A stick of length 2L and negligible mass has a point mass m affixed to each end. The stick is arranged so that it pivots in a horizontal plane about a friction less vertical axis through its center, thus calculated values are:
(A) Torque (τ) = F × L cosθ
(B) α = - Kθ/2m
(C) d²θ/dt² = -Kθ/2m
(D) θ = cos ⁻¹ ( kθ t/2Lm)
What is torque?The term "torque" refers to the twisting force that an engine exerts as it works, and it expresses the rotating force. Everyday activities like turning a doorknob, opening a drink bottle, using a tool, or peddling a bicycle all involve torque.
(A)
General equation of torque (τ) will be product of force and perpendicular distance.
τ = F × (perpendicular distance)
In this case, restoring torque. Hence, force F we taken as restoring force, and the perpendicular distance x-axis is component of L.
or, τ = F × [tex]L_{x-axis}[/tex]
or, τ = F × L cosθ
(B)
Angular acceleration general equation is:
α = Δω/Δt
here, Δω = change in angular velocity.
Δt = change in time
As we know, acceleration = force/mass
now, angular acceleration = angular force / mass of body
so, α = - Kθ/2m
(C)
f(t) = α × (t²/2)
or, f(t) = - Kθ/2m × (t²/2)
or, f(t) = - Kθ t² / 4m
now differentiating with respect to t:
or, dθ/dt = -Kθ t/2m
differentiating again with respect to t:
d²θ/dt² = -Kθ/2m
(D)
The expression for angular displacement:
cos θ = adjacent side / opposite side
or, cos θ = adjacent side / 2L
or, θ = cos ⁻¹ ( kθ t/2Lm)
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Light of wavelength 550 nm illuminates a double slit, and the interference pattern is observed on a screen behind the slit. The third maximum is measured to be 2.8 cm from the central maximum. The slits are then illuminated with light of wavelength 440 nm.
How far is the fourth maximum from the central maximum?
The fourth maximum is measured to be 2.97 cm from the central maximum when slits are illuminated with light of wavelength 440 nm.
The interference or bending of waves passing through an aperture around the corner of an obstacle or into the geometric shadow area of the obstacle/aperture is defined as diffraction. Secondary sources of propagating waves are diffractive objects or apertures. It is the process by which a ray or other wave system expands as it passes through a narrow aperture or edge. It is usually the result of interference between the generated waveforms.
Given,
[tex]\lambda = 550 nm=550*10^{-9}n[/tex]
[tex]y_3 = 2.8 cm = 2.8 * 10^{-2} m[/tex]
The third maximum from the central maximum can be determined by
[tex]y_3 = \frac{3D \lambda}{d}\\\\2.8*10^{-2}=\frac{3*550*10^{-9}D}{d}\\\\\frac{2.8*10^{-2}}{3*550*10^{-9}}=\frac{D}{d}\\\\0.0169*10^6=\frac{D}{d}[/tex]
Now, the fourth maximum from the central maximum can be determined by,
Now, [tex]\lamda = 440 nm =440 *10^{-91} m[/tex]
[tex]y_4 = \frac{4D \lambda }{d}\\\\y_4=4*0.0169*10^6*440*10^{-9}\\\\y_4=29.744*10^{-3}m=2.97 cm[/tex]
Thus, the fourth maximum is measured to be 2.97 cm from the central maximum
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determine whether each geologic feature is being caused by tensional, compressional, or shear stresses by analyzing the directions of the forces being applied. drag the appropriate items to their respective bins. each item may be used only once.
Tensional faults—normal faults—cause each geologic feature, whereas shear faults—strike–slip faults—cause the remaining three.
What names do rocks above and below a fault go by?The hanging wall, also known as the headwall or footwall, is the higher or overlaying block along the fault plane as rocks slide past one another during faulting. The footwall is the lower block.
What error occurs when the footwall shifts uphill while the hanging wall shifts downward?A typical fault occurs when the rocks above the fault plane, also known as the hanging wall or footwall, move downward in relation to the rocks below the fault plane, or the footwall. A reverse fault occurs when the footwall slides up relative to the hanging wall.
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a lunar eclipse occurs when group of answer choices the earth lies directly between the sun and the moon the sun lies directly between the earth and the sun the moon lies directly between the sun and the earth
A lunar eclipse occurs when the earth is directly between the sun and the moon, thus, its shadow creates a lunar eclipse. Correct answer: letter A.
A lunar eclipse occurs when the Earth's shadow is cast on the surface of the Moon. This can only happen if the Sun, Earth, and Moon are all aligned in a straight line. In this configuration, the Earth is directly between the Sun and the Moon, blocking the light from the Sun and casting a shadow onto the Moon.
What is a lunar eclipse?Is an astronomical event that occurs when the Moon passes directly behind Earth and into its shadow. During a lunar eclipse, the Moon will appear dark as it is illuminated by only the faint, red-orange light of the sun scattered by Earth's atmosphere.
