To solve this problem we must use the perioricity equations given as a function of the mass and spring constant. Mathematically this can be expressed as:
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
m = mass
k = Spring constant
Re-arrange to find the mass we have
[tex]m = \frac{T^2k}{4\pi^2 }[/tex]
Replacing with our values we have that
[tex]m = \frac{2.4^2*10}{4\pi^2}[/tex]
[tex]m = 1.459kg[/tex]
D) Mass is independent of acceleration due to gravity (as you can see at the equation previously given) for this reason the mass suspended on mars is given as the same found. Therefore the mass would be
m = 1.459kg
Final answer:
The mass suspended from the spring that would result in a period of 2.4 s is 0.142 kg.
Explanation:
To find the mass suspended from the spring that would result in a period of 2.4 s, we can use the formula for the period of oscillation:
T = 2π √(m/k)
Where T is the period, m is the mass, and k is the force constant of the spring.
Let's rearrange the formula to solve for m:
m = (T^2 · k) / (4π^2)
Substituting the given values:
m = (2.4^2 · 10) / (4π^2)
m = 0.142 kg
A gas pressure of 65,000.0 Pa (gage pressure) is applied to one leg of a U-tube manometer, which contains a fluid of density of 1905.0 kg/m3. The other leg is open to atmosphere. What is the manometer reading? Draw the U-tube manometer and label all relevant parameters. Hint: You may assume that air density is much less than the fluid density
Answer:
[tex]h=3.4817\ m[/tex]
Explanation:
Given:
Pressure of a gas, [tex]P=65000\ Pa[/tex]density of the fluid in the manometer, [tex]\rho=1905\ kg.m^{-3}[/tex]We know that:
[tex]P=\rho.g.h[/tex]
putting respective values
[tex]65000-101325=1905 \times 9.8\times h[/tex]
[tex]h=3.4817\ m[/tex] is the reading in terms of the fluid column height on the manometer.
Considering the rise in the level of fluid this fluid is not suitable for such huge pressures for a practically convenient manometer.
Atomic hydrogen produces a well-known series of spectral lines in several regions of the electromagnetic spectrum. Each series fits the Rydberg equation with its own particular n1 value. Calculate the value of n1 that would produce a series of lines in which the highest energy line has a wavelength of 821 nm.
Answer:
n₁ = 3
Explanation:
The energy of the states in the hydrogen atom is explained by the Bohr model, the transitions heal when an electron passes from a state of higher energy to another of lower energy,
ΔE = [tex]E_{nf}[/tex] - E₀ = - k²e² / 2m (1 / [tex]n_{f}[/tex]²2 - 1 / n₀²)
The energy of this transition is given by the Planck equation
E = h f = h c / λ
h c / λ = -k²e² / 2m (1 / no ²- 1 / no²)
1 / λ = Ry (1/ [tex]n_{f}[/tex]² - 1 / n₀²)
Let's apply these equations to our case
λ = 821 nm = 821 10⁻⁹ m
E = h c / λ
E = 6.63 10⁻³⁴ 3 10⁸/821 10⁻⁹
E = 2.423 10⁻¹⁹ J
Now we can use the Bohr equation
Let's reduce to eV
E = 2,423 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹) = 1,514 eV
[tex]E_{nf}[/tex] - E₀ = -13.606 (1 / [tex]n_{f}[/tex]² - 1 / n₀²) [eV]
Let's look for the energy of some levels
n [tex]E_{n}[/tex] (eV) [tex]E_{nf}[/tex] - E[tex]E_{ni}[/tex] (eV)
1 -13,606 E₂-E₁ = 10.20
2 -3.4015 E₃-E₂ = 1.89
3 -1.512 E₄- E₃ = 0.662
4 -0.850375
We see the lines of greatest energy for each possible series, the closest to our transition is n₁ = 3 in which a transition from infinity (n = inf) to this level has an energy of 1,512 eV that is very close to the given value
A binary star system consists of two stars of masses m1 and m2. The stars, which gravitationally attract each other, revolve around the center of mass of the system. The star with mass m1 has a centripetal acceleration of magnitude a1. Note that you do not need to understand universal gravitation to solve this problem.Find a2, the magnitude of the centripetal acceleration of the star with mass m2.
Answer:
a₂ = m₁ / m₂ a₁
Explanation:
For this exercise we note that the attraction between the two stars is an action and reaction force, therefore it has the same magnitude, but it is applied to each of the bodies
Let's apply Newton's second law on the star 1
F₁ = m₁ a₁
Newton's second law in star 2
F₂ = m₂ a₂
| F₁ | = | F₂ |
m₁ a₁ = m₂ a₂
a₂ = m₁ / m₂ a₁
What temperature increase is necessary to increase the power radiated from an object by a factor of 8?
a. 8 Kb. 2 Kc. 100%d. about 68%
To solve this problem it is necessary to apply the concepts related to the Power defined from the Stefan-Boltzmann equations.
The power can be determined as:
[tex]P = \sigma T^4[/tex]
Making the relationship for two states we have to
[tex]\frac{P_1}{P_2} = \frac{T_1^4}{T_2^4}[/tex]
Since the final power is 8 times the initial power then
[tex]P_2 = 8P_1[/tex]
Substituting,
[tex]\frac{1}{8} = \frac{T_1^4}{T_2^4}[/tex]
[tex]T_2 = T_1 8*(\frac{1}{4})[/tex]
[tex]T_2 = 1,68T_1[/tex]
The temperature increase would then be subject to
[tex]\Delta T = T_2-T_1[/tex]
[tex]\Delta T = 1.68T_1 -T_1[/tex]
[tex]\Delta T = 0.68T_1[/tex]
The correct option is D, about 68%
Final answer:
The correct option is (c) 100%. To increase the power radiated from an object by a factor of 8, the temperature must be doubled, corresponding to a 100% increase. This is based on the Stefan-Boltzmann law.
