The expansion rate of universe is measured in terms of hubble's constant, the expansion rate of universe has increased and at some point in future, it will stop expanding and start to shrink.
There was a lot of uncertainty about the universe's expansion rate before the Hubble telescope was launched. This value is required to calculate the universe's age, estimate its evolution over billions of years, and comprehend the forces that drive it. Astronomers were initially ecstatic when the expansion estimate was reduced to 10% accuracy. They are now approaching one percent accuracy after much perseverance and precise observations.
The cosmic clock can be rewound and the age of the universe calculated by precisely determining the expansion rate, known as the Hubble constant. However, Edwin Hubble's estimates of the universe's expansion implied that it was younger than the Earth and Sun. As a result, Hubble concluded that the redshift phenomenon was a previously unknown property of space rather than a measurement of true space velocity. Redshift was later discovered to be a result of the expansion of space itself, as predicted by Einstein's theory of special relativity.
Thus, by the current data it is noted that the universe is expanding quickly than though before, and if this continues , in the future at some point the universe will stop expanding and it will start shrinking.
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a 90.0 g mass is attached to the bottom of a vertical spring and set vibrating. assume that the maximum speed of the mass is 25.0 cm/s and the period is 0.450 s.
a. Spring constant which has a maximum mass velocity of 25.0 cm/s and a period of 0.450 s = 17.49 N/m.
b. The amplitude of the motion of the spring = 0.0179 m.
c. Frequency of oscillation of the spring = 2.22 Hz.
The angular velocityBriefly, angular velocity is the angular speed accompanied by the direction. The unit of angular speed is rad/s or rad/minute or rad/hour.
Some of the equations that are often used are:
ω = [tex]\sqrt{\frac{k}{m}}[/tex]
ω = 2π/T
ω = angular velocity (rad/second)
k = the spring constant
m = mass of the object
f = frequency (rev/second)
T = period (second)
The question is incomplete, it should be:
Find the
a. constant of the spring?
b. amplitude of the motion?
c. frequency of oscillation?
We have,
Mass of the object = 90.0 g
The maximum speed = 25.0 m/s
The period = 0.450 s
Determine the angular velocity first,
ω = 2π/T
= 2π/0.450
= 4.44 π rad/s
So,
a. Spring constant:
ω = [tex]\sqrt{\frac{k}{m}}[/tex]
k = ω²m
= (4.44 π rad/s)² (0.09)
= 17.49 N/m.
b. The amplitude:
vm = (xm) (ω)
So, xm = vm/ω
= 0.25/4.44 π
= 0.0179 m
c. The frequency:
f = 1/T
= 1/0.450
= 2.22 Hz.
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explain the effect of the added mass on the pressure and the relationship between the fluid levels on the two sides.
Pressure and mass flow relationship It follows that increasing the pressure supplied to the intake section will increase the pressure differential between the inlet and outlet valves.
More people will attempt to hurry past the segment as a result. As a result, we may state that pressure and mass flow rate are directly related (gradient). Mass flow rate is the amount of a substance that moves per unit of time in physics and engineering. In SI units, it is measured in kilograms per second, and in pounds or slugs per second in US customary units. Although occasionally (Greek lowercase mu) is used, the typical symbol is (, pronounced "m-dot"). The force that is delivered perpendicular to an object's surface and expressed as a symbol (p or P) is known as pressure.
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Based on current science which of the following is the most reasonable range for possible values of the number of habitable planets, NHP, in our galaxy?
More than 1 billion is a credible estimate of the number of habitable planets, or NHPs, in our galaxy based on current scientific knowledge.
Out of the eight planets, three (Venus, Earth, and Mars) may be able to support life. One in five planets outside of our Solar System are thought to be capable of supporting life, according to recent findings of extrasolar planets: Planetary lifespan on average. Scientists estimate that the closest such planet may be 12 light-years away. There have been discovered 59 possible inhabited exoplanets as of June 2021. On K2-18b, an estimated 110 light years away super-Earth that may be habitable, water vapor has been found. The exoplanet was previously discovered by NASA's Kepler satellite in 2015, but data processing has uncovered new information that has never been seen on a super-Earth.