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define the principal focus of a concave mirror
The principal focus of a concave mirror is a point on the axis of the mirror at which parallel rays of light are brought to focus after being reflected by the mirror. A concave mirror is a mirror with a curved surface that is shaped like the inside of a bowl, with the center of the curve facing outward. When light rays are reflected by a concave mirror, they are brought to a focus at the principal focus of the mirror. The distance from the center of the mirror to the principal focus is known as the focal length of the mirror. The principal focus of a concave mirror is an important concept in optics and is used in the design and operation of various optical instruments, such as telescopes and cameras.
Your consulting firm is asked to design an activated carbon adsorption system to treat contaminated groundwater containing 10 mg/L of benzene with a treatment goal of 99% reduction in benzene concentration before injection into groundwater. The extraction well pumps groundwater at 100 gpm. The activated carbon removal capacity is expressed as the following Freundlich equation: 9 (mg/kg) = 0.2 CO.2, where Cis concentration (mg/L) of benzene after adsorption equilibrium. (a) What is the amount of activated carbon (kg) needed per day? (b) If the packing density of activated carbon is 1.2 g/L, how many 55-gallon tank it would fill? Approximate to nearest number exceeding the amount of activated carbon required.
the daily need for activated carbon in kg is 4319.327 kg of carbon. It would fill a 55-gallon tank, which is 18.
groundwater containing 10 mg/L of benzene
treatment goal of 99%
Flow rate of ground water =100 Gpm
=378.541 Liter/min
now weight of benzene in water
=378.541 Liter/min*10 mg/lit
= 3.78541 gram/min
now per day weight of benzene
=3.78541 g/min*(24*60)min=5450.9904 gram
now frewndlich equation derives
q(mg/g)=2 c[tex]^{0.2}[/tex]
c=0.01*10=0.1 mg/L
q=2*[tex](0.1)^{0.2}[/tex] = 1.262
q=1.262 (mg/g)
it means 1.262 mg of benzene adsorps on one gram of activated carbon
a)now per day mass=5450.9904*10³ mg
1.262 mg needs 1g of carbon
1mg needs 1/1.262 g of carbon
5450.9904*10³ mg needs
((1/1.262)*5450.9904*10³ ) grams
=4319.327 Kg of carbon /day
b) packing density of activated carbon = 1.2 g/L
tank size =55 gallon
now 1.2g fills in 1 lit
1g fills in (1/1.2)litre
4319.327*10³ g fills in(1/1.2)*4319.327*10³ liter
=3600 liter volume needed
1 tank volume =55gallon
=208.2 lit
number of tanks needed =3600 liter/208.2
=17.29≅18 tanks
∴18 such tanks are needed
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A cyndrical flask of cross sectional area A is fitted with an airtight piston that is free to slide up and down .contained within the flask is an ideal gas. Initially the pressure applied by the piston is 140kpa and the base of the flask is 22cm. When additional mass is added to the piston the pressure increases to 380kpa. Assuming the system is always at the temperature of 290k find the new height of the piston
Assuming the system is always at the temperature of 290k, the new height of the piston be 16.21 cm.
What is pressure?The physical force applied to an object is referred to as pressure. Per unit area, a perpendicular force is delivered to the surface of the objects. F/A is the fundamental formula for pressure (Force per unit area). Pascals are a unit of pressure (Pa).
Absolute, atmospheric, differential, and gauge pressures are different types of pressure.
As height of the piston is proportional to volume of ideal gas. From Boyle's law of gases at constant temperature we can write:
PV = pv
⇒PH = ph
⇒ 140 kpa × 22 cm = 290 kpa × h
⇒ h = 22 × 140/190 cm = 16.21 cm.
Hence, the new height of the piston be 16.21 cm.
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Suppose that current flows in a horizontal conducting loop such that the magnetic flux produced by the current points upward.
As viewed from above, in which direction does this current flow? a. A
b. B
The current will be flowing in an anti-clockwise direction.
We consider a current carrying loop according to the question lying horizontally, say X-Y plane , the magnetic flux has a direction pointing upwards.According to Lenz law, the direction of induced emf is such that it always opposes the cause of effect.In order to find the direction of the current flowing in the loop we can use the screw / right hand thumb rule by curling our fingers in the direction of the magnetic flux and the thumb of the right hand will point towards the the direction of the current flowing in the coil.According to the situation, the direction of the current comes out to be anti - clockwise when viewed from the above of the coil . From below of the coil the direction of the current will be reversed and it will turn clockwise.To know more about magnetism visit:
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Problem 2: Students of College Physics use an experimental set-up like the one shown, except, of course, the cut-away section to aid visualization is only in the drawing. A cylinder with a d=11.43cm inner diameter and an hmax=15.56cm height is placed on a heating plate. A weighted circular disc fits snugly into the cylinder, creating a gas-tight seal; the combined mass of the disc and the weights it supports is 19.07kg.50% Part (a) The disc initially rests on the bottom of the cylinder, but argon gas is introduced through a valve until the bottom of the disc is at a height h0 which is 41.9% of hmax. How much argon gas, in moles, is in the cylinder if it has a temperature of 22.1∘C? (Assume that the gas in the cylinder approximates an ideal gas.)50% Part (b) Suppose an additional mass of 5.56kg is placed on the disc, but at the same time, the heating plate is used to heat the gas in the cylinder. If the height of the disc above the cylinder bottom is now 87.8% of hmax, then what is the change in the temperature, in degrees Celsius, of the gas?
