{9.444 \, \text{m}^2/\text{s}^2} \][/tex]
[tex]\[ v_f \approx 3.073 \, \text{m/s} \][/tex]
A rotating wheel accelerates at a constant rate from an angular speed of 25 rad/s to 37 rad/s in a time interval of 8 s. (a) What is the angular acceleration in rad/s/s? rad/s/s (b) What is the average angular speed? rad/s (c) What is the angle in radians through which the wheel rotates? rad (d) What is this angle in degrees?
The value of all options are mathematically given as
a) a=1.5 rad/s^2
b)theta=248 rad
c) theta '=248 radians
d) angles in degree=14207.511
What is the angular acceleration in rad/s/s?Question Parameter(s):
Generally, the equation for the final angular speed is mathematically given as
w2=w+at
Therefore
37=25+a 8
a=1.5 rad/s^2
What is the average angular speed?b)
Average angular speed
[tex]\theta =\omega _1t+\frac{1}{2}\alpha t^2[/tex]
[tex]\theta =25\times 8+\frac{1}{2}\times 1.5\times 8^2[/tex]
theta=248 rad
Hence
theta =200+48
theta=248 rad
What is the angle in radians through which the wheel rotates, in rad and degree?In conclusion,
c) theta '=248 radians
d) angles in degree=14207.511
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Final answer:
The angular acceleration is 1.5 rad/s², the average angular speed is 31 rad/s, the angle through which the wheel rotates is 320 rad, and the angle in degrees is 18333 degrees.
Explanation:
(a) To find the angular acceleration, we can use the formula: angular acceleration = (final angular velocity - initial angular velocity) / time interval. Plugging in the given values, we get: angular acceleration = (37 rad/s - 25 rad/s) / 8 s = 1.5 rad/s².
(b) The average angular speed can be found by taking the average of the initial and final angular velocities. So, the average angular speed = (25 rad/s + 37 rad/s) / 2 = 31 rad/s.
(c) The angle through which the wheel rotates can be found using the formula: angle = initial angular velocity * time + (1/2) * angular acceleration * time². Plugging in the given values, we get: angle = 25 rad/s * 8 s + (1/2) * 1.5 rad/s² * (8 s)² = 320 rad.
(d) To convert radians to degrees, we use the conversion factor: 1 radian = 180 degrees / pi. So, the angle in degrees = 320 rad * 180 degrees / pi = 18333 degrees (rounded to the nearest whole number).
A rectangular settling tank is being designed to settle sand particles that have a settling velocity of 2.0 cm/s. The water flow is 10,000 m3/d. The tank length should be four times the width and the width is equal to the depth. Please find the dimensions for the tank in meter.
Answer:
b = 1.2 m
L = 4.8 m
d= 1.2 m
Explanation:
Lets take
Width of the tank = b
Depth = d
length = L
Given that L = 4 b
v= 2 cm/s
Flow rate Q= 10,000 m³/d
We know that 1 d = 24 hr = 24 x 3600 s
1 d= 86400 s
Q= 0.115 m³/s
Flow rate Q
Q= Area x velocity
Q= L . b .v
0.115 = 4 b . b . 0.02
4 b² = 5.78
b = 1.2 m
L = 4.8 m
d= 1.2 m
This is the dimension of the tank.
To design the settling tank with the given water flow and settling velocity, we use conservation of mass and settling speed to calculate the dimensions. We convert the flow rate to m3/s, relate it to the tank's cross-sectional area, and given settling characteristics to find the tank's width, length, and depth.
Explanation:To design a rectangular settling tank for settling sand particles with a settling velocity of 2.0 cm/s and a water flow of 10,000 m3/d, we will apply the principles of conservation of mass and the given settling velocity to determine the dimensions of the tank. The problem states that the tank length (L) should be four times the width (W), and the width should be equal to the depth (D), therefore L = 4W and W = D.
To find the dimensions, we first convert the water flow rate into m3/s: flow rate = 10,000 m3/d × (1 day/86400 s) ≈ 0.1157 m3/s. Since the sand particles settle at 2 cm/s, for a particle to settle before reaching the end of the tank, the residence time should be at least the depth divided by the settling velocity. From here, we use the flow rate (Q), which is also equal to the cross-sectional area (A) times the water velocity (V), and we know that A = W² for this tank because W = D.
