Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
70.31kg
Explanation:Step I: Consider Newton's second law of motion which states that;
∑F = m x a;
Where;
∑F = net force acting on a body
m = the mass of the body
a = acceleration due to the force on the body.
Step II: Now to the question;
Since frictional force is negligible and the only force acting on the block of ice is the applied force by the dockworker, the net force on the body (block of ice) is the constant horizontal force. i.e
∑F = 90.0N
Also;
the block starts from rest and moves a distance (s) of 13.0m in a time (t) of 4.50s. Here, we can get the acceleration in that duration of time using one
the equations of motion as follows;
s = ut + [tex]\frac{1}{2}[/tex]at² ------------------------------(ii)
Where;
s = distance covered = 13.0m
u = initial velocity = 0 [since the block starts from rest]
t = time taken to cover the distance = 4.50s
a = acceleration of the body.
Substitute these values into equation (ii) as follows;
13.0 = 0(4.5) + [tex]\frac{1}{2}[/tex](a)(4.50)²
13.0 = 0 + [tex]\frac{1}{2}[/tex](a)(20.25)
13.0 = [tex]\frac{1}{2}[/tex](a)(20.25)
13.0 = 10.125a
Solve for a;
a = [tex]\frac{13.0}{10.125}[/tex]
a = 1.28m/s²
Step III: Now substitute the values of a = 1.28m/s² and ∑F = 90.0N into equation (i) as follows;
90.0 = m x 1.28
m = [tex]\frac{90.0}{1.28}[/tex]
m = 70.31
Therefore, the mass of the block of ice is 70.31kg
A flat circular loop of radius 0.10 m is rotating in a uniform magnetic field of 0.20 T. Find the magnetic flux through the loop when the plane of the loop and the magnetic field vector are perpendicular.'
Answer:
[tex]\phi=6.28\times10^{-3}\;\;weber[/tex]
Explanation:
Given,
Magnetic field [tex]B=0.2\;\;T\\[/tex]
Radius [tex]r=0.1\;\;m[/tex]
The angle between the area vector and magnetic field is 0 degree, because the direction of area vector is always perpendicular to the plane.
[tex]\phi=BAcos\theta\\\phi=o.2\times\pi (0.1)^2\times cos0^o\\\phi=0.00628\;\;weber\\\phi=6.28\times10^{-3}\;\;weber[/tex]
The magnetic flux through a flat circular loop rotating in a uniform magnetic field is zero when the plane of the loop and the magnetic field vector are perpendicular.
Explanation:To find the magnetic flux through the loop when the plane of the loop and the magnetic field are perpendicular, we will need to use the formula for magnetic flux:
Φm = BA cos θ
where Φm is the magnetic flux through the surface, B is the magnitude of the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field direction and the normal (perpendicular) to the surface. In this case, if the loop and the magnetic field are perpendicular, θ = 90°, and cos 90° = 0. Hence, the magnetic flux will be zero.
However, if the provided problem included a different angle, we would adjust the equation to accommodate that. For example, if the field were parallel to the loop (θ = 0), then cos θ would equal 1 and the magnetic flux would just be the product of B and A (A = πr²)
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a certain engine has a second-law efficiency of 85%. During each cycle, it absorbs 480J of heat form a reservoir at 300C and dumps 300J of heat to a cold termperature reservoir.What is the temperature of the cold reservoir?
Final answer:
The temperature of the cold reservoir for an engine with a second-law efficiency of 85% that absorbs 480J of heat from a hot reservoir at 300C and dumps 300J into the cold reservoir is 358.125 K.
Explanation:
The question asks for the temperature of the cold reservoir for an engine with a second-law efficiency of 85% that absorbs 480J of heat from a hot reservoir at 300C and dumps 300J into the cold reservoir.