(a) The amount of argon gas in the cylinder is approximately [tex]\rm \(4.09 \times 10^{-5}\)[/tex] moles.
(b) The change in temperature of the gas is approximately 976.7°C.
Part (a):
Given data:
Inner diameter of the cylinder (d) = 11.43 cm = 0.1143 m
Height of the cylinder ([tex]h_{max[/tex]) = 15.56 cm = 0.1556 m
Height of the disc above the bottom of the cylinder ([tex]h_0[/tex]) = 41.9% of h_max = 0.419 * 0.1556 m
Combined mass of the disc and weights (m) = 19.07 kg
Temperature of the gas (T) = 22.1°C = 22.1 + 273.15 = 295.25 K
We need to find the amount of argon gas in moles in the cylinder.
Step 1: Calculate the volume of the gas in the cylinder.
[tex]\[ V_{\text{gas}} = \pi \left(\frac{d}{2}\right)^2 \times h_0 \]\\\\\ V_{\text{gas}} = \pi \left(\frac{0.1143}{2}\right)^2 \times 0.419 \times 0.1556 \]\\\\\ V_{\text{gas}} \approx 0.00121 \, \text{m}^3 \][/tex]
Step 2: Calculate the number of moles of argon gas using the ideal gas law.
[tex]\[ PV = nRT \][/tex]
Where P is the pressure of the gas (we assume it is constant), V is the volume of the gas, n is the number of moles, R is the universal gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.
[tex]\[ n = \frac{PV}{RT} \]\\\\\ n = \frac{1 \times 0.00121}{8.314 \times 295.25} \]\\\\\ n \approx 4.09 \times 10^{-5} \, \text{moles} \][/tex]
Part (b):
Given data:
Additional mass placed on the disc ([tex]m_{extra[/tex]) = 5.56 kg
New height of the disc above the bottom of the cylinder (h_new) = 87.8% of h_max = 0.878 * 0.1556 m
We need to find the change in the temperature of the gas.
Step 1: Calculate the new volume of the gas in the cylinder.
[tex]\[ V_{\text{gas\_new}} = \pi \left(\frac{d}{2}\right)^2 \times h_{\text{new}} \]\\\\\ V_{\text{gas\_new}} = \pi \left(\frac{0.1143}{2}\right)^2 \times 0.878 \times 0.1556 \]\\\\\ V_{\text{gas\_new}} \approx 0.00178 \, \text{m}^3 \][/tex]
Step 2: Use the ideal gas law again to find the new temperature of the gas.
[tex]\[ n_{\text{new}} = \frac{P \times V_{\text{gas\_new}}}{RT_{\text{new}}} \]\\\\\\ T_{\text{new}} = \frac{P \times V_{\text{gas\_new}}}{n_{\text{new}} \times R} \]\\\\\ T_{\text{new}} = \frac{1 \times 0.00178}{4.09 \times 10^{-5} \times 8.314} \]\\\\\ T_{\text{new}} \approx 1271.95 \, \text{K} \][/tex]
Change in temperature:
[tex]\[ \Delta T = T_{\text{new}} - T \]\\\\\ \Delta T = 1271.95 - 295.25 \]\\\\\ \Delta T \approx 976.7 \, \text{K} \][/tex]
Result:
(a) The amount of argon gas in the cylinder is approximately [tex]\rm \(4.09 \times 10^{-5}\)[/tex] moles.
(b) The change in temperature of the gas is approximately 976.7°C.
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Each item below belongs either with the population of disk stars or the population of halostars of the Milky Way Galaxy. Match each item to the appropriate population.Disk stars:-high-mass stars-youngest stars-stars that all orbit in nearly the same plane-the SunHalo stars:-globular clusters-oldest stars-stars whose orbits can be inclined at any angle-stars with the smallest abundance of heavy elements
Oldest: Red Giant in globular cluster M13 AND red main-sequence star in globular cluster M13
Middle: the sun
Youngest: Hot, blue main-sequence star in disk
What is globular cluster?
Globular clusters, which include tens of thousands to millions of stars, are stable, firmly bound clusters. They are connected to all different kinds of galaxies. Globular clusters are strongly gravitationally bound and often larger than open clusters.
As a result, the Milky Way's globular clusters (which house extremely old stars) show a spherical component, whilst the open clusters, other young stars, and star-forming areas show a disk-shaped component.
