Answer: Below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves
Explanation:
To avoid the surface waves, a submarine has to submerge below the wave base. It is the position below which the motion of the waves is negligible.
This wave base is equal to half of the wavelength. The equation becomes:
Wave base = [tex]\frac{\text{Wavelength}}{2}[/tex]
We are given:
Wavelength = 24 m
Putting values in above equation, we get:
Wave base = [tex]\frac{24m}{2}=12m[/tex]
Hence, below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves
A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are vertical. String A is attached at a distance d
Answer:
a) T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] , b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]
c) x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex], d) m₂ = m₁ ( [tex]\frac{ L}{2d} -1[/tex])
Explanation:
After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act
a) The tension of string A is requested
The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive
∑ τ = 0
T_A d - W₂ x -W₁ L/2 = 0
T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]
b) the tension in string B
we write the expression of the translational equilibrium
∑ F = 0
T_A - W₂ - W₁ - T_B = 0
T_B = T_A -W₂ - W₁
T_ B = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] - g m₂ - g m₁
T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]
c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system
T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0
at the point that begins to rotate T_B = 0
g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0
m₂ (d-x) = m₁ (0.5 L- d)
m₂ x = m₂ d - m₁ (0.5 L- d)
x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]
d) The mass of the block for which it is always in equilibrium
this is the mass for which x = 0
0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}
[tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]
[tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]
m₂ = m₁ [tex]\frac{0.5 L -d}{d}[/tex]
m₂ = m₁ ( [tex]\frac{ L}{2d} -1[/tex])
Explore the Prisms screen to see how your understanding of refraction applies when light travels through a medium like glass. Give specific examples and images from the simulation to explain how your understanding applies
Explanation:
https://tse2.mm.bing.net/th?id=OGC.b52c959ac810db1177599a161631c917&pid=Api&rurl=https%3a%2f%2fupload.wikimedia.org%2fwikipedia%2fcommons%2fthumb%2ff%2ff5%2fLight_dispersion_conceptual_waves.gif%2f266px-Light_dispersion_conceptual_waves.gif&ehk=TdcWPzr5xGP8xUOSOqZXauGOS1jHDMu7WnxPzkl7esw%3d
•. What is called the error due to the procedure and used apparatuses?
a. Random error
b. Index error
c. Systematic error
d. Parallax error.
Answer:
[tex]c.) \: systematic \: error \\ \\ = > it \: is \: the \: error \: caused \: \\ \\ due \: to \: the \: procedure \\ \\ \: and \: used \: apparatuses \\ \\ \huge\mathfrak\red{Hope \: it \: helps}[/tex]
A 55 g soapstone cube--a whisky stone--is used to chill a glass of whisky. Soapstone has a density of 3000 kg/m3, whisky a density of 940 kg/m3. What is the approximate normal force of the bottom of the glass on a single stone?
Answer:
[tex]N=0.37N[/tex]
Explanation:
Mass [tex]m=55g=>0.055kg[/tex]
Soapstone Density [tex]\rho_s=3000kg/m^2[/tex]
Whisky Density [tex]\rho_w=940kg/m^2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=U+N[/tex]
Therefore
[tex]N=m*g-(\frac{m}{\rho_s})*\rho_w*g[/tex]
[tex]N=0.055*9.81 - {(\frac{0.055}{3000})*940*9.81}[/tex]
[tex]N=0.37N[/tex]
You may have been surprised to learn that Olympic gold medals are not made from solid gold, but instead have a coating of • Saved gold on the outside.
To see a possible reason why, determine the value of the medal the size (not mass) of the Olympic gold medal if it were made of solid gold. Hint: As of mid-2018, the cost of gold is about $40 per gram.
Answer:
A gold medal has the (minimum) dimensions of:
diameter = 60mm
thickness = 3mm
So we will work with those dimensions.