The second-law efficiency η of a heat engine is defined as the ratio of the work output W to the heat input Qh at the high temperature, while for a Carnot engine, the efficiency can also be related to the temperatures of the hot (Th) and cold (Tc) reservoirs as:
η = 1 - (Tc/Th)
Given the heat absorbed (Qh = 480 J) and the heat rejected (Qc = 300 J), we can calculate the work done (W = Qh - Qc) which is 180 J here. We know that Th is the temperature of the hot reservoir in kelvin, which we obtain by converting 300C to kelvin (Th = 573 K). Note that 0 degrees Celsius is equivalent to 273 K.
Using the given second-law efficiency:
η = W / Qh = 180 J / 480 J = 0.375
For a Carnot engine:
η = 1 - (Tc/Th)
0.375 = 1 - (Tc/573 K)
Tc = 573 K * (1 - 0.375)
Tc = 358.125 K
The temperature of the cold reservoir for this engine is therefore 358.125 K.
Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the same time by pushing on box A with a horizontal pushing force FP = 8.9 N. Here, A has a mass mA = 10.2 kg and B has a mass mB = 7.0 kg. The contact force between the two boxes is FC. The coefficient of kinetic friction between the boxes and the floor is 0.04. (Assume FP acts in the +x direction.)
Answer:
Explanation:
The force of friction acting on the system
= .04 x 9.8 ( 10.2 + 7 )
= 6.74 N
Net force = 8.9 - 6.74
= 2.16 N
Acceleration in the system
= 2.16 / ( 10.2 + 7 )
= .12558 m / s ²
Contact force between boxes = FP
Considering force on box A
Net force = 8.9 - FP
Applying Newton's law on box A
8.9 - FP = 10.2 x .12558
= 1.28
FP = 8.9 - 1.28
= 7.62 N
Two children stand on a platform at the top of a curving slide next to a backyard swimming pool. At the same moment the smaller child hops off to jump straight down into the pool, the bigger child releases herself at the top of the frictionless slide.
Upon reaching the water, how does the kinetic energy of the smaller child compare with that of the larger child?
Answer:
THE KINETIC ENERGY OF THE SMALLER CHILD IS LESS THAN THAT OF THE BIGGER CHILD
Explanation: Kinetic energy is the energy that is exerted on a body that is in motion, kinetic energy is affected by both the mass of the object and the velocity of the object.
Mathematically,Kinetic energy is represented as follows;
K.E=1/2M[tex]V^{2}[/tex]
Where M represents the mass of the object in kilograms and V represents velocity of the moving object measured in meters per seconds.
The higher the weight of the object the higher the kinetic energy of the object which means the bigger child will have a higher kinetic energy than the smaller child.
Both children start with the same potential energy, which is converted into kinetic energy as they move; hence, both have the same kinetic energy upon landing in the water. However, the distribution of this energy according to the mass and velocity of each child results in a higher velocity for the smaller child and a lower velocity for the larger child.
Explanation:The subject of this question is in the field of Physics, specifically on the concept of conservation of energy. Both children start from the same height, so they both have the same potential energy. When they land in the pool, this potential energy has been converted into kinetic energy.
The kinetic energy of an object in motion is given by the formula [tex]KE = 1/2 mv^2[/tex], where 'm' is the mass of the object, and 'v' is its velocity. Since we're told to ignore factors like friction and air resistance, we can say that when both children land in the pool they have the same kinetic energy, and the only difference is how this energy is distributed between the mass and the velocity.
The smaller child, having less mass, will have to have a greater velocity to account for the same amount of kinetic energy. The bigger child, with more mass, will have a smaller velocity. So in terms of kinetic energy, it is the same for both children when they reach the water.
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The space between two concentric conducting spherical shells of radii b = 1.70 cm and a = 1.20 cm is filled with a substance of dielectric constant ? = 27.0. A potential difference V = 64.5 V is applied across the inner and outer shells.