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An inductor is connected to the terminals of a battery that has an emf of 10.0 V and negligible internal resistance. The current is 4.86 rmmA at 0.940 ms after the connection is completed. After a long time the current is 6.45 mA.
A)
What is the resistance R of the inductor?
B)
What is the inductance
The inductor's resistance, R, is 1550.38.
1.05 H is the inductance L.
What distinguishes an inductor from a capacitor?A capacitance opposes a variation in voltage, while an inductance opposes a changes in current. This is one of the key distinctions between the two components. Additionally, the capacitor and inductor both store energies with in type of either a magnetic charge and a magnetic force, respectively.
Briefing-
Given:
Emf, E=15.5V
Current, i = 4.86mA
Time, t = 0.940 ms
Maximum current, imax Calculation: 6.45mA
a) The resistance is given by,
Ꭱ= V/Imax = 10/6.45×10⁻³= 1550.38Ω
b) The inductance is given by,
L= -Rt/(1-i/imax)= 1550.38×0.940×10⁻³/In(1-4.86/6.45)
The inductance L is 1.05 H.
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A Cassette Player Is Said To Have A Signal-To-Noise Ratio Of 62 DB , Whereas For A CD Player It Is 98 DB. What is the ratio of intensities of the signal and the background noise for each device?
The ratio of intensities of the signal and the background noise for each device is 7943.2 and 316.2 respectively.
The signal-to-noise ratio (SNR) is a measure of the strength of the desired signal relative to the background noise. It is usually expressed in decibels (dB).
A higher SNR indicates a stronger signal relative to the background noise, and therefore a higher quality of sound. To find the ratio of intensities of the signal and the background noise for each device, you can use the formula:
Intensity ratio = 10^(SNR/10)
For the cassette player with an SNR of 62 dB, the intensity ratio would be:
Intensity ratio = 10^(62/10) = 316.2
This means that the intensity of the signal is 316.2 times stronger than the intensity of the background noise.
For the CD player with an SNR of 98 dB, the intensity ratio would be:
Intensity ratio = 10^(98/10) = 7943.2
This means that the intensity of the signal is 7943.2 times stronger than the intensity of the background noise. Therefore, the CD player has a much higher intensity ratio and a higher quality of sound compared to the cassette player.
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mcdougal products is considering the purchase of new equipment to place in its factory. the equipment would cost $365,000, have a ten-year useful life and a salvage value at the end of its useful life of $65,000. the company estimates that annual revenues and expenses associated with the equipment would be as follows: the payback period of the new equipment is closest to:
Form the given annual revenue and expenses the payback period of the new equipment is closest to 4 years .
What is Payback Period ?
Payback Period is defined as amount of time the company will have to wait before it recovers its investment.
the net annual cash flow can be calculated using the formula .
Net annual cash flow = (Net income) + (Non-cash items (i.e. depreciation))
From the table we get , net income = $60000
depreciation = $30000
So ,
Annual Cash Flow = 60000 + 30000 = $90000
We know that , Payback period is = (Initial investment)/(Annual cash flow)
= 365000/90000
= 4.055555 years
≈ 4 years .
Therefore , The payback period of the new equipment is closest to 4 years .
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What is the tension in the string in (Figure 1) ? The volume of plastic ball is 75 cm3 and the density is 840 kg/m3. Express your answer to two sig fig and include appropriate units.
The tension in the string is = 0.118
Given the values in the question,
The volume of the plastic ball = 75 [tex]cm^{3}[/tex]
Density ( ρ ) = 840 kg / [tex]m^{3}[/tex]
Let the tension on the ball = T
Since the density is given in centimeters so convert volume also in meters,
⇒ 75 [tex]cm^{3}[/tex] = 0.000075 [tex]m^{3}[/tex]
The ball is following the principle of buoyant force, so the ball is stable in the water the force that the ball is exerting into the water is equal to the force the water is exerting on the ball.