Therefore, Q = W²V. Using V = L / (residence time) and substituting for L = 4W, we find that Q = W² × (4W / (W/0.02 m/s)) which simplifies to Q = 0.08 m/s × W³. Solving for W gives us W = ∛(Q / 0.08 m/s) and subsequently we find L = 4W and D = W. Finally, we can plug in the flow rate and solve for the dimensions in meters.
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With the switch open, roughly what must be the resistance of the resistor on the right for the current out of the battery to be the same as when the switch is closed (and the resistances of the two resistors are 20 Ω and 10 Ω)?
A) 7ΩB) 15ΩC) 30ΩD) 5Ω
Answer:
The resistance must be 6.67[tex]\Omega[/tex]
Solution:
Resistance, [tex]R_{1} = 20\Omega[/tex]
Resistance, [tex]R_{2} = 10\Omega[/tex]
For the current to be the same when the switch is open or closed, the resistances must be connected in parallel as current is distributed in parallel with the same voltage across the circuit:
Thus in parallel:
[tex]\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}[/tex]
[tex]\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{10}[/tex]
[tex]R_{eq} = 6.67\ \Omega[/tex]
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical surface with radius 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the concave surface.A) Find the position of the image of the arrow formed by paraxial rays incident on the convex surface. (answer is s1 in units cm)B) Find the height of the image formed by paraxial rays incident on the convex surface. (answer is y1 in units mm)C) Is the image erect or inverted?
Answer:
a) q = -9.23 cm, b) h’= 0.577 mm , c) image is right and virtual
Explanation:
This is an optical exercise, where the constructor equation should be used
1 / f = 1 / p + 1 / q
Where f is the focal length, p the distance to the object and q the distance to the image
A) The cocal distance is framed with the relationship
1 / f = (n₂-1) (1 /R₁ -1 /R₂)
In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave
1 / f = (1.60 -1) (1 /∞ - 1 / (-4))
1 / f = 0.6 / 4 = 0.15
f = 6.67 cm
We have the distance to the object p = 24.0 cm, let's calculate
1 / q = 1 / f - 1 / p
1 / q = 1 / 6.67 - 1/24
1 / q = 0.15 - 0.04167 = 0.10833
q = -9.23 cm
distance to the negative image is before the lens
B) the magnification of the lenses is given by
M = h ’/ h = - q / p
h’= - q / p h
h’= - (-9.23) / 24.0 0.150
h’= 0.05759 cm
h’= 0.577 mm
C) the object is after the focal length, therefore, the image is right and virtual
The wavelength range for infrared radiation is 10^-5 m, while that of ultraviolet radiation is 10^-8 m. Which type of radiation has more energy and why?A.) Ultraviolet because it has a higher frequencyB.) Ultraviolet because it has a longer wavelengthC.) Infrared because it has a lower frequencyD.) Infrared because it has a shorter wavelength.
Answer:
A,
Explanation:
To solve the exercise it is necessary to use the concepts and definitions given for electromagnetic energy, in which
[tex]E = \frac{hc}{\lambda}[/tex]
Where,
h = Plank's constant [tex](6.626*10^{-34}J.s)[/tex]
c = Speed of light [tex](3*10^8m/s)[/tex]
[tex]\lambda[/tex] =Wavelength
If we analyze these characteristics both h and c are constant so the energy is inversely proportional to the size of the wave.
The larger the amplitude of the wave, the smaller the energy.
On the other hand we have the frequency value defined as
[tex]f = \frac{c}{\lambda}[/tex]
In this case the frequency is also inversely proportional to the wavelength.
In this case, the amplitude of the largest wave is infrared, so it will have less energy and less frequency. The fact that it has a low frequency by the wavelength, also generates that it has a low energy. But not because it has a large wavelength, on the contrary, because its wavelength is small.
In the case of the ultraviolet wave it will have greater frequency and greater energy. Therefore of all the options, A is the only one valid.