(a) Determine the capacitance of the device.
nF
(b) Determine the free charge q on the inner shell.
nC
(c) Determine the charge q' induced along the surface of the inner shell.
nC
Answer:
a) C = 1.065 * 10^-10 F
b) 7.775 * 10^-9
c) 7.444 * 10^-9 C
Explanation:
A spherical capacitor, with inner radius of a = 1.2 cm and outer radius
of b = 1.7 cm is filled with a dielectric material with dielectric constant of
K = 27 and connected to a potential difference of V = 64.5 V.
(a) The capacitance of a filled air spherical capacitor is given by equation :
C = 4*π*∈o*(a*b/b-a)
if the capacitor is filled with a material with dielectric constant K, we need
to modify the capacitance as ∈o ---->k∈o , thus:
C = 4*π*∈o*(a*b/b-a)
substitute with the given values to get:
C = 4*π*(27)*(8.84*10^-12)[(1.2*10^-2)*(1.7*10^-2)/(1.7*10^-2)-(1.2*10^-2)*]
C = 1.065 * 10^-10 F
(b) The charge on the capacitor is given by q = CV, substitute to get:
q = (1.065 * 10^-10)*64.5 V
= 7.775 * 10^-9
(c) The induced charge on the dielectric material is given by equation as:
q' = q(1-1/k)
substitute with the given values to get:
q' = (7.775 * 10^-9)*(1-1/27)
= 7.444 * 10^-9 C
note:
calculation maybe wrong but method is correct. thanks
The capacitance is 3.36 nF, the free charge is 216.72 nC and the induced charge is zero.
Given information:
Radius, a = 0.017 m
b = 0.012 m
Potential difference, V = 64.5 V
(a)
The capacitance is given by:
C = (4πε₀ / (1/b - 1/a))
C = (8.85*10⁻²×3.14×4)/(1/0.012-1/0.017)
C = (4π(8.85 x 10^-12) / 293.3)
C = 3.36 nF
Hence, the capacitance is 3.36 nF.
(b)
The free charge can be calculated from the relation of charge, capacitance, and voltage:
q = CV
q = 3.36×64.5
q= 216.72 nC
Hence, the free charge is 216.72 nC.
(c)
The induced charge is given by:
q' = q - C × V
q' = 0 nC
Hence, the charge is 0 nC.
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A space station sounds an alert signal at time intervals of 1.00 h . Spaceships A and B pass the station, both moving at 0.400c0 relative to the station but in opposite directions.
Part A
How long is the time interval between signals according to an observer on A?
Part B
How long is the time interval between signals according to an observer on B?
Part C
At what speed must A move relative to the station in order to measure a time interval of 2.00 hbetween signals?
Answer:
(A). The the time interval between signals according to an observer on A is 1.09 h.
(B). The time interval between signals according to an observer on B is 1.09 h.
(C). The speed is 0.866c.
Explanation:
Given that,
Time interval = 1.00 h
Speed = 0.400 c
(A). We need to calculate the the time interval between signals according to an observer on A
Using formula of time
[tex]\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}[/tex]
Put the value into the formula
[tex]\Delta t=\dfrac{1.00}{\sqrt{1-(\dfrac{0.400c}{c})^2}}[/tex]
[tex]\Delta t=\dfrac{1.00}{\sqrt{1-(0.400)^2}}[/tex]
[tex]\Delta t=1.09\ h[/tex]
(B). We need to calculate the time interval between signals according to an observer on B
Using formula of time
[tex]\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}[/tex]
Put the value into the formula
[tex]\Delta t=\dfrac{1.00}{\sqrt{1-(\dfrac{0.400c}{c})^2}}[/tex]
[tex]\Delta t=\dfrac{1.00}{\sqrt{1-(0.400)^2}}[/tex]
[tex]\Delta t=1.09\ h[/tex]
(C). Here, time interval of 2.00 h between signals.