Explanation:
The question relates to the physics concept of radiation and how it varies with temperature. Specifically, it involves the Stefan-Boltzmann law, which states that the power radiated from an object is proportional to the fourth power of its absolute temperature.
If the power radiated from an object is to increase by a factor of 8, we use the relationship P ≈ T^4, where P is power and T is the temperature in Kelvin. To calculate the necessary temperature increase, if the initial power is P and the final power is 8P, then (T_final)^4 = 8 × (T_initial)^4.
By taking the fourth root on both sides, we find that T_final = 2× T_initial. Therefore, the temperature must be doubled, or the temperature increase is by 100%. Hence, the correct answer is option (c) 100%
Why would a plume of solid silicate rock rising slowly from deep in the mantle begin melting as it neared the base of the lithosphere?
A. The rock heats up and expands at lower pressures, causing it to liquefy
B. Temperatures remain high as lowered pressures decrease melting temperatures.
C. The lowered pressures cause rapid heat loss accompanied by melting.
D. None of the above
Answer:
B. Temperatures remain high as lowered pressures decrease melting temperatures.
Explanation:
Rocks , like ice , contracts on melting so , their melting point decreases on decrease in pressure . When hot solid rocks are pushed towards the surface of the earth, due to decrease in pressure , their melting point decreases , so the solid rocks starts melting without any addition of heat . Their present temperature becomes more than their melting point.
What fraction of the volume of an iceberg (density 917 kg/m3) would be visible if the iceberg floats in (a) the ocean (salt water, density 1024 kg/m3) and (b) in a river (fresh water, density 1000 kg/m3)? (When salt water freezes to form ice, the salt is excluded. So, an iceberg could provide fresh water to a community.)
The fraction of the volume of an iceberg that is visible when it floats in saltwater or freshwater can be calculated using Archimedes' principle. For the given densities, the iceberg would sink in both saltwater and freshwater.
Explanation:When an iceberg floats in water, a fraction of its volume is submerged, which is determined by the density of water and the density of the iceberg. To calculate the fraction of the iceberg that is submerged, you can use Archimedes' principle. Archimedes' principle states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
In this case, the density of the iceberg is given as 917 kg/m³. Let's calculate the fraction of the volume of the iceberg that would be visible when floating in (a) the ocean (salt water, density 1024 kg/m³) and (b) a river (fresh water, density 1000 kg/m³).
(a) For the iceberg floating in saltwater, the fraction of the volume submerged can be calculated as:Fraction submerged = (density of iceberg - density of seawater) / density of iceberg
Using the given densities:
(917 kg/m³ - 1024 kg/m³) / 917 kg/m³ = -0.1166
The negative value indicates that the iceberg would not float in saltwater. In other words, it would sink.
(b) For the iceberg floating in freshwater, the fraction of the volume submerged can be calculated as:Fraction submerged = (density of iceberg - density of fresh water) / density of iceberg
Using the given densities:
(917 kg/m³ - 1000 kg/m³) / 917 kg/m³ = -0.0902
Again, the negative value indicates that the iceberg would sink in freshwater as well.
As part of a safety investigation, two 1600 kg cars traveling at 21 m/s are crashed into different barriers.
A) Find the average force exerted on the car that hits a line of water barrels and takes 1.8 s to stop.
B) Find the average force exerted on the car that hits a concrete barrier and takes 0.14 s to stop.
Answer
given,
mass of the car = 1600 Kg
velocity of the car = 21 m/s
time = 1.8 s
a) momentum = mass x velocity
P = 1600 x 21
P = 33600 kg.m/s
[tex]Force = \dfrac{change\ in \ momentum}{t}[/tex]
[tex]Force = \dfrac{33600-0}{1.8}[/tex]
F = 18666.67 N
b) [tex]Force = \dfrac{change\ in \ momentum}{t}[/tex]
[tex]Force = \dfrac{33600-0}{0.14}[/tex]
F = 240000 N
Two cylinders with the same mass density rhoC = 713 kg / m3 are floating in a container of water (with mass density rhoW = 1025 kg / m3). Cylinder #1 has a length of L1 = 20 cm and radius r1 = 5 cm. Cylinder #2 has a length of L2 = 10 cm and radius r2 = 10 cm. If h1 and h2 are the heights that these cylinders stick out above the water, what is the ratio of the height of Cylinder #2 above the water to the height of Cylinder #1 above the water (h2 / h1)? h2 / h1 =
Answer:
Explanation:
Given
density of cylinder is [tex]\rho _c=713 kg/m^3[/tex]
Length of first cylinder is [tex]L_1=20 cm[/tex]
radius [tex]r_1=5 cm[/tex]
For cylinder 2 [tex]L_2=10 cm [/tex]
[tex]r_2=10 cm[/tex]
[tex]h_1[/tex] and [tex]h_2[/tex] are the height above water
E
as object is floating so its weight must be balanced with buoyant force
[tex]\rho _c\frac{\pi }{4}d_1^2L_1g=\rho _w\frac{\pi }{4}d_1^2(L_1-h_1)g----1[/tex]
For 2nd cylinder
[tex]\rho _c\frac{\pi }{4}d_2^2L_2g=\rho _w\frac{\pi }{4}d_2^2(L_2-h_2)g----2[/tex]
Dividing 1 and 2 we get
[tex]\frac{L_1}{L_2}=\frac{L_1-h_1}{L_2-h_2}[/tex]
[tex]\frac{20}{10}=\frac{20-h_1}{10-h_2}[/tex]
[tex]2h_2=h_1[/tex]
[tex]\\\Rightarrow\frac{h_2}{h_1}=\frac{1}{2}[/tex]
The process of vision involves 4 basic steps in an appropriate order. Of the options listed below, what is correct order in the process of vision? a) gathering light, within the eye processing, transduction, and brain processing b) without the eye processing, gather light, transfusion, and brain processing c) photo-production, gathering light, within the eye processing, and transduction d) gathering light, photo-production, transfusion, and within the eye processing e) light collection, photo-production, hoarding light, and brain processing
Answer:
Option (a) is correct.