The forces could be represented as -
∑ [tex]F_{y}[/tex] = [tex]F_{b}[/tex] - T - [tex]W_{b}[/tex] = 0
T = [tex]F_{b}[/tex] - [tex]W_{b}[/tex]
T = ρ x g x [tex]V_{w}[/tex] - ρ x g x [tex]V_{b}[/tex]
T = ( 1000 x 9.8 x 0.000075 ) - ( 840 x 9.8 x 0.000075 )
T = 0.735 - 0.617
T = 0.118
Therefore, The tension in the string is = 0.118
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answer to this required:
Answer:
Explanation:
1) Conduction - 100% (C)
2) Convection can take place in vacuum is incorrect - 100% (A)
3) Remains same - 100% (D)
A paperweight is placed on a tray and the combination is set into horizontal oscillatory motion with a frequency of f= 1.22 Hz. If the paperweight begins slipping when the amplitude of the motion is 5.00 x 102 m, determine the coefficient of static friction between the paperweight and the tray.
When a paperweight and tray are placed together and put into a horizontal oscillating motion, the coefficient of static friction between them is 0.2
Given frequency of motion (f) = 1.22 Hz.
the amplitude of the motion (A) = 5.00 x 10^-2 m
The coefficients of the normal force applied by the surface are determined as friction force.
Force of friction (Fr) = μ x normal force (F)
coefficient(μ) = F/Fr
we know that Fr = kx = mω^
2A anf F = mg
mω^
2A = μmg
μ = ω^
2A/g and ω = 2πf
μ = (2x3.14x1.22)^2 x 5x10^-2/9.8 = 0.2
Hence the coefficient = 0.2
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A viewing window 30 cm in diameter is installed 3 m below the surface of an aquarium tank filled with sea water. The force the window must withstand it approximately.
a. 22 N
b. 218 N
c. 2140 N
d. 8562 N
A viewing window 30 cm in diameter is erected 3 m below the surface of an aquarium tank filled with sea water. The force the glass must endure is roughly 8562 N.
Given,
Diameter = 30 cm
= 0.3 m
radius = 0.3/2
= 0.15
Total pressure = 1atm + hρg
= 10⁵ N/m + 1000 × 9.8 × 3
= 101325 + 29400
= 130725 Pa
therefore Force = 130725 × π × 0.15²
= 9235.721 N that is approximately 8562 N
An object experiences a push or pull as a result of interacting with another item. Each item is subject to a force whenever two things interact. The two items no longer feel the force after the interaction ends. Only by interaction do forces come into being.
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suppose that a plate is immersed vertically in a fluid with den- sity and the width of the plate is at a depth of meters beneath the surface of the fluid. if the top of the plate is at depth and the bottom is at depth , show that the hydrostatic force on one side of the plate is w (x) at a depth of meters
beneath the surface of the fluid. If the top of the plate is at
depth and the bottom is at depth , show that the hydrostatic
force on one side of the plate is
The force that results when a liquid under pressure acts on surfaces that are submerged is known as a hydrostatic force. Fundamental concepts in fluid mechanics include the calculation of the hydrostatic force and the location of the centre of pressure.
suppose that a plate is immersed vertically in a fluid with den- sity and the width of the plate is at a depth of meters beneath the surface of the fluid. if the top of the plate is at depth and the bottom is at depth ,[tex]$$F=\int_a^v \rho g x w(x) d x,$$[/tex]
where [tex]$g$[/tex] stands for the acceleration of gravity.
Solution. Pick a point [tex]$x \in[a, b]$[/tex] and consider a horizontal section of the plate through [tex]$x$.[/tex]
Suppose this section defines a thin rectangle of height [tex]$\Delta x$[/tex] and width [tex]$w(x)$[/tex]whose area is [tex]$w(x) \Delta x$[/tex]. The pressure on this thin rectangle is then [tex]$\rho g x$[/tex], and therefore the hydrostatic force is given by
[tex]$$F=P A=\rho g x w(x) \Delta x$$[/tex]
The total force is then approximated by adding over different horizontal sections:
Taking the[tex]limit Delta x 0 yields[/tex]
[tex]F=\int_a^b \rho g x w(x) d x .$$[/tex]
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The free-fall acceleration on the surface of the Moon is about one sixth that on the surface of the Earth. The radius of the Moon is about 0.250 RE (RE = Earth's radius = 6.4 106 m). Find the ratio of their average densities, rhoMoon/rhoEarth.