The medal is then a cyinder of diameter
D = 60mm = 6cm
and height:
H = 3mm = 0.3cm
Remember that the volume of a cylinder is:
V = pi*(D/2)^2*H
where pi = 3.14
Then the volume of a medal is:
V = 3.14*(6cm/3)^2*0.3cm = 3.768 cm^3
The density of the gold in g/cm^3 is:
d = 19.3 g/cm^3
And remember that:
density = mass/volume
So, if the volume is 3.768 cm^3
Then the mass will be:
mass = density*volume = 19.3 g/cm^3*3.768 cm^3 = 72.7 g
So, a single gold medal would weight 72.7 grams
And each gram of gold costs $40
Then the total cost of the gold medal would be:
value = $40*72.7 = $2,908
Now, if yo think that in the Olympics there are 35 sports (a lot with a large number of players) and near 50 disciplines, they need a lot of gold medals.
And each gold medal costs $2,908
So the total cost (only for the gold medals, ignoring the others) would be to high.
This is why the gold medals are made mostly of silver.
Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running start and jumps with an initial horizontal velocity of 25 m/s. Neither person experiences any significant air resistance. Compare the time it takes each of them to reach the lake below.
a. Alice reaches the surface of the lake first
b. Tom reaches the surface of the lake first
c. Alice and Tom will reach the surface of the lake at the same time.
The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part to 80 dB during a loud part. Approximately how many times louder is the latter sound
Answer:
[tex]\alpha=-3.01dB[/tex]
Explanation:
From the question we are told that:
Sound level intensity
[tex]\triangle I=40dB-80dB[/tex]
Generally the equation for intensity level is mathematically given by
[tex]\alpha=10log_{10}(I/I_x)dB[/tex]
Where
I= Intensity measured
[tex]I_x=Threshold\ of\ audibility[/tex]
[tex]I_x= 10-12 W / m2[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_x} - 10 log10 \frac{}I_2{I_x}[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_2}[/tex]
[tex]\alpha=10 log10\frac{40}{80}[/tex]
[tex]\alpha=-3.01dB[/tex]
Which level of government relies the most on income tax?
OA.
federal
state
OC.
local
Answer:
Its the Federal government
How many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1180 kHz
Answer:
254 m
Explanation:
Applying,
v = λf............... Equation 1
Where v = velocity of radio wave, λ = wave length, f = frequency
make λ the subject of the equation
λ = v/f............ Equation 2
From the question,
Given: f = 1180 kHz = 1180000 Hz
Constant: v = 3×10⁸ m/s
Substitite into equation 2
λ = 3×10⁸/1180000
λ = 2.54×10²
λ = 254 m
Q 26.12: Assume current flows in a cylindrical conductor in such a way that the current density increases linearly with radius, from zero at the center to 1.0 A/m2 at the surface of the conductor. If the conductor has a cross sectional area of 1.0 m2, what can you say about the current in this conductor
Answer:
The current is 0.67 A.
Explanation:
Density, J = 1 A/m^2
Area, A = 1 m^2
Let the radius is r. And outer is R.
Use the formula of current density
[tex]I = \int J dA = \int J 2\pi r dr\\\\I = \int_{0}^{R}\frac{2\pi r^2}{R} dr\\\\I = \frac{2 \pi R^2}{3}.... (1)Now A = \pi R^2\\\\1 =\pi R^2\\\\R^2 = \frac{1}{\pi}\\\\So, \\\\I = \frac{2\pi}{3}\times \frac{1}{\pi}\\\\I = 0.67 A[/tex]
If a car generates 22 hp when traveling at a steady 100 km/h , what must be the average force exerted on the car due to friction and air resistance
Answer:
The average force exerted on the car is 590.12 N.
Explanation:
Given that,
The power generated, P = 22 hp = 16405.4 W
Speed of the car, v = 100 km/h = 27.8 m/s
We need to find the average force exerted on the car due to friction and air resistance.
We know that,
Power, P = F v
Where
F is force exerted on the car
[tex]F=\dfrac{P}{v}\\\\F=\dfrac{16405.4}{27.8}\\\\F=590.12\ N[/tex]
So, the average force exerted on the car is 590.12 N.
a nano second is what
Answer:
one thousand-millionth of a second.
A nanosecond is an SI unit of time equal to one billionth of a second, that is, ¹⁄₁ ₀₀₀ ₀₀₀ ₀₀₀ of a second, or 10⁻⁹ seconds. The term combines the prefix nano- with the basic unit for one-sixtieth of a minute. A nanosecond is equal to 1000 picoseconds or ¹⁄₁₀₀₀ microsecond.