Explanation:
For us to be able to visualize an object, there are basic processes that takes place in a split second.
First Light which strike the object reflect into our eye. This reflected light passes through the pupil as the iris gives way for it. This light moves through other layers and impinges on the retina and stimulates it. This is the process of within the eye processing. The light is then converted to a signal by the optic nerve and sent to the brain for interpretation. The image produced by the retina is inverted. The brain helps to interpret this image as upright. This is the reason why we see objects just the way they are
17 of 24 Constants In an experiment to simulate conditions within an automobile engine, 0.165 mol of air at a temperature of 700 K and a pressure of 3.40×106 Pa is contained in a cylinder of volume 280 cm3 . Then 690 J of heat is transferred to the cylinder. Part APart complete If the volume of the cylinder is constant while the heat is added, what is the final temperature of the air? Assume that the air is essentially nitrogen gas. T = 901 K Previous Answers Correct Part B If instead the volume of the cylinder is allowed to increase while the pressure remains constant, find the final temperature of the air.
Answer:
a. T₂ = 901.43 ° K
b. T₂ = 843.85 ° K
Explanation:
Given
η = 0.165 mol , Q = 690 J , V = 280 cm³ , P = 3.40 x 10 ⁶ Pa , T₁ = 700 ° K
Using the equation that describe the experiment of heat
Q = η * Cv * ΔT
a.
Nitrogen Cv = 20.76 J / mol ° K
ΔT = T₂ - T₁
T₂ = [ Q / ( η * Cv) ] + T₁
T₂ = [ 690 J / ( 0.165 mol * 20.76 J / mol ° K ) ] + 700 ° K
T₂ = 901.43 ° K
b.
Nitrogen Cp = 29.07 J / mol ° K
T₂ = [ Q / ( η * Cv) ] + T₁
T₂ = [ 690 J / ( 0.165 mol * 29.07 J / mol ° K ) ] + 700 ° K
T₂ = 843.85 ° K
The Final temperature of the air when the volume of the cylinder is constant is 700 K.
To find the final temperature of the air when the volume of the cylinder is constant, we can use the ideal gas law equation:
P₁V₁/T₁ = P₂V₂/T₂
Since the volume is constant, V₁ = V₂.
The initial pressure (P₁) and temperature (T₁) are given, and the final pressure (P₂) is also given (3.40×106 Pa).
We can rearrange the equation to solve for T₂:
T₂ = (P₂T₁) / P₁
Substituting the given values, we have:
T₂ = (3.40×106 Pa * 700 K) / (3.40×106 Pa)
T₂ = 700 K
So, the final temperature of the air is 700 K when the volume of the cylinder is constant.
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How long does it take a 750-W coffeepot to bring to a boil 0.85L of water initially at 15?C?
Assume that the part of the pot which is heated with the water is made of 360 g of aluminum, and that no water boils away. Ignore the heat loss to the surrounding environment.
The value of specific heat for water is 4186 J/kg?C? and for aluminum is 900 J/kg?C?.
To calculate the time it takes for the coffee pot to bring the water to a boil, we need to determine the amount of heat required and use the power of the coffee pot. By calculating the heat required for both the water and the aluminum pot, and using the power of the coffee pot, we can determine that it takes approximately 96.077 seconds for the coffee pot to bring the water to a boil.
Explanation:To determine how long it takes for the coffee pot to bring the water to a boil, we need to calculate the amount of heat required. The heat required to raise the temperature of the water from 15?C to 100?C is given by the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature. Using the values given, we can calculate the heat required.
The heat required for the water is Q = (0.85 kg)(4186 J/kg?C)(100?C - 15?C) = 43707.5 J. The heat required for the aluminum coffee pot is Q = (0.36 kg)(900 J/kg?C)(100?C - 15?C) = 28350 J. Adding the two heats together gives us a total heat of 72057.5 J.
The power of the coffee pot is given as 750 W. Power is defined as the rate at which work is done or the rate at which energy is transferred. So, we can use the formula P = E/t, where P is power, E is energy, and t is time. Rearranging the formula, we can solve for t:
t = E/P = 72057.5 J / 750 W = 96.077 seconds.
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A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9.8 m/s2, 2.7 m, 6 kg, 31C. What is the speed of the ball when it is in circular motion? Answer in units of m/s.