A bullet is shot straight up in the air with an initial velocity of v=1000 feet per second. If the equation describing the bullets height from the ground is given by f(t)=−16t2+1000t+7, over what interval is the bullet speeding up, and over what interval is it slowing down?
Answer:
It slows down from 0-31.25s
It speeds up from 31.25-62.507s
Explanation:
If we find the maximum of the equation, we will know the moment when it changes direction. It will slow down on the first interval, and speed up on the second one.
[tex]Y = -16t^2+1000t+7[/tex]
[tex]Y' = -32t+1000 = 0[/tex] Solving for t:
t = 31.25s
The bullet will slow down in the interval 0-31.25s
Let's now find the moment when it hits ground:
[tex]Y = 0 = -16t^2+1000t+7[/tex] Solving for t:
t1 = -0.0069s t2 = 62.507s
The bullet will speed up in the interval 31.25-62.507s
Four point masses, each of mass 1.3 kg are placed at the corners of a rigid massles square of side 1.1 m. Find the moment of inertia of this system about an axis that is perpendicular to the plane of the square and passes through one of the masses.
Answer:
I= 6.292 kg.m²
Explanation:
Given that
m = 1.3 kg
Side of square a= 1.1 m
The distance r
[tex]r=\sqrt{{a^2}+{a^2}}[/tex]
[tex]r={a}{\sqrt 2}[/tex]
[tex]r={1.1}{\sqrt 2}[/tex]
The moment of inertia I
The axis passes through one of the mass then the distance of the that mass from the axis will be zero.
I = m a² + m a² + m r²
By putting the values
I = m a² + m a² + m r²
I =m( 2 a² + r²)
I =1.3( 2 x 1.1² + 2 x 1.1²)
I = 1.3 x 4 x 1.1² kg.m²
I= 6.292 kg.m²
A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be completed by geothermal water available at 160°C at a mass flow rate of 2 kg/s. The inner tube is thin-walled and has a diameter of 1.5 cm. If the overall heat transfer coefficient of the heat exchanger is 640 W/m2K, determine the length of the heat exchanger required to achieve the required heating.
Answer:L=109.16 m
Explanation:
Given
initial temperature [tex]=20^{\circ}C[/tex]
Final Temperature [tex]=80^{\circ}C[/tex]
mass flow rate of cold fluid [tex]\dot{m_c}=1.2 kg/s[/tex]
Initial Geothermal water temperature [tex]T_h_i=160^{\circ}C[/tex]
Let final Temperature be T
mass flow rate of geothermal water [tex]\dot{m_h}=2 kg/s[/tex]
diameter of inner wall [tex]d_i=1.5 cm[/tex]
[tex]U_{overall}=640 W/m^2K[/tex]
specific heat of water [tex]c=4.18 kJ/kg-K[/tex]
balancing energy
Heat lost by hot fluid=heat gained by cold Fluid
[tex]\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)[/tex]
[tex]2\times (160-T)=1.2\times (80-20)[/tex]
[tex]160-T=36[/tex]
[tex]T=124^{\circ}C[/tex]
As heat exchanger is counter flow therefore
[tex]\Delta T_1=160-80=80^{\circ}C[/tex]
[tex]\Delta T_2=124-20=104^{\circ}C[/tex]
[tex]LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}[/tex]
[tex]LMTD=\frac{80-104}{\ln \frac{80}{104}}[/tex]
[tex]LMTD=91.49^{\circ}C[/tex]
heat lost or gain by Fluid is equal to heat transfer in the heat exchanger
[tex]\dot{m_c}c(80-20)=U\cdot A\cdot[/tex] (LMTD)
[tex]A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2[/tex]
[tex]A=\pi DL=5.144[/tex]
[tex]L=\frac{5.144}{\pi \times 0.015}[/tex]
[tex]L=109.16 m[/tex]
The length of a counterflow double-pipe heat exchanger required can be calculated by first determining the heat needed to raise the water temperature, then calculating the necessary surface area, and finally dividing by the inner circumference of the pipe to find the length.
