Answer:
Options A, D and E....make up cell theory
Rudolf Virchow was the first to propose the concept of Omnis cellula-e cellula in relation to cell division.
What is the cell theory?It is a scientific theory that was proposed in the mid-nineteenth century.
Who proposed it?It was proposed by Theodore Schwann a British Zoologist and Matthias Schleiden a German botanist.
What are the three principles of the cell theory?The three principles are:
All living things are composed of cells.Cells are the basic unit of structure and function in all living things.All cells are produced from pre-existing cells.Hence options A, D, E are correct.
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The density of mercury is 13.546 g/cm3 . Calculate the pressure exerted by a column of mercury 76 cm high. Give your answer in Pascals and lbf/in2 . 2. The density of water is 62.43 lbm/ft3 . Calculate the pressure exerted by a column of water 25 ft high. Give your answer in Pascals and lbf/in2 3. What is the power required to pump 10 kg/s water from a height of 5 meters to a height of 30 meters? Report the power in Watts and hp. 4. Water is being pumped by the application of pressure at point 1, see below, up to a height of 50 m at a rate of 1 kg/s. At the top, point 2 the pressure is nearly one atmosphere (105 N/m2 )
Answer:
The required solution is 100890 Pa and 14.3lb/in²
Explanation:
See attached image
Answer:
1. p = 14.63 lb/in² or 100890.608 Pa
2. p = 74676 Pa or 10.83 lb/in²
3. P = 2450 W or 3.28 hp
4. [tex]p_{1}[/tex] = 490105 N/m²
Explanation:
1. Let's begin by listing out the given parameters:
density of mercury = 13.546 g/cm³ = 13546 kg/m³,
height of column = 76 cm = 0.76 m, acceleration due to gravity = 9.8m/s²
Using Pressure = density * acceleration due to gravity * height of column
p = ρ g h = 13546 * 9.8 * 0.76
p = 100890.608 Pa
To get the answer in lb/in², divide by 6895
p = 100890.608 ÷ 6895 = 14.632
p = 14.63 lb/in²
2. Let's list out the parameters given:
density of water = 62.43 lbm/ft³ = 62.43 * 16.018 = 1000kg/m³,
height of column = 25 ft = 25 ÷ 3.281 = 7.62 m,
acceleration due to gravity = 9.8m/s²
Using Pressure = density * acceleration due to gravity * height of column
p = ρ g h = 1000 * 9.8 * 7.62
p = 74676 Pa
To convert from Pa to lb/in², divide by 6895
p = 74676 ÷ 6895
p = 10.83 lb/in²
3. Let's list out the parameters given:
mass flow rate (ṁ) = 10 kg/s, [tex]h_{1}[/tex] = 5 m, [tex]h_{2}[/tex] = 30 m, Δh = 30 - 5 = 25 m, g = 9.8 m/s²
Using Power = Energy (Potential Energy) ÷ Time
Energy (Potential Energy) = m g h
Power = mgΔh ÷ t; m÷ t = ṁ
Substitute ṁ into the equation
P = ṁ g h = 10 * 9.8 * 25
P = 2450 W
To convert from W to hp, divide by 746
P = 2450 ÷ 746 = 3.284
P = 3.28 hp
4. Let's list out the parameters given:
height (Δh) = 50 m, ṁ = 1 kg/s, g = 9.8 m/s²,
p2 = 105N/m², ρ = 1000 kg/m³
Using Bernoulli's Equation,
p1 + ½ρ([tex]v_{1}[/tex])² + ρgh1 = p2 + ½ρ([tex]v_{2}[/tex])² + ρgh2
Assuming steady state flow; [tex]v_{2}[/tex] = [tex]v_{1}[/tex] ⇒ [tex]v_{2}[/tex] - [tex]v_{1}[/tex] = 0
[tex]p_{1}[/tex] - [tex]p_{2}[/tex] = ½ρ([tex]v_{2}[/tex] - [tex]v_{1}[/tex])² + ρg([tex]h_{2}[/tex] - [tex]h_{1}[/tex])
[tex]p_{1}[/tex] - [tex]p_{2}[/tex] = ρgΔh
[tex]p_{1}[/tex] - 105 = 1000 * 9.8 * 50
[tex]p_{1}[/tex] = 490000 + 105 = 490105
[tex]p_{1}[/tex] = 490105 N/m²
The theory of nuclear astrophysics is that all the heavy elements like uranium are formed in the interior of massive stars. These stars eventually explode, releasing these elements into space. If we assume that at the time of the explosion there were equal amount of U-235 and U-238, how long ago did the star(s) explode that released the elements that formed our Earth? The present U-235/U-238 ratio is 0.00700. [The half-lives of U-235 and U-238 are 0.700 × 109 yr and 4.47 × 109 yr.]
Answer:
t = 5.94x10⁹ years.