The speed of the metal ball in circular motion is calculated by using the equation for centripetal acceleration (v = √(rg)) and substituting the given values for radius and gravity. After calculation, the speed of the ball is determined to be approximately 5.1 m/s.
Explanation:In this physics problem, we can calculate the speed of the metal ball by using the provided values and the concept of centripetal acceleration. Centripetal acceleration results from the force that makes the metal ball move in a circular pattern, which in this case is provided by the tension in the thread and gravity. In a scenario where the ball is in uniform circular motion, with the tension at an angle, forming a cone around the vertical axis, the component of tension providing the centripetal force equals the gravitational force or weight of the object.
A key equation that can be used here is: ac = v² /r, where 'ac' is the centripetal acceleration, 'v' is the velocity (or speed), and 'r' is the radius. Given that ac = g (acceleration due to gravity = 9.8 m/s²), we can modify the equation to: v = √(rg), where 'r' is the radius or the length of the thread (2.7 m).
Implementing these values into our equation we find that: v = √[(2.7 m)(9.8 m/s²)] = √26.46 = 5.1 m/s. Therefore, the speed of the metal ball when it is in circular motion is approximately 5.1 m/s.
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A 2.3 kg block executes SHM while attached to a horizontal spring of spring constant 150 N/m. The maximum speed of the block as it slides on a horizontal frictionless surface is 2.3 m/s. What are (a) the amplitude of the block's motion, (b) the magnitude of its maximum acceleration, and (c) the magnitude of its minimum acceleration? (d) How long does the block take to complete 4.1 cycles of its motion?
Answer
given,
mass of block = m = 2.3 Kg
spring constant = k = 150 N/m
speed = 2.3 m/s
a) we know,
[tex]\omega = \sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega = \sqrt{\dfrac{150}{2.3}}[/tex]
ω = 8.08 rad/s
v = A ω
2.3 = A x 8.08
A = 0.285 m
b) maximum acceleration
a = A ω²
a = 0.285 x 8.08²
a = 18.58 m/s²
c) acceleration is minimum at mean position the minimum value of acceleration is zero.
d) time period
[tex]T = \dfrac{2\pi}{\omega}[/tex]
[tex]T = \dfrac{2\pi}{8.08}[/tex]
T = 0.778 s
t = 4.1 x 0.7778
t = 3.188 s
A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.75 ms from an initial speed of 2.25 m/s . What is the magnitude of the average contact force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.65 kg ?
Answer:
1350N
Explanation:
[tex]2.75 ms = 2.75*10^{-3}s[/tex]
The force exerted on the hand would be the momentum divided by the duration of contact.
As the hand is coming to rest, final velocity would be 0
[tex]F = \frac{\Delta P}{\Delta t} = \frac{m(0 - v)}{\Delta t} = \frac{1.65*(2.25 - 0)}{2.75 * 10^{-3}} = -1350 N[/tex]
The magnitude of the force would be 1350N
A standard 14.16-inch (0.360-meter) computer monitor is 1024 pixels wide and 768 pixels tall. Each pixel is a square approximately 281 micrometers on each side. Up close, you can see the individual pixels, but from a distance they appear to blend together and form the image on the screen.
If the maximum distance between the screen and your eyes at which you can just barely resolve two adjacent pixels is 1.30 meters, what is the effective diameter d of your pupil? Assume that the resolvability is diffraction-limited. Furthermore, use lambda = 550 nanometers as a characteristic optical wavelength.
The effective diameter d of your pupil should be considered as the 0.0031 when maximum distance should be 1.30 meters.
Calculation of the effective diameter:Since
y = Length of pixel = 281 μm
L = Distance to screen = 1.3 m
Wavelength = 550 nm
d = Pupil diameter
So,
[tex]tan = y / l\\\\= tan^-1 * 281*10^-6/1.3[/tex]
Now
sin = 1.22 wavelength / d
d = 1.22 * wavelength / sin
[tex]= 1.22*550*10^-9/ sin ( tan^-1 * 281*10^-6/1.3)[/tex]
= 0.0031 m
hence, The effective diameter d of your pupil should be considered as the 0.0031 when maximum distance should be 1.30 meters.
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The effective diameter of the pupil of one's eye when barely able to resolve two adjacent pixels on a computer screen from a certain distance, where the point of resolution is diffraction limited, is approximately 1.7 mm.
Explanation:This question involves the use of the formula for the resolution of an optical system, which is the Rayleigh’s criterion. The formula can be written as follows: θ = 1.22 * λ/D, where θ is the smallest angular separation that can be resolved, λ is the wavelength of light, and D is the diameter of the aperture (i.e., the pupil in this case).
In this question, the pixel’s size and the distance from the screen allow us to calculate θ. The tangent of θ is approximately equal to the ratio of the pixel size to the viewing distance. Hence, θ = arctan((281 * 10^-6 m) / 1.3 m).
By substituting θ, and λ (550 * 10^-9 m) into the formula and solve for D, the effective diameter of the pupil, we get: D = 1.22 * λ / θ. The result is around 0.0017 meters or 1.7 mm.
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Point A has coordinates of (2,5,1) and point D has coordinate of (9, 6, 10). Calculate the unit vector of (landa AD)
Answer:(69,69,69)
Explanation:
The ratio of the intercortex of the inner circle has an area of 98:89. This is due to the fact the vector of the circle is no less than the ratio times the pie squared. Then pie squared the circumference will end up approximately at 68.83957 where you then round up to all 69's.