Explanation:To determine the length of the heat exchanger required to heat water from 20°C to 80°C using geothermal water at 160°C, we can apply the basic principles from thermodynamics and heat transfer. The equation that relates the heat transfer Q with the overall heat transfer coefficient U, the surface area A, and the temperature difference ΔT is Q = U * A * ΔT. We can determine the heat needed to raise the temperature of water by using the specific heat of water and the mass flow rate of water.
First, the amount of heat needed to raise the temperature of water (Q) can be calculated using the specific heat capacity of water (c_p, which is approximately 4.18 kJ/kgK), the mass flow rate of water (m_dot), and the temperature rise (ΔT = T_f - T_i), using the formula Q = m_dot * c_p * ΔT.
Next, we can rearrange the heat transfer equation to solve for the surface area A required for the heat exchanger: A = Q / (U * ΔTLMTD), where ΔTLMTD is the logarithmic mean temperature difference, which can be calculated for a counterflow heat exchanger knowing the inlet and outlet temperatures of both the hot and cold fluids.
Finally, we can find the length of the heat exchanger (L) by dividing the surface area A by the inner circumference of the pipe (π * d), with d being the inner pipe diameter.
By calculating the heat necessary for the temperature rise of the water and applying the given overall heat transfer coefficient, we can determine the required length of the exchanger.
Three polarizing filters are stacked, with the polarizing axis ofthe second and third filters at angles of 22.2^\circ and 68.0^\circ, respectively, to that of thefirst. If unpolarized light is incident on the stack, the light hasan intensity of 75.5 W/cm^2 after it passes through thestack.
a) If the incident intensity is kept constant, what is theintensity of the light after it has passed through the stack if thesecond polarizer is removed?
Answer:
I₂ = 25.4 W
Explanation:
Polarization problems can be solved with the malus law
I = I₀ cos² θ
Let's apply this formula to find the intendant intensity (Gone)
Second and third polarizer, at an angle between them is
θ₂ = 68.0-22.2 = 45.8º
I = I₂ cos² θ₂
I₂ = I / cos₂ θ₂
I₂ = 75.5 / cos² 45.8
I₂ = 155.3 W
We repeat for First and second polarizer
I₂ = I₁ cos² θ₁
I₁ = I₂ / cos² θ₁
I₁ = 155.3 / cos² 22.2
I₁ = 181.2 W
Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized
I₁ = I₀ / 2
I₀ = 2 I₁
I₀ = 2 181.2
I₀ = 362.4 W
Now we remove the second polarizer the intensity that reaches the third polarizer is
I₁ = 181.2 W
The intensity at the exit is
I₂ = I₁ cos² θ₂
I₂ = 181.2 cos² 68.0
I₂ = 25.4 W
The Bullet Train The Shinkansen, the Japanese "bullet" train, runs at high speed from Tokyo to Nagoya. Riding on the Shinkansen, you notice that the frequency of a crossing signal changes markedly as you pass the crossing. As you approach the crossing, the frequency you hear is f; as you recede from the crossing the frequency you hear is 2f/3.What is the speed of the train?
Answer:
68.6 m/s
Explanation:
v = Speed of sound in air = 343 m/s
u = Speed of train
[tex]f_1[/tex] = Actual frequency
From the Doppler effect we have the observed frequency as
When the train is approaching
[tex]f=f_1\frac{v+u}{v}[/tex]
When the train is receeding
[tex]\frac{2f}{3}=f_1\frac{v-u}{v}[/tex]
Dividing the above equations we have
[tex]\frac{f}{\frac{2f}{3}}=\frac{f_1\frac{v+u}{v}}{f_1\frac{v-u}{v}}\\\Rightarrow \frac{3}{2}=\frac{v+u}{v-u}\\\Rightarrow 3v-3u=2v+2u\\\Rightarrow v=5u\\\Rightarrow u=\frac{v}{5}\\\Rightarrow u=\frac{343}{5}\\\Rightarrow u=68.6\ m/s[/tex]
The speed of the train is 68.6 m/s
A 10.0 cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.330 ohms . Pulling the wire at a steady speed of 4.00 m/s causes 4.40 W of power to be dissipated in the circuit.a) How big is the pulling force?b) What is the strength of the magnetic field?