Explanation:
The time of the explosion can be calculated using the decay equation:
[tex] N_{t} = N_{0}e^{-\lambda t} [/tex]
Where:
N(t): is the quantity of the element at the present time
N(0): is the quantity of the element at the time of explosion
λ: is the decay constant
t: is the time
Knowing that the present U-235/U-238 ratio is 0.00700 and that at the time of the explosion there were equal amount of U-235 and U-238, we have:
[tex]\frac{N_{U-235}}{N_{U-238}} = \frac{N_{U-235_{0}}e^{-\lambda_{U-235} t}}{N_{U-238_{0}}e^{-\lambda_{U-238} t}}[/tex] (1)
The decay constant is equal to:
[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex]
For the U-235 we have:
[tex] \lambda_{U-235} = \frac{ln(2)}{0.700 \cdot 10^{9} y} = 9.90 \cdot 10^{-10} y^{-1} [/tex]
For the U-238 we have:
[tex] \lambda_{U-238} = \frac{ln(2)}{4.47 \cdot 10^{9} y} = 1.55 \cdot 10^{-10} y^{-1} [/tex]
By introducing the values of [tex]\lambda_{U-235}[/tex] and [tex]\lambda_{U-238}[/tex] into equation (1) we have:
[tex]0.00700 = \frac{e^{-9.90 \cdot 10^{-10} t}}{e^{-1.55 \cdot 10^{-10} t}}[/tex]
[tex]0.00700 = e^{(-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}) t}[/tex]
[tex]ln(0.00700) = (-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}) t[/tex]
[tex]t = \frac{ln(0.00700)}{-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}} = 5.94 \cdot 10^{9} y[/tex]
Therefore, the star exploded 5.94x10⁹ years ago.
I hope it helps you!
The explosion of the star that released the uranium that formed Earth happened approximately 6 billion years ago, determined using the half-lives of U-235 and U-238 and the current U-235/U-238 ratio.
The question regarding the age of the elements from a star explosion can be addressed using the concept of radioactive decay and isotope half-lives, specifically of Uranium-235 (U-235) and Uranium-238 (U-238). Using the current U-235/U-238 ratio of 0.00700 and knowing the half-lives of U-235 (0.700 × 109 years) and U-238 (4.47 × 109 years), we can calculate that the star exploded approximately 6 billion years ago based on the change in the U-235/U-238 ratio from an assumed equal initial amount. This estimate is consistent with the age of the solar system and the time it would take for such materials to coalesce into a planetary body like Earth.
Tamsen and Vera imagine visiting another planet, planet X, whose gravitational acceleration, gX, is different from that of Earth's. They envision a pendulum, whose period on Earth is 2.243 s, that is set in motion on planet X, and the period is measured to be 2.000 s. What is the ratio of gX/gEarth? Neglect any effects caused by air resistance.
Answer:
1.27
Explanation:
Period of a pendulum T is
T = 2¶(l/g)^0.5
Where g is acceleration due to gravity
l is lenght of pendulum
For earth, T = 2.243 s
2.243 = 2 x 3.142 x (l/9.81)^0.5
0.36 = (l^0.5)/3.13
1.13 = l^0.5
l = 1.28 m
For planet X of the same lenght
Period T = 2.00 s
2 = 2 x 3.142 (1.28/gx)^0.5
0.32 = 1.13 / gx^0.5
3.53 = gx^0.5
gx^0.5 = 3.53
gx = 12.46 m/s^2
gx/gearth = 12.46/9.81 = 1.27
The ratio of the gravitational acceleration on planet X to that on Earth is 1.26, calculated using the periods of a pendulum on both planets and the relationship between period and gravitational acceleration.
Explanation:The period of a simple pendulum is given by the formula T = 2*pi*sqrt(L/g) where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity. To find the ratio gX/gEarth, we can use the formula T_earth/T_x = sqrt(g_x/g_earth) (let's denote T_earth as the period on earth and T_x as the period on planet X).
Since we know the periods, we can substitute these values into the formula: 2.243s/2.000s = sqrt(g_x/g_earth), which gives us 1.1215 = sqrt(g_x/g_earth). To find the ratio g_x/g_earth, square both sides, giving (1.1215)² = g_x/g_earth. Thus, g_x/g_earth = 1.26.
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A 0.468 g sample of pentane, C5H12, was burned in a bomb calorimeter. The temperature of the calorimeter and 1.00 kg of water in it rose from 20.45 °C to 23.65 °C. The heat capacity of the calorimeter by itself is 2.21 kJ/°C and the specific heat capacity of water is 4.184 J/g.°C What is the heat of combustion per mole of pentane?