On the surface of the Moon, the free-fall acceleration is roughly one-sixth that of the Earth. Their average densities are 0.667 apart.
Given the radius of the Moon is (Rm) = 0.250 RE
The radius of the earth (RE) = 6.4 x 10^6 m
Let freefall acceleration of the moon = gm
Let freefall acceleration of the earth = ge
Given gm = ge/6
We know that g = Gm/r^2 where G is the gravitational constant
Let mass of moon = m1 and mass of earth = m2
Gm1/Rm^2 = Gm2/6xRE^2
mass = densityxvolume
let density of moon = d1 and density of earth = d2
d1 x (4/3Rm^3)/Rm^2 = d2x (4/3RE^3)/6xRE^2
d1xRm = d2xRE/6
d1/d2 = RE/6X0.250RE
d1/d2 = 0.667
Hence the ratio of densities of moon and earth is 0.667
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If the tank is a reusable tank, there are no check valves and system refrigerant can enter the tank as long as the system pressure is higher than the pressure in the tank. True or False
It is true that if the tank is a reusable tank, there are no check valves and system refrigerant can enter the tank as long as the system pressure is higher than the pressure in the tank.
A fundamental valve type dispersed generously all through a modern refrigeration framework is a manual shutoff valve. In the totally vacant position, this valve ought to permit a free progression of refrigerant and when shut totally block the stream. The standard capability of the shutoff valve is to confine a part or a segment of the framework.
Air conditioning Valves are such parts without which we have no control over stream in pipes .Valves are required for each medium whether it is water, gas, air or some other fluid. Valves would have been required for solids in the event that they could have capacity to stream. Like different regions, valves track down its broad use in air conditioning.
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a mass m at the end of a spring oscillates with a frequency of 0.92 hz . when an additional 800 g mass is added to m, the frequency is 0.58 hz .
The value of m is 0.534 kg
Given:f1 =0.92 Hz
f2 = 0.58 Hz
Δm = m2 - m1 = 800 g kg
Mass-spring oscillation occurs with a frequency of:
f = (1/2π)√(k/m)
k = m(2πf)²
When we have: we may conclude that k is constant for both trials.
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0. 800 g/ (0.92/0.58)^2 – 1
= 0.534 kg or 0.53kg or 534 g
Does a mass M vibrate when it is connected to a spring?A mass m linked to a spring produces an oscillation with a period of 4 s. A spring's time period lengthens by 1 s when a 2 kg additional mass is added to it.
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the maximum energy a bone can absorb without breaking is surprisingly small. for a healthy human of mass 67 kg , experimental data show that the leg bones can absorb about 200 j .
Maximum height could a person jump and land rigidly upright on both feet without breaking his legs: h = 0.30 m
Briefly:
Potential energy = m g h would be present in a jumper at height h, and as the jumper hits the ground, this potential energy will totally transform into kinetic energy. Now, the human can only absorb a maximum of 200 J of energy.
m = 67 kg
g = 9.8 m/s²
⇒ m g h = 200 J
⇒ h = 200 J / (67 kg × 9.8 m/s²) = 0.30 m
Therefore, a person can only fall from a height of 0.30 m safely and without breaking both of their legs.
Whatever produces the most energy?In metabolic processes, lipids provide the most energy. On reduction, lipids turn into fatty acids. As a result, fat has a higher energy content than both glycogen and proteins together.
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Question:-
the maximum energy a bone can absorb without breaking is surprisingly small. for a healthy human of mass 67 kg , experimental data show that the leg bones can absorb about 200 j . part a from what maximum height could a person jump and land rigidly upright on both feet without breaking his legs?
find the effective spring constant k of the two-spring system. give your answer for the effective spring constant in terms of k1 and k2 .
The spring constant k is (1/k₁ + 1/k₂ + 1/k₃)⁻¹.
The effective spring constant k of the two-spring system will be equal to (1/k₁ + 1/k₂)⁻¹, while the spring constant k′ of the three-spring system will be equal to (1/k₁ + 1/k₂ + 1/k₃)⁻¹.