We need to calculate the speed
Using formula of speed
[tex]\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-(\dfrac{v}{c})^2}}[/tex]
Put the value into the formula
[tex]2.00=\dfrac{1.00}{\sqrt{1-(\dfrac{v}{c})^2}}[/tex]
[tex]\sqrt{1-(\dfrac{v}{c})^2}=\dfrac{1.00}{2.00}[/tex]
[tex]1-(\dfrac{v}{c})^2=(\dfrac{1.00}{2.00})^2[/tex]
[tex](\dfrac{v}{c})^2=\dfrac{3}{4}[/tex]
[tex]v=\dfrac{\sqrt{3}}{2}c[/tex]
[tex]v=0.866c[/tex]
Hence, (A). The the time interval between signals according to an observer on A is 1.09 h.
(B). The time interval between signals according to an observer on B is 1.09 h.
(C). The speed is 0.866c.
Two objects have the same size and shape, but one is much heavier than the other. When they are dropped simultaneously from a tower, they reach the ground at the same time (assuming that there is no air resistance), but the heavier one has a greater
speed
acceleration
none of the above
all of these
Answer:
None of the above.
The correct answer would be momentum
Answer:
Momentum (None of the above)Explanation:
The two objects free-fall at the same rate of acceleration, thus giving them the same speed when they hit the ground. The heavier object however has more momentum since momentum takes into account both the speed and the mass of the object (p=m*v).
Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 7.50 cm apart. You see an object jump from side to side by 0.95 degree as you blink back and forth between your eyes. How far away is the object?
Answer:
Distance of the object from eye is approx 4.52 m
Explanation:
As we know that the object subtend a small angle on both the eyes which is given as
[tex]\theta = 0.995 degree[/tex]
now we know that the distance between two eyes is given as
d = 7.50 cm
so we have
[tex]angle = \frac{arc}{Radius}[/tex]
so here the radius is same as the distance from eye while arc is the distance between two eyes
so we have
[tex]0.95 \frac{\pi}{180} = \frac{7.50}{R}[/tex]
[tex]R = 452 cm[/tex]
a copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)
Answer:0.01678325m
Explanation:
Original length(L1)=27.5m
Temperature rise(@)=35.9°C
Linear expansivity of copper(α)=0.000017
Length increment(L)=?
L=α x L1 x @
L=0.000017 x 27.5 x 35.9
L=0.01678325m
Answer:
0.0168
Explanation:
Remember about significant digits (there must be 3)
What is the phase angle of an AC series circuit that is constructed of a 14.5-Ω resistor along with 16.5-Ω inductive reactance and 9.41-Ω capacitive reactance?
Answer: cosθ = 0.7531
Explanation: the phase angle cosθ is given as
cosθ = R/Z
Where R = resistive reactance = 14.5 ohms
Z = impeadance = √R^2 +(Xl - Xc)^2
Where Xl = inductive reactance = 16.5 ohms and Xc= capacitive reactance = 9.41 ohms
By substituting the parameters, we have that
Z = √14.5^2 + (16.5^2 - 9.41^2)
Z = √210.25 + (272.25 - 88.5481)
Z = √210.25 + 183.7019
Z = √393.9519
Z = 19.85 ohms
Z = 19.85 ohms, R = 14.5 ohms
cosθ = R/Z = 14.5/19.85
cosθ = 0.7531
26.06°
Explanation:Given an RLC circuit [a circuit containing a capacitor, inductor and resistor], the phase angle (Φ), which is the difference in phase between the voltage and the current in the circuit, is given by;
Φ = tan⁻¹ [ [tex]\frac{X_{L} - X_{C}}{R}[/tex]] --------------------------(i)
Where;
[tex]X_{L}[/tex] = inductive reactance of the circuit
[tex]X_{C}[/tex] = capacitive reactance of the circuit
R = resistance of the circuit
From the question;
[tex]X_{L}[/tex] = 16.5 Ω
[tex]X_{C}[/tex] = 9.41 Ω
R = 14.5 Ω
Substitute these values into equation (i) as follows;
Φ = tan⁻¹ [ [tex]\frac{16.5 - 9.41}{14.5}[/tex]]
Φ = tan⁻¹ [ [tex]\frac{7.09}{14.5}[/tex]]
Φ = tan⁻¹ [ 0.4890]
Φ = 26.06°
Therefore the phase angle of the AC series circuit is 26.06°
Light of wavelength 710 nm passes through two narrow slits 0.66 mm apart. The screen is 2.00 m away. A second source of unknown wavelength produces its second-order fringe 1.25 mm closer to the central maximum than the 710-nm light. What is the wavelength of unknown light?