Answer:
E = 3,154.37 KJ
The heat of combustion per mole of pentane is 3,154.37 KJ
Explanation:
Given;
Change in temperature of system ∆T = 23.65-20.45 = 3.2 °C
Mass of water m1 = 1 kg
Specific heat capacity of water C1 = 4.184J/g°C = 4184J/kg °C
Heat capacity of calorimeter mC2 = 2.21 kJ/°C = 2210J/°C
Heat gained by both calorimeter and water is;
H = (m1C1 + mC2)∆T
Substituting the values;
H = (1×4184 + 2210)×3.2
H = 20460.8 J
Mass of pentane burned = 0.468 g
Molecular mass of pentane = 72.15g
If 0.468g of pentane releases 20460.8 J of heat,
72.15g of pentane will release;
E = (72.15/0.468) × 20460.8 J
E = 3154373.333333J
E = 3,154.37 KJ
The heat of combustion per mole of pentane is 3,154.37 KJ
Which two elements have the same number of valence electrons?
Elements in the same group on the periodic table, like alkali metals lithium and sodium or alkaline earth metals beryllium and magnesium, have the same number of valence electrons. This similarity is a key factor in their chemical reactivity and the formation of covalent bonds, evidenced through electron dot diagrams.
Explanation:Elements that have the same number of valence electrons are located in the same group or column on the periodic table. For instance, the alkali metals such as lithium (Li) and sodium (Na) each possess one valence electron. Similarly, the alkaline earth metals like beryllium (Be) and magnesium (Mg) each have two valence electrons.
Another example can be found in the halogens group, as elements such as fluorine (F) and chlorine (Cl) each exhibit seven valence electrons. These elements have not only the same valence electron count but also exhibit similar chemical properties due to this shared characteristic. It's important to note that the number of valence electrons contributes to an element's ability to bond with others through the loss, gain, or sharing of these electrons.
Understanding that elements in the same group on the periodic table share the same number of valence electrons can greatly aid in predicting their chemical behavior and reactivity. Electron dot diagrams provide a visual representation of the valence electron distribution in each element and are particularly useful for visualizing how elements bond covalently. For example, elements in the first group have a single dot representing the single valence electron they possess.
Sandra is riding a bicycle at 10.0 meters/second. She slows down to 2.0 meters/second in 10 seconds.
Answer:
Acceleration= final velocity-initial velocity/time taken
so
a= 10-2/10
a=-8/10
a= -.8 meters per second
Answer: 8.0 = 8 meters/seconds.
Explanation:
Started at= 10.0 meters/seconds.
Slows down= 2.0
10-2=8
Sandra is currently riding her bicycle at 8.0 meters/seconds.
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If a light bulb is missing or broken in a parallel circuit, will the other light bulb right ?
A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 mm and a mass of 3.38 grams. Assume that the ends of the rope are held fixed and that there is both this traveling wave and the reflected wave traveling in the opposite direction.
What is the wave function y(x,t) for the standing wave that is produced?
Answer:
Explanation:
y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]
The angular velocity ω = 742 rad /s
wave function k = 6.98 rad /m
Amplitude A = 2.3 mm
y(x,t) = A cos( ωt + kx )
equation of wave reflected wave
y(x,t) = A cos( ωt - kx )
resultant standing wave
= y₁ +y₂
= A cos( ωt + kx ) +A cos( ωt - kx )
y(x,t) = 2 A cosωt cos kx
= 2 x 2.3 cos 742t .cos6.98x
= 4.6 mm . cos 742t .cos6.98x
A solid copper sphere hangs at the bottom of a steel wire of negligible mass. The top end of the wire is fixed. When the wire is struck, it vibrates with a fundamental frequency of 300 Hz. The copper object is then submerged in water so that half its volume is below the water line. Determine the new fundamental frequency. The density of water, copper, and steel is 1000kg/m³, 8960kg/m³, and 8050kg/m³ respectively
Answer:
291.509 Hz
Explanation:
Fundamental frequency, often referred to simply as the fundamental, is defined as the lowest frequency of a periodic waveform.
It is a vital concept in musical instruments and many aspects .
See the attached file for the solution to the given problem.
Consider three starships that pass by an observer on Earth. Starship A is traveling at speed v=c/3v=c/3 relative to Earth and has a clock placed aboard. Starship B is traveling at speed v=c/3v=c/3 relative to Earth and in the same direction as Starship A. Starship C is traveling at speed v=c/3v=c/3 relative to Earth, but in the opposite direction as Starship A. In which reference frame can a time interval be measured that equals the time interval measured by the clock aboard Starship A?
An observer on Starship B will measure the same time interval as that on Starship A, because they are both travelling at the same speed and in the same direction relative to Earth. Starship C's measurement would differ due to its opposite direction of travel relative to A and B.
Explanation:To determine which reference frame measures a time interval equal to the time interval measured by the clock aboard Starship A, we need to consider the principles of the theory of special relativity. Since Starship B is traveling at the same speed and in the same direction as Starship A relative to Earth, the time dilation effect will be the same for both starships. Therefore, an observer on Starship B will measure the same time interval on their clock as that measured on Starship A's clock.
In contrast, Starship C is moving at the same speed but in the opposite direction relative to Earth, and its relative velocity to Starships A and B is not zero. As a result, the time interval measured on Starship C will not match the time interval measured on Starship A's clock.