Since only one force F acts, the force on spring with spring constant k₁ is F = k₁x₁ where x₁ is its extension
the force on spring with spring constant k₂ is F = k₂x₂ where x₁ is its extension
Let F = kx be the force on the equivalent spring with spring constant k
We must consider the variable x as the extension of each spring.
The total extension , x = x₁ + x₂
x = F/k = F/k₁ + F/k₂
1/k = 1/k₁ + 1/k₂
k = (1/k₁ + 1/k₂)⁻¹
B
The force on spring with spring constant k₃ is F = k₃x₃ where x₃ is its extension
Let F = kx be the force on the equivalent spring with spring constant k and extension x.
The total extension , x = x₁ + x₂ + x₃
x = F/k = F/k₁ + F/k₂ + F/k₃
1/k = 1/k₁ + 1/k₂ + 1/k₃
k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹
Therefore, the value of spring constant k is (1/k₁ + 1/k₂ + 1/k₃)⁻¹.
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[NOTE: THIS IS AN INCOMPLETE QUESTION. THE COMPLETE QUESTION IS: Consider two massless springs connected in series. Spring 1 has a spring constant k1, and spring 2 has a spring constant k2. A constant force of magnitude F is being applied to the right. When the two springs are connected in this way, they form a system equivalent to a single spring of spring constant k.
(A) What is the effective spring constant k of the two-spring system? Express the effective spring constant in terms of k1 and k2.
(B) Now consider three springs set up in series as shown. (Figure 2) The spring constants are k1, k2, and k3, and the force acting to the right again has magnitude F. Find the spring constant k′ of the three-spring system. Express your answer in terms of k1, k2, and k3.]
Glycerin is poured into an open U-shaped tube until the height in both sides is 22 cm. Ethyl Alcohol is then poured into one arm until the height of the alcohol column is 18 cm. The two liquids do not mix. What is the difference in height between the top surface of glycerin and the top surface of alcohol?
The distinction in top among the pinnacle floor of glycerin and the pinnacle floor of ethyl alcohol is 0.0432 meter or 4.32 centimeter the difference in height between the top surface of glycerin and the top surface of alcohol is 4.32 centimeter.
Given the subsequent data:
Height of ethyl alcohol = 25cm to n = 0.25I mHydrostatic top = 20 cm to m = 0.2 m.Scientific data:Density of ethyl alcohol = 790 kg/m³Density of glycerin = 1260 kg / (m ^ 3)To calculate the distinction in top among the pinnacle floor of glycerin and the pinnacle floor of ethyl alcohol:The system for hydrostatic strain.Where:p is the density.g is the acceleration because of gravity.h is the top.At steady temperature, the strain on the pinnacle floor of glycerin withinside the open U-formed tube is identical to the strain on the pinnacle floor of ethyl alcohol:rho_*h_ = rho_*h_Substituting the given parameters into the system, we have;1260h_ = 790 * 0.251260h_ = 197.5h_ = 197.5/1260Height of glycerin = 0.1568 meters.Now, we are able to locate the distinction in top:Height of glycerin = 0.1568 meters.Now, we are able to locate the distinction in top:Difference = 0.2 - 0.1568Difference = 0.0432 meter or 4.32 centimeter.Read greater on hydrostatic strain here:
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you need to push the couch 4 m to the other side of the room. after the initial push to get it going you push on the 90 kg couch with a steady horizontal force of 600 n. the coefficient of kinetic friction between the couch and the floor is 0.6.
Force needed to push the couch 4 m to the other side of the room is 70.8 N.
Kinetic friction is described as a force that acts among transferring surfaces. A body transferring at the surface reports a pressure within the contrary direction of its movement. The value of the pressure will depend on the coefficient of kinetic friction among the two materials.
Calculation:-
mass = 90 Kg
force = 600 N
coefficient of the friction is 0.6
F = μN
= 0.6 × 90 × 9.8
= 529.2 N
force needed = 600 - 529.2
= 70.8 N
F = ma
a = f/m
= 70.8/90
= 0.78 m/s²
Static friction is what maintains the box from transferring without being driven, and it need to be conquer with a sufficient opposing pressure before the field will pass. Kinetic friction (additionally known as dynamic friction) is the force that resists the relative motion of the surfaces as soon as they may be in motion.