Magnets produce _________ in the spaces surrounding them
Answer:
magnetic field
Explanation:
The outer surface of a spacecraft in space has an emissivity of 0.44 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.
Answer:
[tex]T=326.928K[/tex]
Explanation:
From the question we are told that:
Emissivity [tex]e=0.44[/tex]
Absorptivity [tex]\alpha =0.3[/tex]
Rate of solar Radiation [tex]R=0.3[/tex]
Generally the equation for Surface absorbed energy is mathematically given by
[tex]E=\alpha R[/tex]
[tex]E=0.3*950[/tex]
[tex]E=285W/m^2[/tex]
Generally the equation for Emitted Radiation is mathematically given by
[tex]\mu=e(\sigmaT^4)[/tex]
Where
T=Temperature
[tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]
Therefore
[tex]\alpha*E=e \sigma T^4[/tex]
[tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]
[tex]T=326.928K[/tex]
At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 mls2, and friction is negligible. The coupling bars have negligible mass. By how much would the tension in each of the coupling bars A, B, and C change if 39 kg of luggage were removed from car 2 and placed in (a) car I and (b) car 3
Answer:
a) ΔT₁ = -4.68 N, ΔT₂ = 4.68 N, b) ΔT₂ = 4.68 N, ΔT₁ = 4.68 N
Explanation:
In this exercise we will use Newton's second law.
∑F = m a
Let's start with the set of three cars
F_total = M a
F_total = M 0.12
where the total mass is the sum of the mass of each charge
M = m₁ + m₂ + m₃
This is the force with which the three cars are pulled.
Now let's write this law for each vehicle
car 1
F_total - T₁ = m₁ a
T₁ = F_total - m₁ a
car 2
T₁ - T₂ = m₂ a
T₂ = T₁ - m₂ a
car 3
T₂ = m₃ a
note that tensions are forces of action and reaction
a) They tell us that 39 kg is removed from car 2 and placed on car 1
m₂’= m₂ - 39
m₁'= m₁ + 39
m₃ ’= m₃
they ask how much each tension varies, let's rewrite Newton's equations
The total force does not change since the mass of the set is the same F_total ’= F_total
car 1
F_total ’- T₁ ’= m₁’ a
T₁ ’= F_total - m₁’ a
T₁ ’= (F_total - m₁ a) - 39 a
T₁ '= T₁ - 39 0.12
ΔT₁ = -4.68 N
car 2
T₁’- T₂ ’= m₂’ a
T₂ ’= T₁’- m₂’ a
T₂ '= (T₁'- m₂ a) + 39 a
T₂ '= T₂ + 39 0.12
ΔT₂ = 4.68 N
b) in this case the masses remain
m₁ '= m₁
m₂ ’= m₂ - 39
m₃ ’= m₃ + 39
we write Newton's equations
car 3
T₂ '= m₃' a
T₂ ’= (m₃ + 39) a
T₂ '= m₃ a + 39 a
T₂ '= T₂ + 39 0.12
ΔT₂ = 4.68 N
car 1
F_total - T₁ ’= m₁’ a
T₁ ’= F_total - m₁ a
car 2
T₁' -T₂ '= m₂' a
T₁ ’= T₂’- m₂’ a
T₁ '= (T₂'- m₂ a) + 39 a
T₁ '= T₁ + 39 0.12
ΔT₁ = 4.68 N
The tension in each of the coupling bars A, B, and C of the luggage carrier changes as,
When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.What is tension force?Tension is the pulling force carried by the flexible mediums like ropes, cables and string.
Tension in a body due to the weight of the hanging body is the net force acting on the body.
At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 m/s², and friction is negligible.