So, the time interval that equals the time interval measured by the clock aboard Starship A can be measured in the reference frame of Starship B, which is moving at the same speed and in the same direction as Starship A relative to Earth.
A parallel-plate capacitor of capacitance 20 µF is fully charged by a battery of 12 V. The battery is then disconnected. A dielectric slab of K = 4 is slipped between the two plates of the capacitor:
(a) Find the change in potential energy of the capacitor.
(b) Does the potential energy increase or decrease? Explain
Answer:
Explanation:
capacitance = 20 x 10⁻⁶ F .
potential V = 12 V
charge = CV
= 20 x 10⁻⁶ x 12
Q = 240 x 10⁻⁶ C
energy = Q² / 2C
= (240 x 10⁻⁶ )² / 2 x 20 x 10⁻⁶
= 1440 x 10⁻⁶ J
b )
In this case charge will remain the same but capacity will be increased 4 times
new capacity C = 4 x 20 x 10⁻⁶
= 80 x 10⁻⁶
energy = Q² / 2C
= (240 x 10⁻⁶ )² / 2 x 4 x 20 x 10⁻⁶
= 360 x 10⁻⁶ J .
potential energy will decrease from 1440 x 10⁻⁶ J to 360 x 10⁻⁶ J
There are two possibilities for final stage of extremely massive stars. The first is a
neutron star and the second is a
Answer: Black hole.
Explanation:
As the massive star "compacts" under its own gravity, it triggers a massive supernova, after this point the remains of the star can become a neutron star, which is a very compact star made primarily, as the name says, of neutrons. The other possibility is a black hole, which is a finite region of space wherein it's interior there is a big concentration of mass, which creates a gravitational field strong enough that there is no particle that can escape it.
Which of the following do not make their own energy through nuclear fusion? Select all that apply.
A. giant star
B. protostar
C. dwarf star
D. black hole
E. neutron star
F. main sequence star
A large diameter closed top tank is filled with a depth of 3 meters of a fluid (density 1200 kg/m3). A small pipe leading from the bottom of the tank must carry the fluid to a height of 5 meters (2 meters above the top level of the tank fluid level). a) What pressure must be maintained in the space above the fill line in the tank to provide an exit speed of 5 m/sec for the fluid out of the pipe? (the pressure at the exhaust end of the pipe is 1 atmosphere or about 1 x 105 pascal) 15 pt If the pipe diameter is 2 cm, what will be the initial volume rate of flow (m3 /sec) 5pt
Answer:
1.57×10^-3 m^3/s
Explanation:
Please see the attached filw for a detailed and step by step solution of the given problem.
The attached file explicitly solves it.
A simple experiment to measure the speed of sound doesn't involve a stopwatch. You can fill up along tube with water and put a tuning fork over the opening. The sound waves that travel down into the tube will reflect from the surface of the water and come back to the tuning fork. This is like a half open pipe with a fixed end condition on the water side and an open end condition on the openThe condition for resonance is if the depth of the air column is an odd integer multiple of lamda/4. The speed of sound is 345 m/s.A) If a tuning fork was used for this experiment, what is the depth of the air column that will satisfy the resonance condition for the fundamental mode?B) What is the depth of L3the air column for the 3rd harmonic resonance?C) Is there a depth that will result in a 2nd harmonic resonance? Explain
Answer:
Explanation:
In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.
To find the fundamental mode you use:
[tex]f_n=\frac{nv_s}{4L}[/tex]
n: mode of the sound
vs: sound speed
L: length of the column of air in the tube.
A) The fundamental mode id obtained for n=1:
[tex]f_1=\frac{v_s}{4L}[/tex]
B) For the 3rd harmonic you have:
[tex]f_3=\frac{3v_s}{4L}[/tex]
C) For the 2nd harmonic:
[tex]f_2=\frac{2v_s}{4L}[/tex]
A wedge-shaped air film is made by placing a small slip of paper between the edges of two thin plates of glass 12.5 cm long. Light of wavelength 600 nm in air is incident normally on the glass plates. If interference fringes with a spacing of 0.200 mm are observed along the plate, how thick is the paper? This form of interferometry is a very practical way of measuring small thicknesses.
Answer:
The thickness of the paper is [tex]t = 188\mu m[/tex]
Explanation:
From the question we are told that
The length of the wedge-shaped air film [tex]L = 12.5 cm = \frac{12.5}{100} = 0.125m[/tex]
The wavelength of the light is [tex]\lambda = 600nm = 600* 10^{-9}m[/tex]
The spacing of the interference fringe is [tex]D = 0.200mm = \frac{0.200}{1000} = 0.2*10^{-3} m[/tex]
For destructive interference the thickness is mathematically represented as
[tex]t =\frac{\lambda * L}{2 * D }[/tex]
Substituting values
[tex]t = \frac{600 * 10^{-9} * 0.125 }{2 * 0.2 *10^{-3}}[/tex]
[tex]t = 188\mu m[/tex]
8. A tuning fork vibrating with a sonometer wire 20cm long produces 5 beats per second. Given
that the beat frequency does not change if the length of the wire is changed to 21cm, calculate
the frequency of the tuning fork and frequency of sonometer wire when changed to 21cm.