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The capacitors in are initially uncharged and are connected as in the diagram with switch S open. The applied potential difference is Vab=+360V.a. What is the potential difference Vcd
b What is the potential difference across each capacitor after switch S is closed?
c. How much charge will flow through the switch after it is closed?
There is a flow of electrons from charged to uncharged whenever a charged capacitor is connected to an uncharged capacitor, and this flow continues until a steady state is reached. As a result, there is no further charge transfer and both capacitors are in equilibrium.
What does a capacitor being discharged mean?When we say a capacitor is uncharged, we imply that there are an equal amount of positively and negatively charged ions and electrons on each of its plates.
The capacitor's starting voltage is zero.
The capacitor's charge Q and voltage across it are both 0 before the switch is closed.
The charge has not had a chance to accumulate before the switch is closed.
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a fixed quantity of gas at a constant temperature exhibits a pressure of 737 torr and occupies a volume of 20.5 l. use boyle's law to calculate the volume the gas will occupy if the pressure is increased to 1.80 atm.
Considering the Boyle's law, if the pressure is increased to 1.80 atm, the gas will ocuppy a volume of 11.04 L.
Boyle's lawBoyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant.
Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant: if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.
Mathematically, Boyle's law states that if the amount of gas and the temperature remain constant, the product of the pressure and the volume always has the same value:
P× V= k
where:
P is the volume.V is the volume.k is the constant.Considering the initial state 1 and the final state 2, it is fulfilled:
P₁× V₁= P₂× V₂
New volumeIn this case, you know:
P₁= 737 torrV₁= 20.5 LP₂= 1.80 atm= 1368 torr (being 1 atm= 760 torr)V₂= ?Replacing in Boyle's law:
737 torr× 20.5 L= 1368 torr× V₂
Solving:
(737 torr× 20.5 L)÷ 1368 torr= V₂
11.04 L= V₂
Finally, the new volume is 11.04 L.
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during a knee extension exercise there are two forces that are producing torque at the knee joint. one is the quadriceps muscle and the other is the weight at the ankle joint. the quad is capable of producing 500 newtons of force, while the weight at the ankle is 10 kg. the distance from the knee joint to the quad force is 0.05 meters and the distance from the knee joint to the weight is 0.38 meters (these are not the moment arms). the knee is flexed to 125 degrees. draw and calculate the moment arms for each force on the picture below and decide if the joint flexes or extends.
The moment arm for each force is 0.218 m and 0.0287 m , and the joint is a flex .
In the question ,
it is given that ,
quad force is ([tex]F_{q}[/tex]) = 500 N
weights at the ankle is wₐ ,
So , wₐ = (10 Kg)(9.8 m/s²)
= 98 N
given
the distance from knee joint to weight(a) = 0.38 m
the distance from knee joint to quad force (q) = 0.05 m
From the diagram , we can see that ,
the moment arm for the weight is ([tex]a_{t}[/tex]) = a*cos55° = (0.38)*cos55° = 0.218 m
and
the moment arm for equal force([tex]q_{t}[/tex]) = q*cos55° = (0.05)*cos55° = 0.0287 m
Now , torque (net) = ([tex]F_{q} \times q_{t}[/tex]) - ([tex]w_{a} \times a_{t}[/tex])
After substituting the values , we get
Net Torque = 500×0.0287 - 98×0.218
= 14.35 - 21.364
we can see that as the net torque is negative , its weight will pull down the leg ,
So , the joint will flex not extend .
Therefore , The joint will Flex .
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a double pendulum consists of two simple pendula, with one pendulum suspended from the bob of the other. if the two pendula have equal lengths, $l$, and have bobs of equal mass, $m$, and if both pendula are confined to move in the same vertical plane, find lagrange's equations of motion for the system. use $\theta $ and $\phi$--the angles the upper and lower pendulums make with the downward vertical (respectively)--as the generalized coordinates. do not assume small angles.
The Lagrange's equations of motion for the system is d(ml³Ф2 + ml²Ф1 cos (Ф1-Ф2)) − (−ml³Ф1Ф2 sin (Ф1-Ф2)-mgl sinФ2 )/dt = 0.