The acceleration is the same, Tension due to the horizontal component of the forces for car 1, 2 and 3 can be given as,
[tex]\sum F_{1h}=T_A-T_B=m_1a\\\sum F_{2h}=T_B-T_C=m_2a\\\sum F_{3h}=T_C=m_3a[/tex]
On solving the above 3 equation, we get the values of tension in each bar as,
[tex]T_A=(m_1+m_2+m_3)a\\T_B=(m_3+m_2)a\\T_C=m_3a[/tex]
Case 1- When 39 kg of luggage were removed from car 2 and placed in car IThe tension is A and C does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,
[tex]\Delta T_B=39\times0.12\\\Delta T_B=4.68\rm \;N[/tex]
Case 2- When 39 kg of luggage were removed from car 2 and placed in car IIIThe tension is A and B does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,
[tex]\Delta T_C=39\times0.12\\\Delta T_C=4.68\rm \;N[/tex]
Hence, the tension in each of the coupling bars A, B, and C of the luggage carrier changes as,
When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.
Learn more about the tension here;
https://brainly.com/question/25743940
Un objeto de 0.5kg de masa se desplaza a lo largo de una trayectoria rectilínea con aceleración constante de 0.3m/s2. Si partió del reposo y la magnitud de su cantidad de movimiento en kg*m/s después de 8s es:
Answer:
p = 1.2 kg-m/s
Explanation:
The question is, "An object of mass 0.5kg is moving along a rectilinear path with constant acceleration of 0.3m / s2. If it started from rest and the magnitude of its momentum in kg * m / s after 8s is".
Mass of the object, m = 0.5 kg
Acceleration of the object, a = 0.3 m/s²
We need to find the momentum after 8 seconds.
We know that,
[tex]p=F\times t[/tex]
i.e.
p = mat
So,
[tex]p=0.5\times 0.3\times 8\\\\p=1.2\ kg-m/s[/tex]
So, the momentum of the object is 1.2 kg-m/s.
If the cornea is reshaped (this can be surgically done or with contact lenses) to correct myopia, should its curvature be made greater or smaller? Explain. Also, explain how hyperopia can be corrected?
Answer:
Myopia curvature of the cornea if it is negative the curvatures are positive,
hypermetry,
Explanation:
Myopia is the visual defect that does not allow to see distant objects, which is why it is corrected with a divergent lens so that the image is formed on the retina, therefore, by reforming the curvature of the cornea if it is negative
therefore the curvature must decrease
To correct hypermetry, the curvatures are positive, consequently the curvature of the lens must increase
A merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg-m2
and is rotating at 10 rev/min. A 25-kilogram child at rest jumps onto the edge of the merry-go-round. What is the new angular speed of the merry-go-round?
Answer:
dont be lose because the person who lose will win the match
Oxygen is obtained through various methods. Which of the following methods involves a chemical
change?
1. Electrolysis of water
2. Distillation of liquid air
3. Heating of KCIO,
02
1 and 2
1 and 3
Answer:
1
Explanation:
Electrolysis is the passing of an current through a conducting solution, when the occurs, a chemical reaction takes place.
Heating a chemical will always cause a chemical reaction, which is why 3 is also correct
Some information as to why 2 is NOT correct.
2 is NOT a chemical reaction, but rather a process of physical separation. It uses selective boiling and condensation, but is not considered a chemical reaction.
as with 3, heating is not considered a chemical reaction, but rather a physical temperature change. This is always what it is considered to be (e.g boiling water is a physical temperature change, not a chemical reaction)
Hope this helps.
Hope this helps.
You have three identical metallic spheres, A, B and C, fixed to isolating pedestals. They all start off uncharged. You then charge sphere A to +32.0 uC. You use rubber gloves to move sphere A so that it briefly touches sphere B, and then is separated. Next, sphere A briefly touches sphere C, and again is separated. Finally, sphere A touches sphere B a second time, and is again separated. What will be the final charge of sphere B?
Answer:
Charge on B is 12 uC.