Answer:
Explanation:
length of wire is proportional to frquency of sound produced by it.
n₁ /n₂ = l₂ / l₁
= 21 / 20
If n be the frequency of tuning fork
n₁ = n + 5 ; n₂ = n - 5 ( no of beat / s = frequency difference )
n + 5 / n - 5 = 21 / 20
20n + 100 = 21n - 105
n = 205 Hz
frequency of tuning fork = 205
the frequency of the tuning fork and frequency of sonometer wire when changed to 21cm. = n - 5 = 200 Hz .
Given the beat frequency and the inverse relationship between the length of the sonometer wire and frequency, we determine the tuning fork frequency. Beat frequency remains constant despite the change in wire length. Calculation involves solving system of linear equations with fundamentals of wave physics.
To find the frequency of the tuning fork and the frequency of the sonometer wire for both lengths, we utilize the concept of beat frequency. The beat frequency is given by the difference in frequency between the tuning fork and the sonometer wire.
Let’s denote the frequency of the tuning fork as ft and the frequency of the sonometer wire for 20 cm as f1. Given that the beat frequency is 5 Hz, we have:
|ft - f1| = 5 Hz
When the length of the wire is changed to 21 cm, assume the new frequency of the sonometer wire to be f2. Beat frequency remains unchanged:
|ft - f2| = 5 Hz
The frequency of a vibrating string is inversely proportional to its length:
f1 * L1 = f2 * L2
Given L1 = 20 cm and L2 = 21 cm,
f1 * 20 = f2 * 21 which simplifies to:
f2 = (20/21) * f1
Since |ft - f1| = 5 and |ft - f2| = 5, solving these equations requires knowing the possible frequencies of the sonometer.
Determine f1 and f2 using the equations:
Solve both scenarios for ft + 5 = f1 and ft - 5 = f2:
If ft= f1 ± 5 Hz,
Scenario 1: for f1 = (n/20), f2 = (20/21) * (n/20) = n/21, verify both.
You are spinning a rock, of mass 0.75 kg, at the end of a string of length 0.86m in a vertical circle in uniform circular motion. What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path? If the string's maximum tension is 45 N, what is the maximum speed the rock can have so that the string does not break? At what point in the vertical circle does this maximum value occur?
Answer:
v (minimum speed) = 2.90 m/sec.
[tex]\\ \\ maximum speed (v)= 6.57 m/sec.\\[/tex]
Maximum value of speed will occur at lowest point of vertical circle.
Explanation:
a) What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?
Using the force balance expression at the top of the circle,
Gravitational Force + Tension force = Centrifugal force
[tex]m*g + T = m*v^2/R[/tex]
Given that : T = 0
R = length of string = 0.86 m
mass of the spinning rock = 0.75 kg
[tex]v = \sqrt{g*R}[/tex]
[tex]v = \sqrt{9.81*0.86}[/tex]
v (minimum speed) = 2.90 m/sec.
b) what is the maximum speed the rock can have so that the string does not break?
Here the force balance at bottom of circle is represented by the illustration:
[tex]T = m*g + m*v^2/R[/tex]
Given that:
maximum tension T = 45 N
maximum speed v = ??
mass m = 0.75 kg
∴
[tex]45 - 0.75*9.81 = 0.75*\frac{v^2}{0.86} \\\\v^2 = 0.86*(45 - 0.75*9.81)/0.75 \\ v = \sqrt{0.86*(45 - 0.75*9.81)/0.75\\ maximum speed (v)= 6.57 m/sec.\\[/tex]
c)
At what point in the vertical circle does this maximum value occur?
Maximum value of speed will occur at lowest point of vertical circle.
This is so because at the lowest point; the tension in string will be maximum.
The Mariana trench is located in the Pacific Ocean at a depth of about 11 000 m below the surface of the water. The density of seawater is 1025 kg/m3. (a) If an underwater vehicle were to explore such a depth, what force would the water exert on the vehicle's observation window (radius = 0.280 m)? (b) For comparison, determine the weight of a jetliner whose mass is 2.44 × 105 kg.
Answer:
A) 27209506.5 N
B) 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
Explanation:
Force du to depth of water is
F = pghA
P = density of salt water = 1025 kg/m3
g = acceleration due to gravity 9.81 m/s2
h = depth of water 11000 m
A = area pressure acts
Area = ¶r^2 = 3.142 x 0.280^2 = 0.246 m^2
Therefore
F = 1025 x 9.81 x 11000 x 0.246
= 27209506.5 N
Weight of a jetliner with mass 2.44 × 10^5 kg is,
2.44×10^5 x 9.81 = 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
The force exerted by the water on the sea underwater vehicle's window at the depth of the Mariana Trench is approximately 2.73 x 10⁷ N. For comparison, the weight of a jetliner with a mass of 2.44 x 10⁵ kg is 2.39 x 10⁶ N. Thus, the force exerted on the underwater vehicle at the Mariana's depth is an order of magnitude greater than the weight of the jetliner.