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First-order analysis of a seismogram record allows seismologists to do all but which of the following?
First-order analysis of a seismogram record enables seismologists to do all except to develop a Modified Mercalli Intensity map.
First-order analysis of a seismogram record identifies the different kinds of seismic waves reaching to the earth's surface by the fault movement. This analysis records P waves.
The P wave will be the first wave to reach the surface. P waves are the fastest seismic waves. So, they will usually be the first ones that are recorded by a seismograph. The next seismic waves to record on a seismogram are the S waves.
A seismograph is the main instrument used to measure earthquakes. The seismograph produces a digital graphic recording of the ground motion caused by the reaching seismic waves on the earth surface. The digital recording is known as a seismogram.
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when a digital fluoroscopy unit's magnification button is used, the resulting magnification is due to a decrease in the size of the: input phosphor. output phosphor. photocathode. focusing lenses.
When a digital type of fluoroscopy unit's gets magnification button and used it, then the resulting magnification decrease in size due to input phosphor. Option (a) is correct.
The front phosphor (or enter phosphor) is a element of the photo intensifier in fluoroscopic structures that converts the strength from x-rays into mild photons. It consists of a fluorescent cloth including cesium iodide activated with sodium (CsI:Na) and coats the doorway floor of the photo intensifier.
Output phosphor consists of a fluorescent compound (referred to as P20) fabricated from silver-activated zinc cadmium sulphide (ZnCdS:Ag) particles. In addition, caesium iodide has a crystalline structure, and the enter phosphor may be synthetic in order that the slim needle-like crystals (about five μm in diameter) are laid down perpendicular to the screen.
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Correct Question:
when a digital fluoroscopy unit's magnification button is used, the resulting magnification is due to a decrease in the size of the:
a). input phosphor.
b). output phosphor.
c). photocathode.
d). focusing lenses.
Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 68.0 N, m1 = 12.0 kg, m2 = 18.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.100.
(a) Draw a free-body diagram for each block.
(b) Determine the tension T and the magnitude of the acceleration of the system.
The tension T is 27.2N and the magnitude of the acceleration of the system is 1.29m/s^2 when two blocks connected by a rope of negligible mass are being dragged by a horizontal force.
Given force exerted on block F = 68.0 N
Mass of first block m1 = 12.0 kg
Mass of second block m2 = 18.0 kg
The coefficient of kinetic friction is 0.100.
(a) Block 1's tension force on block 2 is equal to object 2's tension force on object 1, in terms of size. A light string's tension, which remains constant along its length, indicates how firmly the string is pulling on the objects at both ends.
(b) The second law of Newton is applied using the free-body diagrams.
On m1 we have net force acting = T-f = ma
Fnet = N1-m1g = 0 then N= m1g
the coefficient of friction gives f1 = μN1
f1 = 0.100x12x9.8 = 11.8N
On m2 we have net force acting = F1 - T- f = ma
We have from the y component N2 = m2g
then f2 = 0.100x18x9.8 = 17.6N
F−f1 −f2 =m2a+m1a
68-11.8-17.6 = a(12+18)
a = 1.29m/s^2
The acceleration of the system is 1.29m/s^2
we have tension T = m1a+f1 = 12x1.29+11.8 = 27.2N
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An elevator, suspended by a cable, is moving upward at constant speed. The correct relationship between the magnitude of force of gravity on the elevator and tension force in the cable is: O A) the magnitude of force of gravity equals the magnitude of tension force B) the magnitude of force of gravity is greater than the magnitude of tension force C) the magnitude of force of gravity is less than the magnitude of tension force D) the magnitude of force of gravity is less than or equal to the magnitude of tension force E) the magnitude of force of gravity is equal to or greater than the magnitude of tension force
An elevator is traveling steadily upward while being hung by a cable. When this happens, the tension force's strength is equal to the gravitational force's strength.
The elevator, which is hung by a cable and traveling upward at a steady pace, is principally being acted upon by two forces. One of the forces is gravitational force, which is the weight of gravity acting on the elevator. Another force is tension in the cable, which is holding the elevator in place. Acceleration is the same as the velocity's time rate of change.
Where v is the end velocity and u is the beginning velocity, an is equal to v-u/t. However, the speed here is constant. Thus, there will be no acceleration. As a result, the net force acting on this elevator will also become zero utilizing Newton's Second Law of Motion.
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A grinding wheel of radius 0.350 m rotating on a frictionless axle is brought to rest by applying a constant friction force tangential to its rim. The constant torque produced by this force is 76.0 N m. Find the magnitude of the friction force.
The magnitude of the constant friction force applied tangential to the rim of the grinding wheel is 217.14 N.
What is the constant friction?Friction force is a type of force that opposes the motion of two surfaces that are in contact.
This type of force tends to stop an object in motion from moving or sometimes it tends to stop an object at rest from moving.