Answer:
The wavelength is 503 nm
Explanation:
Considering constructive interference , this means that route(path) difference is equal to the product of order of fringe and wavelength of the light
i.e dsinθ = m[tex]\lambda[/tex]
Where [tex]\lambda[/tex] is the wavelength of light and m is the order of the fringe
Looking at θ to be very small , sin θ can be approximated to θ
and [tex]\theta \approx \frac{x}{l}[/tex]
Substituting this into the above equation
[tex]d[\frac{x}{l} ] =m\lambda[/tex]
making x the subject
[tex]x =\frac{m\lambda l}{d}[/tex]
This above equation will give the value of the distance of the [tex]m^{th}[/tex] order fringe of the wavelength [tex]\lambda[/tex] from the central fringe
Replacing with the value given in the question we have
[tex]\lambda[/tex] = 710 nm m = 2 d =0.66 mm , l = 2.0 m
[tex]x = \frac{(2)(710nm)(2.0m)[\frac{10^9}{1m} ]}{(0.66mm)(\frac{10^6}{1mm} )}[/tex]
[tex]=(4.303*10^6nm)[\frac{\frac{1}{10^6}mm }{1nm} ][/tex]
[tex]=4.303mm[/tex]
The separation of the second fringe from central maximum is 4,303 mm
To obtain the separation of the second order fringe of the unknown light from central maximum
[tex]x' = 4.303mm - 1.25 mm = 3.053mm[/tex]
Now to obtain the wavelength of this second source
from [tex]x = \frac{m\lambda l}{d}[/tex]
[tex]\lambda' = \frac{x'd}{ml}[/tex]
Now substituting 3,053 mm for [tex]x'[/tex] 2.0 mm for l , 0.66 mm for d and 2 for m in the above formula
[tex]\lambda' =\frac{(3.053mm)(0.66mm)}{(2)(2.0)(\frac{10^3mm}{1m} )}[/tex]
[tex]= (503.7*10^{-6}mm)(\frac{10^6nm}{1mm} )[/tex]
[tex]=503.7nm[/tex]
Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer connection.17 pointsb. Calculate the efficiency of the transformer in this connection when it is supplying its rated load at unity power factor. 17 points
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The High Voltage Rating for Auto - Transformer is 86kV
The Low Voltage Rating for Auto - Transformer is 78kV
The MVA rating is 268.75[tex]MVA[/tex]
b
The efficiency is 99.4%
Explanation:
From the question we are given are given that
The transformer has Mega Volt Amp rating of 25MVA
The frequency is 60-Hz
Voltage rating 8.0kV : 78kV
The short circuit test gives : 453kV,321A,77.5kW
The open circuit test gives : 8.0kV, 39.6A, 86.2kW
This can be represented on a diagram shown on the second uploaded image
From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the Low Voltage Rating for Auto - Transformer is 78kV
Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have
[tex]\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}[/tex]
The volt will cancel each other
[tex]\frac{25*10^6}{8*10^3} = 3125\ A[/tex]
Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.