Explanation:To calculate the force that the water exerts on the observation window of an underwater vehicle, you first need to find the pressure at the depth of the Mariana Trench. The pressure is given by the formula P = ρgh, where P is pressure, ρ is density, g is acceleration due to gravity, and h is height (in this case, depth).
Substituting the given values, the pressure at a depth of 11,000 m is about 1.11 x 10⁸ Pa. The force exerted on the window can be found using F = PA, where A is the area of the window. With a radius of 0.280 m, the area of the window (A = πr²) is 0.246 m². Thus, the force applied by the water on the window is F = (1.11 x 10⁸ Pa) (0.246 m²) = 2.73 x 10⁷ N approx.
For the second part of the question, the weight (force) of the jetliner is given by the equation W = m × g. Using the given mass (2.44 x 10⁵ kg) and the acceleration due to gravity (9.8 m/s²), the weight of the jetliner is 2.44 x 10⁵ kg × 9.8 m/s² = 2.39 x 10⁶ N, which is an order of magnitude less than the force exerted on the underwater vehicle at the depth of the Mariana Trench.
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In a double‑slit interference experiment, the wavelength is λ = 452 nm λ=452 nm , the slit separation is d = 0.190 mm d=0.190 mm , and the screen is D = 49.0 cm D=49.0 cm away from the slits. What is the linear distance Δ x Δx between the eighth order maximum and the third order maximum on the screen?
Answer:
Δx = 5.82mm
Explanation:
To find the distance between the eight maximum and the third one you use the following formula:
[tex]x_m=\frac{m\lambda D}{d}[/tex] (1)
λ: wavelength = 452*10^-9 m
m: order of the fringes
D: distance to the scree = 0.49m
d: distance between slits = 0.190*10^-3 m
you use for m=8 and m=3, then you calculate x8 - x3:
[tex]x_8=\frac{8(452*10^{-9}m)(0.49m)}{0.190*10^{-3}m}=9.32*10^{-3}m\\\\x_3=\frac{3(452*10^{-9}m)(0.49m)}{0.190*10^{-3}m}=3.49*10^{-3}m\\\\\Delta x_{8,3}=5.82*10^{-3}m=5.82mm[/tex]
hence, the distance between these fringes is 5.82mm
Outside my window a squirrel is scurrying up and down a tree. Its position function is given by s(t) = t 3 − 12t 2 + 36t for the seven seconds that I’m watching it (so from t = 0 to t = 7). (a) What is the velocity function, v(t), for the motion of the squirrel? (b) What is the acceleration function, a(t), for the motion of the squirrel? (c) At the four second mark, is the squirrel moving up the tree or down the tree? Justify your answer.
Answer:
Explanation:
Given
Position of squirrel is given by
[tex]s(t)=t^3-12t^2+36t[/tex]
Velocity is given by
[tex]v(t)=\frac{ds(t)}{dt}=\frac{d(t^3-12t^2+36t)}{dt}[/tex]
[tex]v(t)=3t^2-12\times 2t+36[/tex]
[tex]v(t)=3t^2-24t+36[/tex]
(b) acceleration is given by
[tex]a(t)=\frac{da(t)}{dt}=\frac{d(3t^2-24t+36)}{dt}[/tex]
[tex]a(t)=6t-24[/tex]
(c)at [tex]s(3)=3^3-12(3)^2+36(3)[/tex]
[tex]s(3)=27\ m[/tex]
at [tex]s(4)=4^3-12(4)^2+36(4)[/tex]
[tex]s(4)=16\ m[/tex]
at [tex]t=3\ s[/tex] Position is [tex]27\ m[/tex] and at [tex]t=4\ s[/tex] position is [tex]16\ m[/tex]
therefore squirrel is moving down
The instructions for an electric lawn mower suggest that an A gauge extension cord ( cross sectional area = 4.2 x 10-7 m2 ) should only be used for distances up to 35 m. The resistivity of copper (used in the extension cord) is 1.72 x 10-8 .m at 20oC and the temperature coefficient of resistivity of copper is 0.004041 (oC)-1. What is the resistance of a A type extension cord of length 35 m at 40oC
Answer:
1.43 Ω
Explanation:
Given that
Cross sectional area of the cord, A = 4.2*10^-7 m²
Distance meant to be used, L = 35 m
Resistivity of copper, p = 1.72*10^-8 Ωm
Temperature of copper, t = 20° C
Temperature coefficient of copper, t' = 0.004041 °C^-1
To solve this, we would use the resistance and resistivity formula
R = pL / A, on substituting
R = (1.72*10^-8 * 35) / 4.2*10^-7
R = 6.02*10^-7 / 4.2*10^-7
R = 1.43 ohm
Therefore, the resistance of the extension cord is 1.43 ohm
The light pattern shown is a result from a beam of light being passed through a single slit. The pattern is created from constructive and destructive interference. This phenomenon is due to
A) the wave nature of light.