Answer
given,
length of wire = 10 cm = 0.1 m
resistance of the wire = 0.330 ohms
speed of pulling = 4 m/s
Power = 4.40 W
a) Force of pull = ?
[tex]P = F_{pull} v[/tex]
[tex]F_{pull} =\dfrac{P}{v}[/tex]
[tex]F_{pull} =\dfrac{4.40}{4}[/tex]
[tex]F_{pull} =1.1\ N[/tex]
b) using formula
[tex]P = \dfrac{B^2l^2v^2}{R}[/tex]
where B is the magnetic field
v is the pulling velocity
R is the resistance of the wire
[tex]B =\sqrt{\dfrac{PR}{l^2v^2}}[/tex]
[tex]B =\sqrt{\dfrac{F_{pull} \times v R}{l^2v^2}}[/tex]
[tex]B =\sqrt{\dfrac{F_{pull} \times R}{l^2v}}[/tex]
[tex]B =\sqrt{\dfrac{1.1 \times 0.33}{0.1^2\times 4}}[/tex]
[tex]B =\sqrt{9.075}[/tex]
B = 3.01 T
The pulling force in this magnetic field is equal to 1.1 Newton.
Given the following data:
Length of wire = 10.0 cm.Total resistance = 0.330 Ohms.Speed = 4.00 m/s.Power = 4.40 Watts.How to calculate the pulling force.Mathematically, the pulling force in a magnetic field is given by this formula:
[tex]F = \frac{Power}{Speed} \\\\F=\frac{4.40}{4.00}[/tex]
F = 1.1 Newton.
To determine the strength of the magnetic field, we would apply this formula:
[tex]B=\sqrt{\frac{PR}{L^2V^2} }\\\\B=\sqrt{\frac{FVR}{L^2V^2} }\\\\B=\sqrt{\frac{FR}{L^2V} }[/tex]
Where:
B is the magnetic field.P is the power.V is the velocity.L is the length.R is the resistance.F is the pulling force.Substituting the given parameters into the formula, we have;
[tex]B=\sqrt{\frac{1.1 \times 0.330}{0.1^2 \times 4.00} } \\\\B=\sqrt{\frac{0.363}{0.01 \times 4.00} }\\\\B=\sqrt{9.075}[/tex]
B = 3.012 T.
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A. Situation A: A 20-kg block is pressed against a relaxed spring on a frictionless surface. The spring is compressed by ∆x = 10 cm from its equilibrium position x = x 0 e . The block is released and the spring uncompresses. The speed of the block at x = x 0 e is va,f = 5.0 m/s.
B. Situation B: The same 20-kg block is pressed against two relaxed springs that are attached in series on a frictionless surface. (These two springs are both identical to the spring in Situation A). The entire spring assembly is compressed by ∆x = 10 cm from its equilibrium position x = x 0 e . The block is released and the spring assembly uncompresses. The speed of the block at x = x 0 e is vb,f Question:
What is the value of vb,f ?
(A) 3.5 m/s
(B) 2.5 m/s/s
(C) 5.0 m/s
(D) 6.4 m/s
(E) 7.0 m/s.
Final answer:
In Situation B, the speed of the block at x = x0e is 3.5 m/s.
Explanation:
In Situation A, a 20-kg block is pressed against a single spring and released. The speed of the block at x = x0e is va,f = 5.0 m/s. In Situation B, the same 20-kg block is pressed against two identical springs in series. The entire spring assembly is compressed by Δx = 10 cm. The question asks for the speed of the block at x = x0e, which we can find using the principle of conservation of mechanical energy.
The speed of the block in Situation B, vb,f, can be found as follows:
Calculate the potential energy stored in the spring assembly when it is compressed by Δx = 10 cm. This is done using the formula U = (1/2)k(Δx)^2, where k is the spring constant.Use the principle of conservation of mechanical energy to relate the potential energy of the spring assembly to the kinetic energy of the block at x = x0e. This can be written as (1/2)mvb,f^2 = U.Solve for vb,f to find the speed of the block at x = x0e.Solving this equation gives us vb,f = 3.5 m/s.