The magnitude of the constant friction force applied tangential to the rim of the grinding wheel is calculated by applying the formula for torque as shown below.
τ = F r
where;
τ is the torque produce by the force of frictionr is the radius of the grinding wheelF = τ / r
F = ( 76 N m ) / ( 0.35 m )
F = 217.14 N
Thus, the magnitude of the friction force applied tangential to the rim of the grinding wheel is a function of the torque produced and the radius of the wheel.
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7. All elements are composed of
Answer:
atoms
Explanation:
atoms the smallest unit of matter which can be identified as an element
Miguel (m=72) and Fernando (89 kg) board the bumper cars at the local carnival. Miguel is moving at a velocity of 4 m/s when he rear-ends Fernando who is at rest in his path. Fernando and his 125-kg car lunge forward at 2 m/s. Determine the post-collision velocity of Miguel and his 125-kg car.
Miguel will be travelling at 1.2 m/s inside the opposite direction following the collision, according to this.
Define collision.
A collision in physics is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.
The conservation of momentum law allows for the calculation of Miguel's post-collision speed.
Mass times velocity (p=mv), a vector quantity, equals momentum.
Before the collision, Miguel and Fernando's car had a momentum of 782 kg*m/s (m1=72 kg) (v1=4 m/s) + (m2=125 kg) (v2=0 m/s).
The momentum of the system following the collision is equivalent to (m1=72kg)(v1=?m/s) + (m2=125kg)(v2=2m/s) = 782 kg*m/s.
Miguel's comment velocity is -1.2 m/s, as determined by solving for v1. Miguel will be travelling at 1.2 m/s inside the opposite direction following the collision, according to this.
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T/F what is the semantic (meaning) of the following sentence? [0.5 pt] show detailed steps to decompose sentence into cnf format [0.5 pt]
The significance or understanding of a word or sentence is referred to as semantic. The way a statement in a multi-page report is understood; the semantic significance of the sentence—is an example of semantics.
What is semantics?
Semantics is the study of the significance of language. It could as well be used on entire texts or just a single word.
For instance, despite the fact that "objective" and "final stop" have identical meanings, semantics novices analyze their somewhat different shades of importance.
Example:
1) Sam was tired [ Adj. Description]
2) My interview was yesterday [Adv. Time].
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figure 21-5b shows the concentration profile of a band of material after it has traveled through a column for 2 min and for 26 min. why is the band broader after 26 min? (b) explain why a chromatographic separation normally has an optimum flow rate that gives the best separation.
The broader band because of mobility of ions and because of peaks developed it has optimum flow rate.
There are two characteristics of the concentration profile that are important in determining the efficiency of a column and its ability to separate or separate subsequent solute zones. The first peak maximum refers to the position of maximum concentration in the peak. To obtain sufficient resolution, the maxima of two adjacent peaks must be separated from each other. Such separations depend on the nature of the solutes and the selectivity of the stationary and mobile phases.
A second characteristic that is important for efficiency and resolution is peak width. Peaks with distant maxima are still broad and may not fully resolve the solutes. For this reason, peak width is very important in chromatography.
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(a) what is the energy in joules of an x-ray photon with wavelength 1.00x10-10 m? (b) convert the energy to electron volts. part (b) answers listed below:
The energy in joules of an x-ray photon with wavelength 1.00x10-10 m is 1.99* 10^-15J.
What is energy ?
Scientists define energy as the ability to do work. Modern civilization is possible because people have learned how to change energy from one form to another and then use it to do work.
The energy to electron volts is 1.24*10^4 eV.
Wavelength [tex]$\lambda=1.00 \times 10^{-10} \mathrm{~m}$[/tex]
Speed of light [tex]$c=3.0 \times 10^8 \mathrm{~m} / \mathrm{s}$[/tex]
P1anck's constant [tex]$h=6.626 \times 10^{-34} \mathrm{Js}$[/tex]
a) The energy of an X-rays is
[tex]$$\begin{aligned}E & =\frac{h c}{\lambda} \\& =\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{\left(1.00 \times 10^{-10} \mathrm{~m}\right)} \\& =1.9878 \times 10^{-15} \mathrm{~J} \\& =1.99 \times 10^{-15} \mathrm{~J}\end{aligned}$$[/tex]
b)
Convert energy of an electron [tex]$\mathrm{J}$[/tex] to[tex]$\mathrm{eV}$[/tex],
[tex]$$\begin{aligned}E & =1.9878 \times 10^{-15} \mathrm{~J}\left(\frac{1 \mathrm{eV}}{1.60 \times 10^{-19} \mathrm{~J}}\right) \\& =1.24 \times 10^4 \mathrm{eV}\end{aligned}$$[/tex]
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basilisk lizards can run across the top of a water surface (fig. 9-52). with each step, a lizard first slaps its foot against the water and then pushes it down into the water rapidly enough to form an air cavity around the top of the foot. to avoid having to pull the foot back up against water drag in order to complete the step, the lizard withdraws the foot before the water can flow into the air cavity. if the lizard is not to sink, the average upward impulse on the lizard during this full action of slap, downward push, and withdrawal must match the downward impulse due to the gravitational force. suppose the mass of a basilisk lizard is 90.0 g, the mass of each foot is 2.53 g, the speed of a foot as it slaps the water is 1.08 m/s, and the time for a single step is 0.417 s. (a) what is the magnitude of the impulse on the lizard during the slap? (assume this impulse is directly upward.) (b) during the 0.417 s duration of a step, what is the downward impulse on the lizard due to the gravitational force? (clearly the push, not the slap, provides the primary support for the lizard.)