Which of the following best describes a substance in which the temperature remains constant while at the same time it is experiencing an inward heat flow?
a. gasb. liquidc. solidd. substance undergoing a change of state
Answer:
The right option is (d) substance undergoing a change of state
Explanation:
Latent Heat: Latent heat is the heat required to change the state of a substance without change in temperature. Latent heat is also known as hidden heat because the heat is not visible. The unit is Joules (J).
Latent heat is divided into two:
⇒ Latent Heat of fusion
⇒ Latent Heat of vaporization.
Latent Heat of fusion: This is the heat energy required to convert a substance from its solid form to its liquid form without change in temperature. E.g (Ice) When ice is heated, its temperature rise steadily until a certain temperature is reached when the solid begins to melts.
Latent Heat of vaporization: This is the heat required to change a liquid substance to vapor without a change in temperature. The latent heat depend on the mass of the liquid and the nature of the liquid. E.g When water is heated from a known temperature its boiling point (100°C) When more heat is supplied to its boiling temperature, it continue to boil without a change in temperature.
From The above, Latent heat brings about a change of state of a substance at a steady temperature.
The right option is (d) substance undergoing a change of state
The substance that best represents a temperature remaining constant despite an inward heat flow is one that is undergoing a change of state, such as during the melting of ice or boiling of water. So, the correct option is D.
The best description of a substance in which the temperature remains constant while it is experiencing an inward heat flow is a substance undergoing a change of state (option d). This occurs during a process called phase transition.
Here's an example: when ice melts into water, it absorbs heat from the surroundings but the temperature stays constant at 0 degrees Celsius until all of the ice has melted.
Similarly, when water is boiling, its temperature holds steady at 100 degrees Celsius until all the water has transitioned into steam, despite the continued input of heat.
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What two elements are used with other metals to lower their melting points?
A) antimony and carbon
B) bismuth and carbon
C) antimony and bismuth
D) arsenic and antimony
Answer:
C) antimony and bismuth
Explanation:
Answer:
its C
Explanation:
An external pressure applied to an enclosed fluid in a hydraulic lift. Piston 1 has a radius that is 1/200 the radius of piston 2 (r1 = r2/200). If the bottom of both pistons start from the same vertical height, how much force needs to by applied to the first piston in order to begin lifting a weight of 20,000 N positioned atop the second piston?
Answer:
0.5 N
Explanation:
[tex]r_{1}[/tex] = radius of piston 1
[tex]r_{2}[/tex] = radius of piston 2
Given that :
[tex]r_{2} = 200 r_{1}[/tex]
[tex]F_{1}[/tex] = Force applied on piston 1
[tex]F_{2}[/tex] = Force applied on piston 2 = Weight being lifted = 20000 N
Using pascal's law
[tex]\frac{F_{1}}{\pi r_{1}^{2} } = \frac{F_{2}}{\pi r_{2}^{2} } \\\frac{F_{1}}{r_{1}^{2} } = \frac{20000}{(200)^{2}r_{1}^{2} } \\\\F_{1}= \frac{20000}{(200)^{2}} } \\F_{1}= 0.5 N[/tex]
A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point where the spring is not stretched. Then the mass is released and begins to oscillate. The lowest point in the oscillation is 18 cm below the point where the mass was released. What is the oscillation frequency?
Answer:
The oscillation frequency of the spring is 1.66 Hz.
Explanation:
Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :
[tex]E_s=\dfrac{1}{2}kx^2[/tex]
k is the spring constant
x is the distance to the lowest point form the initial position.
When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.
[tex]E_P=mgx[/tex]
Equating these two energies,
[tex]\dfrac{1}{2}kx^2=mgx[/tex]
[tex]\dfrac{k}{m}=\dfrac{2g}{x}[/tex].............(1)
The expression for the oscillation frequency is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}}[/tex] (from equation (1))
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}[/tex]
f = 1.66 Hz
So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.
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The diagram shows models of both prokaryotic and eukaryotic cells. The diagram shows models of both prokaryotic and eukaryotic cells. Prokaryotic cells are considered to be the more primitive of the two. What is a similarity between eukaryotes and prokaryotes as illustrated by the models? A) Both contain ribosomes. B) Both contain mitochondria. C) Both are surrounded by a capsule. D) Both contain a membrane-bound nucleus.
Answer:
both have plasma membrane, ribosomes, cytoplasm, and DNA
Explanation:
Both eukaryotic and prokaryotic cells contain ribosomes, which are responsible for protein synthesis. Eukaryotic cells are more complex with a true nucleus, found in animals and plants, while prokaryotic cells, found in bacteria, are simpler and lack a true nucleus.
Explanation:Option A is the correct answer. Both eukaryotic and prokaryotic cells contain ribosomes. The ribosomes in both types of cells serve the same function: protein synthesis, or assembly of amino acids to form proteins. Eukaryotic cells are those with a true nucleus and complex structures, found in animals, plants, fungi, and protists, while prokaryotic cells are smaller, simpler cells without a true nucleus, found in bacteria and archaea.
Unlike eukaryotic cells, prokaryotic cells don't contain a membrane-bound nucleus or mitochondria. A capsule is found in some bacteria, not all prokaryotes and no eukaryotes.