Explanation:
Initial charge on A = 32 uC
Initial charge on B and C = 0
Now A touches to B, so the charge on A and B both is
q = (32 + 0) / 2 = 16 uC
Now A touches to C, so the charge on A and C both is
q' = (16 + 0) / 2 = 8 uC
Now again A touches to B so the charge on B is
q''= (8 + 16) / 2 = 12 uC
A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant pressure until 60 percent of it, by mass, condenses. Determine (a) the work (W) done during the process. (b) What-if Scenario: What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
Answer:
a) the work (W) done during the process is -2043.25 kJ
b) the work (W) done during the process is -2418.96 kJ
Explanation:
Given the data in the question;
mass of water vapor m = 10 kg
initial pressure P₁ = 550 kPa
Initial temperature T₁ = 340 °C
steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4
from superheated steam table
specific volume v₁ = 0.5092 m³/kg
so the properties of steam at p₂ = 550 kPa, and dryness fraction
x = 0.4
specific volume v₂ = v[tex]_f[/tex] + xv[tex]_{fg[/tex]
v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )
v₂ = 0.1377 m³/kg
Now, work done during the process;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.1377 - 0.5092 )
W = 5500 × -0.3715
W = -2043.25 kJ
Therefore, the work (W) done during the process is -2043.25 kJ
( The negative, indicates work is done on the system )
b)
What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
x₂ = 100% - 80% = 20% = 0.2
specific volume v₂ = v[tex]_f[/tex] + x₂v[tex]_{fg[/tex]
v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )
v₂ = 0.06939 m³/kg
Now, work done during the process will be;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.06939 - 0.5092 )
W = 5500 × -0.43981
W = -2418.96 kJ
Therefore, the work (W) done during the process is -2418.96 kJ
The voltage across a membrane forming a cell wall is 74.0 mV and the membrane is 9.20 nm thick. What is the electric field strength in volts per meter
Answer:
7.60× 10^6 V/m
Explanation:
electric field strength can be determined as ratio of potential drop and distance, I.e
E=V/d
Where E= electric field
V= potential drop= 74.0 mV= 0.07 V
d= distance= 9.20 nm = 9.2×10^-9 m
Substitute the values
E= 0.07/ 9.2×10^-9
= 7.60× 10^6 V/m
A horizontal, mass spring system undergoes simple harmonic motion. which of the following statements is correct reguarding the mass in the system when it is located at its maximum distance from the equilibrium position?
a. The acceleration of the mass is zero.
b. The potential energy of the spring attached to the mass is at a minimum.
c. The total mechanical energy of the mass is zero.
d. The kinetic energy of the mass is at a maximum.
e. The speed of the mass is zero.
Answer:
Option (e)
Explanation:
A body executing SHM moves to and fro or back and forth about its mean position.
When the particle is at mean position, its velocity is maximum and when it is at extreme position its velocity is zero.
So, when it is at maximum distance:
a.
The acceleration is maximum.
b.
The potential energy is maximum.
c.
The total mechanical energy is non zero.
d.
The kinetic energy is zero.
e. The speed is zero. Correct
Helium gas at 20 °C is confined within a rigid vessel. The gas is then heated until its pressure is doubled. What is the final temperature of the gas?
Answer:
586 K
Explanation:
Let P is the initial pressure.
Initial temperature, T₁ = 20°C = 293 K
Final pressure, P₂ = 2P
We need to find the final temperature of the gas.
The relation between the pressure and the temperature is as follows
[tex]P\propto T\\\\or\\\\\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}[/tex]
Put all the values,
[tex]\dfrac{P}{2P}=\dfrac{293}{T_2}\\\\\dfrac{1}{2}=\dfrac{293}{T_2}\\\\T_2=2\times 293\\\\T_2=586\ K[/tex]
So, the final temperature of the gas is 586 K.
As a roller coaster car crosses the top of a 48.01-m-diameter loop-the-loop, its apparent weight is the same as its true weight. What is the car's speed at the top?
Answer:
The speed of the car, v = 21.69 m/s
Explanation:
The diameter is = 48.01 m
Therefore, the radius of the loop R = 24.005 m
Weight at the top is n = mv^2/R - mg
Since the apparent weight is equal to the real weight.
So, mv^2/R - mg = mg
v = √(2Rg)
v = √[2(24.005 m)(9.8 m/s^2)]
The speed of the car, v = 21.69 m/s
Answer:
The speed is 15.34 m/s.
Explanation:
Diameter, d = 48.01 m
Radius, R = 24.005 m
Let the speed is v and the mass is m.
Here, the weight of the car is balanced by the centripetal force.