The magnitude of impulse acting directly upwards during the slap is -0.00273 N.s.
The magnitude of the impulse that is downward on the lizard due to the gravitational force is 0.375 kg.m/s.
Given that, mass of the basilisk lizard = 90 g = 90/1000 kg = 0.09 kg
Mass of each foot = 2.53 g = 0.00253 kg
Vf = 0
Vi = 1.08 m/s
Time taken for a single step is 0.417 s
I = Δp = Mf * (Vf - Vi ) = 0.00253* (0 - 1.08) = -0.00273 N.s
On each step, the basilisk lizard must provide an upward force that counters its weight which is equal to W= mg.
We know the formula for impulse acting downwards is, I = F Δt
It can also be written as I = W Δt = Ml* g Δt
Substituting the values in the above formula,
Ml*g Δt = 0.09* 10* 0.417 = 0.375 kgm/s
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the forearm in the figure(figure 1) accelerates a 3.2 kg ball at 7.5 m/s2 by means of the triceps muscle, as shown. a) Calculate the torque needed. Ignore the mass of the arm. Answer in m*Nb) Calculate the force that must be exerted by the triceps muscle. Answer in N
The calculated answer for the torque is 7.812 Nm and the force is 25.2 N
In the given question, we have to find the torque needed and the force exerted by the triceps muscle.
We know that, when there is a net torque on the system, it causes angular acceleration and the angular momentum is no more conserved.
Here let be the arm is massless.
The torque can be written as
τ=Iα
Where I is the moment of inertia.
α is the angular acceleration.
So,
τ=mR^2 α
τ=mRa
Here we have used the relationship between angular and tangential acceleration, which is: α=a/R.
The position vector is: R = 0.31
The mass is: m =3.6
The tangential acceleration is: a =7 m/s2
So, the torque will be,
τ=3.6*0.31*7
=7.812
So, the required torque is 7.812 Nm.
And we know that torque is: τ=F.R
So, force is F = τ/r
= 7.812/0.31
=25.2 N
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A 75 kg person stands on a metric bathroom scale that reads in Newtons while in an elevator. The
elevator accelerates upward at 3 m/s/s. Determine the person's actual weight and the reading of the
scale under these conditions.
Answer: 735 N (actual), 960 N (accelerating up)
Explanation:
Actual weight = W = mg = (75 kg)(9.8 m/s²) = 735 N
Scale reading = N (normal force acting on the person by the elevator) = mg + ma = 735 N + (75 kg)(3.0 m/s²) = 735 N + 225 N = 960 N
That's why you feel heavier when you are accelerating up in an elevator
in the instant of fig. 11-41, two particles move in an xy plane. particle p1 has mass 6.5 kg and speed v1 2.2 m/s, and it is at dis- tance d1 1.5 m from point o. particle p2 has mass 3.1 kg and speed v2 3.6 m/s, and it is at distance d2 2.8 m from point o. what are the (a) magnitude and (b) direction of the
The magnitude and direction of net angular momentum of particles is [tex]52.7kg.m^2/s[/tex] out of the page.
Given,
particle 1:-
Mass ,m1 = 6.5 kg
speed , v1 = 2.2 m/s
distance from point O, d1 = 1.5 m
Particle 2:-
Mass ,m2 = 3.1 kg
speed, v2 = 3.6 m/s
distance from point O, d2 = 2.8 m
Angular momentum can be determined by formula,
p=mvd
angular momentum of particle-1, [tex]\rho_1=6.5*2.2*1.5=21.45 kg.m^2/s[/tex]
angular momentum of particle-2, [tex]\rho_2=3.1*3.6*2.8=31.248 kg.m^2/s[/tex]
Then, the net angular momentum is
[tex]\rho=\rho_1+\rho_2=21.45+31.248=52.69kg.m^2/s[/tex]
The direction of angular momentum of particle is always perpendicular to the motion of particle, as the particles are moving on xy-plane then the their direction will be perpendicular to xy-plane i.e. z-plane, it means out of the page.
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Your question is incomplete, here is the complete question.