Newton's method of developing the equations of motion requires element decomposition. If the forces on the connections are not the primary concern, it is more advantageous to consider the energies in the system to derive the equations of motion.
A double pendulum exhibits simple harmonic motion when the non-equilibrium displacement is small. However, when large displacements are imposed, the behavior of nonlinear systems becomes dramatically chaotic indicating that deterministic systems are not always predictable. There are several possible variations of the double pendulum.
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The electric potential V in the space between two flat parallel plates 1 and 2 is given (in volts) by V = 1270x2, where x (in meters) is the perpendicular distance from plate 1. What is the magnitude and direction of the electric field at x = 3.5 cm? (Take the direction perpendicular to and away from plate 1 to be positive.)
The magnitude and direction of the electric field at x = 3.5 cm is 85.75V/m and is towards the plate 1.
Given the electric potential in the space between two flat parallel plates 1 and 2 is (V) = 1270x^2
the perpendicular distance from plate 1 is = xm = 3.5cm
We know that changing electric field is given as :
E(x) = -dV/dx
By differentiating we get :
E(x) = -(d(1270x^2)/dx)
E(x) = - (2*1270*x)
E(x) = -2540x
Here given x = 3.5cm = 0.035m
Then E(x) = -2450*0.035 = -85.75V/m
Hence the magnitude and direction of the electric field is 85.75V/m and is towards the plate 1
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A potentiometer wire is 10 m long and has a resistance of `18Omega`. It is connected to a battery of emf 5 V and internal resistance `2Omega`. Calculate the potential gradient along the wire.
The potential gradient along the wire is 0.208Vm⁻¹
We are given that,
Resistance of wire = R = 20Ω
Internal resistance = r = 18Ω
length of wire = L = 10m
E.M.F = ∈ =5V
Thus to find the value of potential gradient of the wire, we can use the equation,
i = (∈)/(R+r)
i = 5v/ (20Ω +18Ω)
i = 0.104 A
Vr = iR =(0.104 A)(20Ω) =
Vr = 2.08 volt
Thus the potential gradient of the wire is given as,
potential gradient = V/L = (2.08volt)/(10m)
potential gradient = 0.208Vm⁻¹
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a small sphere radious 0.05 surrounds the point (2, 3, -1). the flux of a vector g into this sphere 0.00004pi. estimate div g at the point (2, 3, -1).
The flux of a vector g into this sphere 0.00004pi. estimate div g at the point (2, 3, -1) is the divergence at the point is 0.03.
Calculation :
Divergence of a vector field G at a point p is defined as:
div(G)p=Fluxp/Volumep
A point p=(3,3,−1) has a radius of 0.1 units and the flux entering p is 0.00004π
.Volume of point p = (4/3)π 0.1³=0.0043π
Therefore, the divergence at the point is given by:
div(G)p=−0.00004π/(0.004/3)π=−0.03
divergence, in mathematics, the differential operator applied to a vector-valued function in three dimensions. The result is a function representing the rate of change. The divergence of vector v is given by . where v1, v2, and v3 are the vector components of v, typically the fluid flow velocity field.
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Throughout the reflection, make sure
you have a copy of the Student Guide and your data tables.
In this experiment, the
was intentionally manipulated. This was the independent variable
The dependent variable measured was the
Communications satellites generally occupy a special orbit in which they go completely around the Earth once a day; since the Earth also turns once a day, the satellite is therefore always over the same point on the ground. Your satellite dish can then always be pointed at the same location in the sky.
A typical electromagnetic signal from a communications satellite might have an intensity of 0.750 pW/m2 at the surface of the Earth. Your satellite dish collects the energy of the electromagnetic signal from the satellite; then your dish passes that energy to its central receiver, where the signal can be processed by a circuit. The area of a satellite dish can be calculated from pi (3.1416) times the radius squared. If your satellite dish has a radius of 0.38 m, and if it is receiving the signal described above, how much electromagnetic energy is reflected to the central receiver in 3.5 minutes? Give your answer in picojoules.
In 3.5 minutes, the electromagnetic energy of the signal stated above, 71.45 pJ, is reflected to the main receiver.