B) the particle nature of light.
C) the resonance effect of light.
D) the photoelectric effect of light.
Answer:
A
Explanation:
Certain neutron stars (extremely dense stars) are believed to be rotating at about 0.83 rev/s. If such a star has a radius of 40 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation? Number Enter your answer in accordance to the question statement Units Choose the answer from the menu in accordance to the question statement
Answer:
The mass of the star would be [tex]M = 2.644*10^{24} \ kg[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w = 0.83\ rev/s = 0.83 * 2 \pi = 1.66 rad/s[/tex]
The radius of the star is [tex]r = 40km = 40 *1000 = 40 * 10^{3} m[/tex]
Generally the minimum mass of the start is mathematically evaluated as
[tex]M = \frac{r^3 w^2}{G}[/tex]
Where is the gravitational constant with a values of [tex]G = 6.67*10^{-11} N \cdot m^2 /kg[/tex]
[tex]M = \frac{(40*10^3)^3 * 1.66^2}{6.67*10^{-11}}[/tex]
[tex]M = 2.644*10^{24} \ kg[/tex]
Buffy is rolling along in her 10.4 kg wagon at 3.4 m/s (in the positive direction) when she jumps off the back. She continues to move forward at 1.7 m/s relative to the ground. This causes her wagon to go speeding forward at 8.04 m/s relative to the ground. How much does Buffy weigh
Answer:
Buffy weigh's 28.39 kg
Explanation:
Given;
mass of Buffy and wagon, M = 10.4 kg
final velocity of Buffy - wagon system, v = 3.4 m/s
Buffy's velocity relative to the ground, u₁ = 1.7 m/s
Wagon's velocity relative to the ground, u₂ = 8.04 m/s
Buffy's velocity = 1.7 - 3.4 = - 1.7 m/s
Wagon's velocity = 8.04 - 3.4 = 4.64 m/s
Apply the principle of conservation of linear momentum;
1.7 x m = 4.64 x 10.4
where;
m is mass of wagon
1.7 m = 48.256
m = 48.256 / 1.7
m = 28.39 kg
Therefore, Buffy weigh's 28.39 kg
An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick. What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, m proton = 1.67 × 10-27 kg, = 1.055 × 10-34 J · s, h = 6.626 × 10-34 J · s) a. 7.7 × 10-10 % b. 0.11% c. 1.1% d. 0.011% e. 1.1 × 10-4 %
Answer:
d) 0.011%
Explanation:
The probability for tunneling the barrier is given by the following formula:
[tex]P=exp(-2d\sqrt{\frac{2m_e(U_o-E)}{\hbar ^2} }\ )[/tex] ( 1 )
me: mass of the electron
Uo: energy of the barrier
E: energy of the electron
d: thickness of the barrier
By replacing the values of the parameters in (1), you obtain:
[tex]P=exp(-2(0.20*10^{-9}m)\sqrt{\frac{2(9.11*10^{-31}kg)(100eV-80eV)(1.60*10^{-19}J)}{(1.055*10^{-34}Js)^2}})\\\\P=e^{-9.15}=1.08*10^{-4}\approx0.011\%[/tex]
hence, the probability is 0.011% (answer d)
A ring is attached at the center of the underside of a trampoline. A sneaky teenager crawls under the trampoline and uses the ring to pull the trampoline slowly down while his 67-kg mother is sleeping on it. When he releases the trampoline, she is launched upward. As she passes through the position at which she was before her son stretched the trampoline, her speed is 3.0m/s.
How much elastic potential energy did the son add to the trampoline by pulling it down? Assume the interaction is nondissipative.
Answer:
E = 301.5 J
Explanation:
We have,
Mass of mother, m = 67 kg
Here, a sneaky teenager crawls under the trampoline and uses the ring to pull the trampoline slowly down. As she passes through the position at which she was before her son stretched the trampoline, her speed is 3 m/s.
It is required to find the elastic potential energy the son add to the trampoline by pulling it down. It is based on the conservation of energy.
The elastic potential energy of the mother = the elastic potential energy the son add to the trampoline.
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2[/tex]
So, the elastic potential energy is :
[tex]E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 67\times 3^2\\\\E=301.5\ J[/tex]
So, the elastic potential energy of 301.5 J the son add to the trampoline by pulling it down.