In the instant shown in the diagram, two particles move in an xy-plane. Particle P1 has mass 6.5 kg and speed v1 = 2.2 m/s, and it is at distance d1 = 1.5 m from point O. Particle P2 has mass 3.1 kg and speed v2 = 3.6 m/s, and it is at distance d2 = 2.8 m from point O. What is the magnitude and direction of net angular momentum of the two particles about O?
A) 52.7 kg · m2/s out of the page B) 52.7 kg · m2/s into the page C) 21.5 kg · m2/s into the page D) 9.8 kg · m2/s into the page E) 9.8 kg · m2/s out of the page
When block is placed on a wedge as shown in figure, the block starts sliding down and the wedge also starts sliding on ground. All surfaces are rough. The centre of mass of (wedge + block) system will move
A. leftward and downward
B. rightward and downward
C. leftward and upwards
D. only downward
Rightward and downwards. option B is the correct answer
Internal friction between the wedge and the block means that it won't affect the center of mass' motion. The external friction force exerted on the wedge by the ground causes the center of mass to shift to the right. The center of mass of a block moves downward due to the gravitational force (mg) acting on it.
Frictional Force:
When surfaces of two objects come into contact, the force called friction is what stops motion. In other words, friction weakens a machine's mechanical advantage and decreases the output to input ratio. In a car, reducing friction requires one-fourth of the energy required. Friction in the clutch and tires, however, also has an impact on how well the car can maintain its place on the road. One of the most significant occurrences in the physical universe is friction, a phenomenon that affects everything from machines to molecular structures to matches.
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a fighter jet is traveling at 525 m/s directly away from a communication antenna that broadcasts at 410 mhz . What change in frequency does the fighter jet observe
The fighter jet notices a frequency change of 718 Hz.
from Doppler's effect, f' = f(1 - [tex]\frac{u}{c}[/tex])
Here, c stands for the speed of light, u for the speed of a fighter jet, and f for the antenna's broadcast frequency.
so, f' = f(1 - [tex]\frac{u}{c}[/tex])
⇒f'/f =( 1 - [tex]\frac{u}{c}[/tex])
⇒(f - f')/f = [tex]\frac{u}{c}[/tex]
⇒∆f = f([tex]\frac{u}{c}[/tex])
so change in frequency, ∆f = f([tex]\frac{u}{c}[/tex])
now f = 410 × 10⁶ Hz , u = 525 m/s and c = 3 × 10 m/s
so, ∆f = (410 × 10⁶ × 525)/(3 × 10⁸)
= 71750 × 10⁻²
= 717.75 Hz ≈ 718 Hz
As a result, the fighter jet recorded a 718 Hz change in frequency.
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A 2.50-kg solid, uniform disk rolls without slipping across a level surface, translating at 3.75 m/s. If the disk’s radius is 0.100 m, find its (a) translational kinetic energy and (b) rotational kinetic energy.
The translational & rotational kinetic energy is :
a) The translational kinetic energy of the disk across the level surface is 17.58J.
b) The rotational kinetic energy of the disk is 8.79J.
How do you find total kinetic energy with translational and rotational?The combined translational and rotational kinetic energies of an item in motion determine its total kinetic energy. 12 mvCM2 is the translational kinetic energy. Rotational kinetic energy is equal to 12 I2. Total kinetic energy is 12mvCM2 plus 12I2.Ek=12mv2, where m is the mass (mass of the molecule, atom, or ion) and v is the velocity of its Centre of mass, is the energy associated with the movement of a chemical entity.There is just one distinction between rotational and translational kinetic energy: translational energy moves in a straight path, but rotational energy does not. When a bicycle is being ridden down a bike path, its tyre provides an illustration of both kinetic and translational kinetic energy.Learn more about translational & rotational kinetic energy refer to :
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Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/ 15. The lasers produce separate interference patterns on a screen a distance 4.40 m away from the slits. Part What is the distance Aymamin between the second maximum of laser 1 and the third minimum of laser 2 on the same side of the central maximum? Express your answer in meters. View Available Hint(s) Ay mas ma 5867 k m Submit Previous Answers x incorrect: Try Again; 3 attempts remaining
the distance Aymamin between the second maximum of laser 1 and the third minimum of laser 2 on the same side of the central maximum is 0.4 m
For part be we can use the equation
ym = m*λ*L/d
so for the first max for each one m =1
so for laser 1
y = (d/20)*(6.0m)/d
y = 6.0m /20 = 0.3
for laser 2 we do the same thing
y = 6.0m/15 = 0.4
y ( laser1 ) < y ( laser2 )
(1) hence first maxima of laser1 is closest to central maxima
Δy = 0.4m -0 .3 m = 0.1m
For part C)
we use the same equation to find the y for laser 1, except m =2
y= 2*6.0 m / 20 = 0.6 m
Now for laser 2 we use:
ym=(m+1/2)*λ*L/d
since there is no central minimum the first minimum is at m = 0.that means that the third minimum is at m = 2
simplifying the equation we get
y = (2.5)*6.0 / 15 = 1m
now we solve
Δy= 1m - 0.6m = 0.4 m
Double Slit Experiment:
The double-slit experiment shows that light and matter can exhibit properties of both classically defined waves and particles. Furthermore, it demonstrates the fundamental stochastic nature of quantum mechanical phenomena.