Intensity = 0.75 x 10⁻¹² W/m²
Intensity = Power / Area
here, Area = A = π r = π (0.38)² = 0.4536 m²
so, Power = 0.75 x 10⁻¹² x 0.4536 = 3.402 x 10⁻¹³ W
Energy = Power x time
here, time = 3.5 minutes = 3.5 x 60 seconds
therefore, energy reflected = 3.402 x 10⁻¹³ x 3.5 x 60 = 71.45 x 10⁻¹² J = 71.45 pJ
this is the energy reflected the central receiver in 3.5 minutes.
Radiation with electromagnetic properties is electromagnetic energy. The energy connected to electromagnetic waves is commonly referred to by this phrase. Energy is what causes these waves to move through any medium. Photons, a unit of light energy that electromagnetic radiation travels in, are free of charge and mass. One of the most significant elements in the cosmos is electromagnetic energy. The force behind the electromagnetic waves is it.
When a force outside of the electrical charge accelerates it, electromagnetic energy is released. As a result of the acceleration, a wave of alternating magnetic and electric fields is created. This wave separates from the charge and flows independently through the medium.
The discrete energy packets known as photons make up electromagnetic waves.
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for lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of italian ham. the slices of ham are weighed on a plate of mass 0.400 kg placed atop a vertical spring of negligible mass and force constant of 200 n/m . the slices of ham are dropped on the plate all at the same time from a height of 0.250 m . they make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (shm). you may assume that the collision time is extremely small. Express your answer numerically in meters and take free-fall acceleration to be g= 9.80m/s^2
What is the period of oscillation T of the scale?
Express your answer numerically in seconds.
The period of oscillation T of the scale is T = 0.37 sec .
In the question ,
it is given that ,
weight of the Italian ham (m₁) = 0.300 Kg = 0.3 Kg
weight of the slices of ham are (m₂) = 0.400 kg = 0.4 Kg
the force constant of the vertical spring is(k) = 200 N/m .
Let oscillation period be = T ,
the plates make a totally inelastic collision ,
We know that , In inelastic collision, momentum is conserved but the kinetic energy is not conserved.
So , when the slice of ham land on the plate, then the kinetic energy of system equal to maximum potential energy of spring.
that means , T = 2π√(m₁ + m₂)/k
Substituting the values ,
we get ,
T = 2π√(0.3 + 0.4)/200
Simplifying further ,
we get ,
T = 0.37 seconds .
Therefore , If the collision is inelastic then the Period of oscillation is 0.37 s .
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11. in which direction are nearly all galaxies moving? 12. what is hubble's law? 13. match the terms with their definitions
Hubble's law is a statement of a direct correlation between the distance to a galaxy and its recessional velocity as determined by the red shift. It states that all galaxies are moving away from the Earth and from each other.
Edwin Hubble discovered that most of the galaxies are moving away from the Earth and away from each other. He also discovered that there is a relationship between the distance to a galaxy and its speed. He stated that the farther away a galaxy is, the faster it is moving away from us. Hence, Hubble’s law, also known as the Hubble–Lemaître law, refers to the observation in physical cosmology that galaxies are moving away from Earth at speeds proportional to their distance.
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5. a particular satellite with a mass of 200kg is put into orbit around ganymede (the largest moon of jupiter) at a distance 300 km from the surface. what is the gravitational force of attraction between the satellite and the moon? (ganymede has a mass of 1.48x1023 kg and a radius of 2631 km.) (4 pts)
The gravitational force of attraction between the satellite and the moon is 1.44 x 10^22 N.
To calculate the gravitational force of attraction between the satellite and the Ganymede moon, you can use the formula:
F = (G * m1 * m2) / r^2where F is the gravitational force, G is the gravitational constant (which is approximately 6.67 x 10^-11 N*m^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between them.
Plugging in the values provided, we get:
F = (6.67 x 10^-11 N*m^2/kg^2) * (200 kg) * (1.48 x 10^23 kg) / (300 km)^2
= 1.44 x 10^22 N
This is the gravitational force of attraction between the satellite and the Ganymede. It is a very strong force, due to the large mass of the moon and the relatively close distance between the satellite and the moon.
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