A soap bubble is 103 nm thick and illuminated by white light incident perpendicular to its surface. What wavelength (in nm) and color of visible light is most constructively reflected, assuming the same index of refraction as water (nw = 1.33)?
Answer:
λ = 547.96 nm
Explanation:
given,
thickness of soap bubble = 103 nm
refractive index of thin film = 1.33
using formula of constructive interference
[tex]2 n t = (m + \dfrac{1}{2})\times \lambda[/tex]
t is thickness of the medium
n is refractive index
m = 0,1,2...
now,
[tex]2 n t = (m + 0.5)\times \lambda[/tex]
at m = 0
[tex]2\times 1.33 \times 103 =0.5\times \lambda[/tex]
λ = 547.96 nm
at m = 1
[tex]2 n t = (m + 0.5)\times \lambda[/tex]
[tex]2\times 1.33 \times 103 =1.5\times \lambda[/tex]
λ = 182.65 nm
the only visible light enhanced by the thin film is of wavelength
λ = 547.96 nm
An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction µ between the object and the surface. The object has speed v when it reaches x=0 and encounters a spring. The object compresses the spring, stops, and then recoils and travels in the opposite direction. When the object reaches x=0 on its return trip, it stops.
1. Find k, the spring constant.
2. Express k in terms of µ, m, g, and v.
Answer:
[tex]k= \frac{8\mu^2mg^2}{v^2}[/tex]
Explanation:
The mass of object is m
The coefficient of friction is μ
the speed of object at x= 0 is v
when the object compress the springs stops so,
The kinetic energy of the spring is equal to and friction work is done
[tex]\frac{1}{2}mv^2 = \frac{1}{2}kx^2+\mu(mg)x[/tex]...........
after that it returns from the rest and finally comes to rest after coming to x=0
position.
So, now the potential energy of the spring is equal to the work done by the friction force.
[tex]\frac{1}{2}kx^2= \mu mgx[/tex]
[tex]\frac{1}{2}kx= \mu mg[/tex]
[tex]x= \frac{2\mu mg}{x}[/tex]
substitute the x value in equation 1
mv^2 = kx^2+2\mu (mg)x
[tex]mv^2= k\frac{2\mu mg}{k}+ 2\mu mg \frac{2\mu mg }{k}[/tex]
solving this we get
[tex]mv^2= \frac{8m^2\mu^2g^2}{k}[/tex]
therefore the force constant
[tex]k= \frac{8\mu^2mg^2}{v^2}[/tex]
Spring Constant is the measure of a spring's stiffness. The value of the spring constant k in terms of µ, m, g, and v is [tex]\dfrac{8m \mu ^2 g^2}{v^2 }[/tex].
What is the spring constant?Spring Constant is the measure of a spring's stiffness.
Given to us
Mass of object = m
The coefficient of friction = μ
The speed of the object at (x= 0) = v
When the object compresses the spring after a distance the spring and the object both stop, therefore, the Kinetic energy of the object transforms to the kinetic energy of the spring during compression and the potential energy when fully compressed, therefore,
The kinetic energy of the object during the compression,
[tex]\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x[/tex]
The kinetic energy of the object after complete compression,
[tex]\dfrac{1}{2}kx^2 = \mu m g x\\\\\dfrac{1}{2}kx = \mu m g \\\\x = \dfrac{2 \mu m g}{k}[/tex]
Substitute the value of x in the kinetic energy of the object during the compression,
[tex]\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x\\\\\dfrac{1}{2} mv^2 = \dfrac{1}{2}k(\dfrac{2 \mu m g}{k})^2 + \mu (mg)(\dfrac{2 \mu m g}{k})\\\\\\mv^2 = \dfrac{8m^2 \mu ^2 g^2}{k}\\\\k= \dfrac{8m^2 \mu ^2 g^2}{mv^2 }\\\\k= \dfrac{8m \mu ^2 g^2}{v^2 }[/tex]
Hence, the value of the spring constant k in terms of µ, m, g, and v is [tex]\dfrac{8m \mu ^2 g^2}{v^2 }[/tex].
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