To apply Problem-Solving Strategy 21.1 Faraday's law. A closely wound rectangular coil of 80 turns has dimensions 25.0 cm by 40.0 cm. The plane of the coil is rotated from a position in which it makes an angle of 37.0 degrees with a magnetic field of 1.10 T to a position perpendicular to the field. The rotation takes 0.0600 s. What is the average emf E induced in the coil
Answer:
58.37 V
Explanation:
Given that
Number of winding on coil, N = 80 turns
Area of the coil, A = 25 cm * 40 cm = 0.25 m * 0.4 m
Magnitude of magnetic force, B = 1.1 T
Time of rotation, t = 0.06 s
See the attachment for calculations
Answer:
ℰ[tex]Ф_{av}[/tex]=58.37V
Explanation:
Given:
Number of winding on coil 'N' = 80 turns
Area 'A' = 25 cm x 40 cm = 0.25 m x 0.4 m =>0.10m²
Magnitude of magnetic force 'B' = 1.1 T
Time ' t' = 0.06 s
The average magnitude of the induced emf is given by:
ℰ[tex]Ф_{av}[/tex]=N|ΔФB/Δt| => N |Ф[tex]Ф_{B,f}[/tex] - Ф[tex]Ф_{B,i}[/tex]| /Δt
The flux through the coil is Ф[tex]Ф_{B[/tex] = BA cos∅
ℰ[tex]Ф_{av}[/tex]=NBA |cos(∅[tex]Ф_{f}[/tex]) - cos(∅[tex]Ф_{i}[/tex])| /Δt
As the initial angle is ∅= 97-37 =>53° and the final angle is ∅=0°
ℰ[tex]Ф_{av}[/tex]=[tex]\frac{(80)(1.1)(0.1)|cos(0)-cos(53)|}{0.06}[/tex]
ℰ[tex]Ф_{av}[/tex]= [tex]\frac{(8.8)|1-0.002|}{0.06}[/tex]
ℰ[tex]Ф_{av}[/tex]=58.37V
The number of protons in the nucleus of an atom determines the species of the atom, i.e., the element to which the atom belongs. An atom has the same number of protons and neutrons. But the electron number cannot be used instead because
A. electrons are not within the nucleus
B. electrons are negatively charged
C. electrons can be removed from or added to an atom
D.electrons are lighter than protons
The number of protons in the nucleus of an atom determines the element that the atom belongs to.
Explanation:The subject of this question is Chemistry. The number of protons in the nucleus of an atom determines the species or element to which the atom belongs. This is because each element has a unique number of protons, known as the atomic number. The other options are not correct because:
Electrons are not within the nucleus: Electrons are found in electron shells outside the nucleus of an atom.Electrons are negatively charged: While electrons are negatively charged, this does not affect their use in determining the species of an atom.Electrons can be removed from or added to an atom: While electrons can be added or removed from an atom, their number does not define the species of the atom.Electrons are lighter than protons: Although electrons are lighter than protons, this is not the reason why their number cannot be used to determine the species of an atom.Learn more about protons in the nucleus here:https://brainly.com/question/30654172
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Assume that a certain location on the Earth reflects 33.0% of the incident sunlight from its clouds and surface. (a) Given that the intensity of solar radiation at the top of the atmosphere is 1368 W/m2, find the radiation pressure on the Earth, in pascals, at the location where the Sun is straight overhead. Pa (b) State how this quantity compares with normal atmospheric pressure at the Earth's surface, which is 101 kPa. Patm Prad
Answer:
a) The radiation pressure on the Earth is 6.065x10⁻⁶Pa
b) The atmospheric pressure is 1.665x10¹⁰ greater than the pressure exerted by the radiation.
Explanation:
Given:
I = intensity of solar radiation = 1368 W/m²
Earth reflects 33%, therefore Earth absorbs 67%
P = pressure = 101 kPa = 1.01x10⁵Pa
c = speed of light = 3x10⁸m/s
Questions:
a) Find the radiation pressure on the Earth, in pascals, at the location where the Sun is straight overhead, P = ?
b) State how this quantity compares with normal atmospheric pressure at the Earth's surface
a) To solve, it is necessary to calculate the pressure exerted by both the reflected light and the light that is absorbed, in this way:
The pressure exerted by the reflected light:
[tex]P_{1} =\frac{2*0.33*I}{c} =\frac{2*0.33*1368}{3x10^{8} } =3.01x10^{-6} Pa[/tex]
The pressure exerted by the absorbed light:
[tex]P_{2} =\frac{0.67*I}{c} =\frac{0.67*1368}{3x10^{8} } =3.055x10^{-6} Pa[/tex]
The radiation pressure on the Earth:
Pt = P₁ + P₂ = 3.01x10⁻⁶ + 3.055x10⁻⁶ = 6.065x10⁻⁶Pa
b) Comparing with normal atmospheric pressure
[tex]Ratio=\frac{P_{atm} }{P_{t} } =\frac{1.01x10^{5} }{6.065x10^{-6} } =1.665x10^{10}[/tex]
According to this result, the atmospheric pressure is 1.665x10¹⁰ greater than the pressure exerted by the radiation.