[tex]MVA \ rating = (86*10^3)(3125) =268.75[/tex]
We need to understand that Iron losses is due to open circuit test which has power = 86.2kW
While copper loss is due to short circuit test which has power = 77.5kW
The the current flowing through the secondary coil [tex]I_2[/tex] as shown in the circuit diagram can be obtained as
[tex]I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321[/tex]
Now the efficiency can be obtained as thus
[tex]\frac{(operational \ MVA )*(Power factor \pf))}{(operational\ MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}[/tex]
=99.941%
If the momentum of a 1000 kg car travelling at 10 m/s was transferred completely to a 20.0 kg traffic barrier, what would the final speed of the barrier be
Answer:
500 m/s
Explanation:
Momentum, p is a product of mass and velocity. From law of conservation of momentum, the initial momentum equals final momentum
[tex]m_1v_1=m_2v_2\\1000*10=20v_2\\v_2=10000/20=500 m/s[/tex]
Here m and v represent mass and velocity respectively and subscripts 1 and 2 represent car and barrier respectively
Therefore, the velocity of barrier will be 500 m/s
When tension is applied to a metal wire of length L , it stretches by Δ L . If the same tension is applied to a metal wire of the same material with the same cross-sectional area but of length 2 L , by how much will it stretch?
Answer:
The metal wire will stretch by [tex]2 \delta L[/tex]
Explanation:
[tex]T = \frac{kA \delta L}{L}[/tex]......................................(1)
Where T = Tension applied
ΔL = Extension
L = length
k = constant
T₁ = T₂ = T
A₁ = A₂ =A
L₁ = L
L₂ = 2L
(ΔL)₁ = ΔL
(ΔL)₂ = ?
From equation (1)
[tex]TL/kA = \delta L[/tex].....................(2)
[tex]TL/kA = (\delta L) ........................(3)\\ 2TL/kA = (\delta L)_{2} ........................(4)[/tex]
Divide (4) by (3)
[tex]\frac{(\delta L)_{2} }{\delta L} =\frac{\frac{2TL}{kA} }{\frac{TL}{kA} } \\\frac{(\delta L)_{2} }{\delta L} = 2\\ (\delta L)_{2} = 2\delta L[/tex]
If the starting length is twice, the extension is doubled as well.
Given that;
Length of metal wire = L
Stretch = ΔL
So,
It will stretched to a length of 2L.
Although when tension is continuous, the length of the extension is proportional to the size of the initials.
As a result, if the starting length is doubled, the extension will be twice as well.
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A rectangular loop of wire with width 0.4 cm and length 0.4 cm is oriented with the normal to the face of the loop making an angle of 30° with respect to the direction of B. The B field has a magnitude of 0.77 T. Find the magnetic flux through the loop.
Answer:
0.001067 Wb
Explanation:
Parameters given:
Magnetic field, B = 0.77 T
Angle, θ = 30º
Width = 0.4cm = 0.04m
Length = 0.4cm = 0.04m
Magnetic flux, Φ(B) is given as:
Φ(B) = B * A * cosθ
Where A is Area
Area = length * width = 0.04 * 0.04 = 0.0016 m²
Φ(B) = 0.77 * 0.0016 * cos30
Φ(B) = 0.00167 Wb
Answer:
1.07×10⁻⁵ Wb
Explanation:
Using
Φ = BAcosθ.................. Equation 1
Where Φ = magnetic Flux, B = magnetic Field, A = Area of the rectangular loop, Angle between the loop and the Field.
But
A = L×W........................ Equation 2
Where L = Length, W = Width.
Substitute equation 2 into equation 1
Φ = BLWcosΦ................ Equation 3
Given: B = 0.77 T, L = 0.4 cm = 0.004 m, W = 0.4 cm = 0.004 m, Ф = 30°
Substitute into equation 3
Ф = 0.77(0.004)(0.004)cos30
Ф = 1.07×10⁻⁵ Wb.
Hence the magnetic Field through the loop = 1.07×10⁻⁵ Wb.