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A projectile of mass m1 moving with speed v1 in the +x direction strikes a stationary target of mass m2 head-on. The collision is elastic. Use the Momentum Principle and the Energy Principle to determine the final velocities of the projectile and target, making no approximations concerning the masses. (Use the following as necessary: m1, m2, and v1.)
Using Momentum Principle and the Energy Principle to determine the final velocities of the projectile and target are respectively;
v1 = (m1•u1 - m2•u1)/(m1 + m2)
v1 = (m1•u1 - m2•u1)/(m1 + m2)v2 = (m1•u1 + m1•u2)/(m1 + m2)
Let us denote the following;
Mass of projectile; m1
Mass of target; m2
Initial velocity of projectile; u1
Final velocity of projectile; v1
Initial velocity of target; u2 = 0 m/s
Final velocity of target; v2
From conservation of momentum, we have;
(m1•u1) + (m2•u2) = (m1•v1) + (m2•v2)
Plugging in the relevant values;
(m1•u1) + (m2•0) = (m1•v1) + (m2•v2)
(m1•u1) = (m1•v1) + (m2•v2) - - - (eq 1)
From energy principle, we have;
u1 + v1 = u2 + v2
Thus;
Since u2 is zero, then we have;
u1 + v1 = v2 - - - (eq 2)
Put (u1 + v1) for v2 in eq 1 to get;
(m1•u1) = (m1•v1) + (m2(u1 + v1))
(m1•u1) = (m1•v1) + (m2•u1 + m2•v1) - - - (eq 3)
Let us make v1 the subject;
m1•u1 - m2•u1 = m1•v1 + m2•v1
v1(m1 + m2) = m1•u1 - m2•u1
v1 = (m1•u1 - m2•u1)/(m1 + m2)
Let us make v1 the subject in eq 2;
v1 = v2 - u2
Put v2 - u2 for v1 in eq 1 to get;
(m1•u1) = (m1(v2 - u2) + (m2•v2)
(m1•u1) = m1•v2 - m1•u2 + m2•v2
m1•u1 + m1•u2 = v2(m1 + m2)
v2 = (m1•u1 + m1•u2)/(m1 + m2)
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The final velocities of the projectile and the target after an elastic collision can be arrived at by solving the equations for Conservation of Momentum and Conservation of Kinetic Energy simultaneously. Here, the target's initial velocity is zero (it's at rest). By substituting this into the equations and solving them, we can find the final velocities.
Explanation:An elastic collision is a type of collision where both momentum and kinetic energy are conserved. For this problem, we can solve the final velocities using the conservation laws.
Conservation of Momentum gives us:
m1*v1 = m1*v1' + m2*v2' -------- (1)
V1' and V2' are the final velocities of the projectile and target, respectively.
Conservation of Kinetic Energy gives us:
1/2*m1*v1^2 = 1/2*m1*(v1')^2 + 1/2*m2*(v2')^2 -------- (2)
While solving these two equations, one should remember that m2 was initially at rest, so the momentum in the Y-direction is zero. The above-mentioned equations can be solved by substitution method or any other algebraic method to get the final velocities v1' and v2'.
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In cloud formation, water vapor turns into water droplets which get bigger and bigger until it rains.
This will cause the temperature of the air in the clouds to:
a. get warmer.
b. get cooler.
c. will not affect the temperature of the air in the clouds.
d. There is no air in clouds.
Answer:
a. get warmer.
Explanation:
When the water vaporous reach the upper layer of the atmosphere they get a cooler air to which they loose their temperature and condense to form clouds as a the temperature of the air increases.
It may be noted that the water looses its high amount of latent heat of vaporization to condense into water this significantly increases the temperature of the air in contact.
Three people are standing on a horizontally rotating platform in an amusement park. One person is almost at the edge, the second one is (3/5)R from the center, and the third is (1/2)R from the center.
1-If the period of rotation of the first person is T1 = T, find in terms of Tthe period of rotation of the second person.
2- If the period of rotation of the first person is T1 = T, find in terms of T the period of rotation of the third person.
3-If the speed of the first person is v1 = v find in terms of v the speed of the second person
4- If the speed of the first person is v1 = v find in terms of v the speed of the third person.
5- If the radial acceleration of the first person is a1 = a, find in terms of athe radial acceleration of the second person.
6- If the radial acceleration of the first person is a1 = a, find in terms of athe radial acceleration of the third person.
Answer:
1- Period is the time for which one full rotation is completed. Regardless of their positions on the platform, periods of all three are the same. T2 = T.
2- Similarly, T3 = T.
3- The platform is making rotational motion. So, the relation between the angular velocity and the linear velocity is
[tex]v = \omega R[/tex]
For all the people, angular velocity is the same. Their linear velocities are different.
[tex]\omega = \frac{v_1}{R} = \frac{v_2}{3R/5}\\v_2 = \frac{3v}{5}[/tex]
4- Similarly,
[tex]\omega = \frac{v_1}{R} = \frac{v_2}{R/2}\\v_2 = \frac{v}{2}[/tex]
5- Radial acceleration in constant circular motion is
[tex]a_{rad} = \frac{v^2}{R}[/tex]
For the second person:
[tex]a_2 = \frac{v_2^2}{3R/5} = \frac{(3v/5)^2}{3R/5} = \frac{9v^2/25}{3R/5} = \frac{3v^2}{5R} = 3a/5[/tex]
6- Similarly,
[tex]a_3 = \frac{v_3^2}{R/2} = \frac{(v/2)^2}{R/2} = \frac{v^2/4}{R/2} = \frac{v^2}{2R} = a/2[/tex]
Explanation:
As a result, the period is same for every object on the rotating platform, as they all complete their revolutions at the same time. Their speed and radial acceleration is different according to their distance to the center.