According to the question
[tex]m g = \frac{mv^2}{R}\\\\v =\sqrt{24.005\times9.8}\\\\v = 15.34 m/s[/tex]
During typical urination, a man releases about 400 mL of urine in about 30 seconds through the urethra, which we can model as a tube 4 mm in diameter and 20 cm long. Assume that urine has the same density as water, and that viscosity can be ignored for this flow.a. What is the flow speed in the urethra?b. If we assume that the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder (a reasonable approximation), what bladder pressure would be necessary to produce this flow? (In fact, there are additional factors that require additional pressure; the actual pressure is higher than this.)
Answer:
Explanation:
Given:
volume of urine discharged, [tex]V=400~mL=0.4~L=4\times 10^{-4}~m^3[/tex]
time taken for the discharge, [tex]t=30~s[/tex]
diameter of cylindrical urethra, [tex]d=4\times10^{-3}~m[/tex]
length of cylindrical urethra, [tex]l=0.2~m[/tex]
density of urine, [tex]\rho=1000~kg/m^3[/tex]
a)
we have volume flow rate Q:
[tex]Q=A.v[/tex] & [tex]Q=\frac{V}{t}[/tex]
where:
[tex]A=[/tex] cross-sectional area of urethra
[tex]v=[/tex] velocity of flow
[tex]A.v=\frac{V}{t}[/tex]
[tex]\frac{\pi d^2}{4}\times v=\frac{4\times 10^{-4}}{30}[/tex]
[tex]v=\frac{4\times4\times 10^{-4}}{30\times \pi (4\times 10^{-3})^2}[/tex]
[tex]v=1.06~m/s[/tex]
b)
The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:
[tex]P=\rho.g.l[/tex]
[tex]P=1000\times 9.8\times 0.2[/tex]
[tex]P=1960~Pa[/tex]
Express the unit of force in terms of fundamental unit
Answer:
The fundamental unit of force is kg.m/s²
Explanation:
According to Newton's second law of motion, force is given as the product of mass and acceleration.
Mathematically, force can be expressed as; F = ma
where;
F is the force
M is mass of the object, unit of mass = kg
a is acceleration of the object, unit of acceleration = m/s²
Force = kg x m/s²
Force = kg.m/s² = Newton [N]
Therefore, the fundamental unit of force is kg.m/s²
when blueshift occurs,the preceived frequency of the wave would be?
Answer:
When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.
Explanation:
As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.
Blueshifts happens when the source of the wave and the observer are moving closer towards one another.
Assume that the wave is of frequency [tex]f\; {\rm Hz}[/tex] at the source. In other words, the source of the wave sends out a peak after every [tex](1/f)\; {\text{seconds}}[/tex].
Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.
The source of this wave sends out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. It would appear to the observer that consecutive peaks arrive every [tex](1/f)\; {\text{seconds}}\![/tex]. That would correspond to a frequency of [tex]f\; {\rm Hz}[/tex].
On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of [tex]v \; {\rm m \cdot s^{-1}}[/tex].
Again, the source of this wave would send out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. However, by the time the source sends out the second peak, the source would have been [tex]v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}[/tex] closer to the observer then when the source sent out the first peak.
When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than [tex](1/f)\; {\text{seconds}}[/tex]. The observed frequency of this wave would be larger than the original [tex]f\; {\rm Hz}[/tex].
A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. 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 his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?
Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
[tex]\frac{1}{p} = \frac{1}{f} -\frac{1}{q}[/tex]
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}[/tex]
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}[/tex]
1 / p = 0.107375
p = 93.13 cm
After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at 19.0 with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits
Answer:
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Explanation:
Given :
The first dark fringe is for m = 0
[tex]$\theta_1 = \pm 19^\circ$[/tex]
Now we know for a double slit experiments , the position of the dark fringes is give by :
[tex]$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$[/tex]
The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :
[tex]$d \sin \theta=\left(\frac{1}{2}\right) \lambda$[/tex] (since, m = 0)
[tex]$d \sin \theta=\frac{\lambda}{2}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$[/tex]
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Therefore, the ratio is [tex]$\frac{1}{1.54}$[/tex] or 1 : 1.54