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A copper rod has a length of 60 cm at room temperature (220C). The coefficient of expansion of copper is 17 x 10-6 /Co, its specific heat is 386 J/(kg K), its melting point is 1356 K and its heat of fusion is 207 kJ/kg. The copper rod is heated so that its length increases to 60.3 cm a. What is the new temperature of the copper rod? T 316 b. If the mass of the rod is 0.52 kg, how much thermal energy was added to the rod? Q 5.9 x104 J c. If the mass of the rod is 0.52 kg, how much total thermal energy will need to be added to the copper if we want to completely melt the copper rod starting at room temperature? Qー ? x105 J
There are 3 questions here that need to be solved.
a. The new temperature of the copper rod is 316°C (rounded)
b. If the mass of the rod is 0.52kg, the thermal energy that was added to the rod is 5.9 * 10^4J
c. If the mass of the rod is 0.52kg, the total thermal energy will need to be added to the copper if we want to completely melt the copper rod starting at room temperature by using coefficient of thermal expansion of copper at 17 * 10^-6 / C° is 3.21 * 10^5J
What is coefficient of thermal expansion?It refers to the rate at which material expands with increase in temperature. More specifically, this coefficient is determined at constant pressure and without a phase change, such as the material is expected to still be in its solid or fluid form.
Now, moving to the question.
a. What is the new temperature of the copper rod?
L₂ = L₁ + L₁ * α * (t₂ - t₁)
Where
L2 is length of the rod that already increased
L1 is the original length of the rod
α is coefficient of expansion of copper
t2 is the new temperature
t1 is the original temperature
60.3cm = 60cm + 60cm * 17 * 10^-6 * (t₂ - 22°C)60.3
= 60 + 0.00102 * (t₂ - 22)0.3 = 0.00102 * (t₂ - 22)t₂ - 22
= 0.3 / 0.00102 t₂ - 22 = 294.12
t₂ = 316°C (rounded)
b. If the mass of the rod is 0.52kg, how much thermal energy was added to the rod?
Q = m * c * Δt
m = mass
c = coefficient
= 0.52 * 386 * (316 - 22)
= 59011.68
= 5.9 * 10^4 J (rounded)
c. If the mass of the rod is 0.52kg, how much total thermal energy will need to be added to the copper if we want to completely melt the copper rod starting at room temperature?
Q = m * c * Δt + m * L fusion
= m * c * (t melt - t₁) + m * L fusion
= 0.52 * 386 * (1356 - (22 + 273)) + 0.52 * 207 * 10^3Q
= 320603.92
Q = 3.21 * 10^5 J
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A car moving at 18 m/s crashes into a tree and stops in 1.14 s. The mass of the passenger inside is 69 kg.Calculate the magnitude of the average force, in newtons, that the seat belt exerts on the passenger in the car to bring him to a halt.
The force the seat belt exerts on a passenger in the car to bring him to a halt is 0.6 × [tex]10^{3}[/tex]
A car moving at 18 m/s crashes into a tree and stops in 1.14 s. calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. the mass. of the passenger. is 69. kg
Generally the equation for the net Force is mathematically given as
Fnet = mdv/dt
net = 69*10/1.14 = 0.6 × [tex]10^{3}[/tex]
When objects interact with one another, push and pull are generated. You can also describe force using words like stretch and squeeze.
Force is described as follows in physics:
A massed item changes its velocity in response to a push or pull.
Force is applied at a location known as the application, and the direction in which it is applied is referred to as the force's direction.
Using a spring balance, one can quantify the Force. Newton is the unit of force in the SI (N) The following list of consequences of the Force is not exhaustive.
A body at rest can move with enough force.
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Now use a handle in the middle to make the pipe narrower in the middle. Record the readings of the flux meter: Flow rate: 5000L/s, area: 1.1m2Which of these quantities stayed the same when you change the size of the pipe and why? Please also move the speed gauge around to see how the speed varies at different locations inside the pipe.
Flow rate stayed the same but why...?
The flow rate is given as 5000L/s, which means that 5000 liters of fluid pass through a cross-sectional area of 1.1m2 per second. If the size of the pipe is changed by using a handle to make it narrower in the middle, the cross-sectional area of the pipe will also change.
However, if the flow rate remains the same, it means that the volume of fluid passing through the pipe per unit of time is still constant.
To maintain the same flow rate, the velocity of the fluid must increase as the cross-sectional area decreases, and vice versa. This can be observed by moving the speed gauge around to different locations inside the pipe and measuring the speed of the fluid at those locations.
As the cross-sectional area of the pipe decreases, the velocity of the fluid will increase, and as the cross-sectional area increases, the velocity of the fluid will decrease.
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