The rotational kinematics relations allow to find the results for the questions about the movement of the three people on the turntable are:
1 and 2) All periods are equal, T₂ = T and T₃ = T.
3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]
4) The linear velocity of the 3rd person is: v₃ = ½ v
5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]
6) The linear acceleration of the 3rd person: a₃ = ½ a
Rotational kinematics studies the rotational motion of bodies looking for relationships between angular position, angular velocity, and angular acceleration.
In the case where the angular accleration is zero, the expression for the velocity is:
[tex]w = \frac{\Delta \theta }{\Delta t}[/tex]
Where w is the angular velocity and Δw and Δt are the variation in angle t over time.
1 and 2)
Indicates that people are on a turntable, the period is when we have a complete rotation θ = 2π rad in time, therefore the period and the angular velocity are related.
[tex]w= \frac{2\pi }{T} \\T = \frac{2\pi }{w}[/tex]
In the apparatus of parks the angular velocity is constant and we see that it does not depend on the radius, therefore the period for all the people is the same.
T = T₁ = T₂
3) They indicate that the speed of the 1 person who is in the position r=R on the plate is v, let's calculate the speed for the 2 person who is in the position r = [tex]\frac{3}{5} \ R[/tex]
Linear and angular variables are related.
v = w r
Let's substitute for the 1st person.
v = w R
For the 2nd person.
v₂= w ( [tex]\frac{3}{5} R[/tex])
We solve these two equations.
[tex]v_2 = \frac{3}{5} \ v[/tex]
4) We carry out the same calculation for the 3rd person.
v₃ = w ½ R
We solve the two equations.
v₃ = ½ v
5) Ask for radial acceleration.
The relationship between radial and angular acceleration is.
a = α R
We substitute for the 1st person.
a = α R
For the second person.
a₂ = α ( [tex]\frac{3}{5} R[/tex])
We solve the two equations
a₂ = [tex]\frac{3}{5} \ a[/tex]
6) Ask the radial acceleration of the 3rd person.
We substitute.
a₃ = α (½ R)
We solve.
a₃ = ½ a
In conclusion, using the rotational kinematics relations we can find the results for the questions about the movement of the three people on the turntable are
1 and 2) All periods are equal, T₂ = T and T₃ = T.
3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]
4) The linear velocity of the 3rd person is: v₃ = ½ v
5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]
6) The linear acceleration of the 3rd person: a₃ = ½ a
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You've always wondered about the acceleration of the elevators in the 101 story-tall Empire State Building. One day, while visiting New York, you take your bathroom scale into the elevator and stand on them. The scales read 140 lb as the door closes. The reading varies between 120 lb and 170 lb as the elevator travels 101 floors.i)What is the maximum acceleration upward?ii)What is the maximum magnitude of the acceleration downward?
To develop this problem we will proceed to convert all units previously given to the international system for which we have to:
[tex]140 lb = 63.5 kg \rightarrow 63.5kg (9.8m/s) =622.3 N[/tex]
[tex]120 lb = 54.4 kg \rightarrow 54.4kg (9.8m/s)= 533 N[/tex]
[tex]170 lb = 77.1 kg \rightarrow 77.1 kg (9.8m/s) =756 N[/tex]
PART A ) From the given values the minimum acceleration will be given for 120Lb and maximum acceleration when 170Lb is reached therefore:
[tex]F = 756 - 622.3[/tex]
[tex]F = 133.7N[/tex]
Through the Newtonian relationship of the Force we have to:
[tex]F= ma[/tex]
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{133.7}{63.5}[/tex]
[tex]a = 2.1m/s^2[/tex]
PART B) For the maximum magnitude of the acceleration downward we have that:
[tex]F = 622.3 - 533[/tex]
[tex]F = 89.3N[/tex]
Through the Newtonian relationship of the Force we have to:
[tex]F= ma[/tex]
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{89.3}{63.5}[/tex]
[tex]a = 2.1m/s^2[/tex]
[tex]a = 1.04 m/s^2[/tex]
Events 1 and 2 are exploding firecrackers that each emit light pulses. In the reference frame of the detector, event 1 leaves a char mark at a distance 3.40 m from the detector, and event 2 leaves a similar mark at a distance 2.10 m from the detector. If the two events are simultaneous in the reference frame of the detector and occur at instant t=0, at what instant of time will each light pulse be detected?
Answer:
[tex]1.13333\times 10^{-8}\ s[/tex] and [tex]0.7\times 10^{-8}\ s[/tex]
Explanation:
Light which travels from the crackers reaches the detector at [tex]c=3\times 10^{8}\ m/s[/tex]
[tex]\Delta x_1[/tex] = Distance at which event 1 leaves a char mark = 3.4 m
[tex]\Delta x_2[/tex] = Distance at which event 2 leaves a char mark = 2.1 m
The speed of light in a medium is a universal constant
[tex]c=\dfrac{\Delta x_1}{\Delta t_1}\\\Rightarrow \Delta t_1=\dfrac{\Delta x_1}{c}\\\Rightarrow \Delta t_1=\dfrac{3.4}{3\times 10^8}\\\Rightarrow \Delta t_1=1.13333\times 10^{-8}\ s[/tex]
[tex]c=\dfrac{\Delta x_2}{\Delta t_2}\\\Rightarrow \Delta t_2=\dfrac{\Delta x_2}{c}\\\Rightarrow \Delta t_2=\dfrac{2.1}{3\times 10^8}\\\Rightarrow \Delta t_2=0.7\times 10^{-8}\ s[/tex]
The pulse will be detected at [tex]1.13333\times 10^{-8}\ s[/tex] and [tex]0.7\times 10^{-8}\ s[/tex]
In the reference frame of the detector, both light pulses will be detected approximately 1.13 x 10^-8 seconds after they occur.
Explanation:In the reference frame of the detector, event 1 leaves a mark at a distance of 3.40 m and event 2 leaves a mark at a distance of 2.10 m. Since the events are simultaneous in the reference frame of the detector, the time it takes for each light pulse to be detected will be the same for both events. To calculate the time, we can use the speed of light as the distance traveled divided by the speed of light. Therefore, the time it takes for each light pulse to be detected is t = distance/speed of light.
For event 1, the distance is 3.40 m and for event 2, the distance is 2.10 m. The speed of light is approximately 3.00 x 10^8 m/s. Plugging in these values, we get:
t = (3.40 m) / (3.00 x 10^8 m/s) = 1.13 x 10^-8 s
Therefore, each light pulse will be detected approximately 1.13 x 10^-8 seconds after the events occur.
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true or false conceptual physics 2 questions from chapter 25.
1. [ ] The electrostatic potential difference between point A and B, ∆V_AB, can be calculated using a path integral, and ∆V_AB does not depend on the integral path used.
2.[ ] A common reference point for electrostatic potential is V = 0 at [infinity].
3. [ ] The magnitude of work required to bring a charge q from point A to point B is |qV_BA|.
4. [ ] eV is a unit for potential.
5. [ ] When a charged particle is accelerated from rest in an electric field, the amount of kinetic energy increased equals the amount of potential energy decreased, i.e., |1/2 mv^2| = |qΔV|
6. [ ] An electric field line points to the direction of higher electric potential.
7. [ ] An equipotential surface is always parallel to electric field lines.
8. [ ] The electrostatic potential generated by multiple charges, is the vector addition of the potentials generated by individual charges.
9. [ ] Inside a conductor, the electric field and potential are both zero.
10. [ ] The electric fields created by a point charge, a conducting sphere, and an insulator sphere are identical, if they all have the same amount of charge and the field is measured outside the spheres.
Answer:
Explanation:
1) TRUE; potential difference can be calculated using path integral. Since the electric field is a conservative, the potential difference can be calculated using any path.
2) TRUE; since potential due to a charge is inversely dependent on distance, at infinity the potential will be almost zero.
3) TRUE, W = q.VBA.
4) FALSE; eV is a unit for work (or) energy.
5) TRUE; since the electric force is conservative force. There will be no loss in energy, the decreased potential energy will be coverted to kinetic energy.
6) FALSE; in the direction of electric field the potential decreases.
7) FALSE; equipotential surface is perpendicular to the electric field lines.
8) FALSE; electrostatic potential is scalar quantity. It depends only on the charge and distance from it.
9) FALSE; Inside a conductor the electric field is zero but the electric potential is constant at the value that is at the surface of the conductor.
10) TRUE; as long as the field is being measured outiside the body the bodies act as point charges. So electric fields due to all types of bodies charged identically will be equal.
A block with mass m =6.7 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.6 m/s.
The block oscillates on the spring without friction.1) What is the spring constant of the spring? ANSWER = (234.5)2) What is the oscillation frequency? ANSWER = (0.942)3) After t = 0.47 s what is the speed of the block?4) What is the magnitude of the maximum acceleration of the block?5)At t = 0.47 s what is the magnitude of the net force on the block?6)Where is the potential energy of the system the greatest?---At the highest point of the oscillation.---At the new equilibrium position of the oscillation.---At the lowest point of the oscillation
The spring constant of the spring is approximately 234.50 N/m.
Explanation:To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. In this case, we know that the mass of the block is 6.7 kg and the displacement of the spring is 0.28 m. Therefore, we can use the equation F = kx, where F is the force, k is the spring constant, and x is the displacement. Plugging in the given values, we get:
F = kx
F = (6.7 kg)(9.8 m/s^2)
F = 65.66 N
65.66 N = k(0.28 m)
k = (65.66 N)/(0.28 m)
k ≈ 234.50 N/m
Therefore, the spring constant of the spring is approximately 234.50 N/m.
The unit of capacitance is the farad (F).
Which of the following combinations of units is equivalent to the farad?
A) N/C
B) V/C
C) C/V
D) J/C
E) N/s
Answer:
C/V
Explanation:
Mathematically it is known that one farad is defined as the capacitance across which, when charged with one coulomb, there is a potential difference of one volt. In other words, the Farad is Coulomb/Volt or C/V
Some amount of ideal gas with internal energy U and initial temperature 1000C was compressed to half of volume meanwhile absolute pressure inside of a container increased twice. We can say that internal energy of this gas after compression in terms of U is (20.2, 20.1, 19.4, 19.5) Group of answer choices
Answer:
U. With no variation.
Explanation:
Note- since temperature remains constant when pressure becomes twice and volume becomes half, and internal energy of ideal gas is function of only temperature so it remains constant. The internal energy is independent of the variables stated in the exercise.