Tarik winds a small paper tube uniformly with 183 turns 183 turns of thin wire to form a solenoid. The tube's diameter is 9.49 mm 9.49 mm and its length is 2.09 cm 2.09 cm . What is the inductance, in microhenrys, of Tarik's solenoid?
143μH
Explanation:The inductance (L) of a coil wire (e.g solenoid) is given by;
L = μ₀N²A / l --------------(i)
Where;
l = the length of the solenoid
A = cross-sectional area of the solenoid
N= number of turns of the solenoid
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
From the question;
N = 183 turns
l = 2.09cm = 0.0209m
diameter, d = 9.49mm = 0.00949m
But;
A = π d² / 4 [Take π = 3.142 and substitute d = 0.00949m]
A = 3.142 x 0.00949² / 4
A = 7.1 x 10⁻⁵m²
Substitute these values into equation (i) as follows;
L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]
L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209
L = 143 x 10⁻⁶ H
L = 143 μH
Therefore the inductance in microhenrys of the Tarik's solenoid is 143
The inductance, in microhenrys, of Tarik's solenoid should be considered as the 143μH.
Calculation of the inductance:Since
The inductance (L) of a coil wire (e.g solenoid) should be provided by
L = μ₀N²A / l --------------(i)
here,
l = the length of the solenoid
A = cross-sectional area of the solenoid
N= number of turns of the solenoid
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
So,
N = 183 turns
l = 2.09cm = 0.0209m
diameter, d = 9.49mm = 0.00949m
So,
A = π d² / 4
= 3.142 x 0.00949² / 4
= 7.1 x 10⁻⁵m²
Now
L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]
L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209
L = 143 x 10⁻⁶ H
L = 143 μH
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A system of two cylinders fixed to each other is free to rotate about a frictionless axis through the common center of the cylinders and perpendicular to the page. A rope wrapped around the cylinder of radius 2.50 m exerts a force of 4.49 N to the right on the cylinder. A rope wrapped around the cylinder of radius 1.14 m exerts a force of 9.13 N downward on the cylinder. What is the magnitude of the net torque acting on the cylinders about the rotation axis? Answer in three decimal places.
Torque is the force's twisting action about the axis of rotation magnitude of the net torque acting on the cylinders about the rotation axis will be 331.402 Nm.
What is torque?Torque is the force's twisting action about the axis of rotation. Torque is the term used to describe the instant of force. It is the rotational equivalent of force. Torque is a force that acts in a turn or twist.
The amount of torque is equal to force multiplied by the perpendicular distance between the point of application of force and the axis of rotation.
In the first situation, force is delivered in the right direction, hence torque is applied upwards.
The value of torque for case 1
[tex]\rm \tau_1=4.49\times 2.50\\\\\rm \tau_1=11.25 Nm[/tex]
The value of torque for case 2
[tex]\rm \tau_2=91.3\times 1.14\\\\\rm \tau_2=104.08[/tex]
The resultant value of torque will be;
[tex]\rm \tau =\sqrt{(11.25)^2+(41.90)^2} \\\\\ \rm \tau =331.402 Nm[/tex]
Hence the magnitude of the net torque acting on the cylinders about the rotation axis will be 331.402 Nm.
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Final answer:
The magnitude of the net torque acting on the system of two cylinders is 21.633 N⋅m.
Explanation:
The question asks us to calculate the net torque on a system of two cylinders with ropes exerting forces at different radii. Torque (τ) is calculated by the product of the radius (r), the force (F), and the sine of the angle (θ) between them, τ = rFsinθ. In this case, as both forces are perpendicular to the radii, sinθ = 1, simplifying the torque to τ = rF.
For the cylinder with radius 2.50 m, the torque exerted by the 4.49 N force is τ = 2.50 m * 4.49 N = 11.225 N⋅m. For the cylinder with radius 1.14 m, the torque from the 9.13 N force is τ = 1.14 m * 9.13 N = 10.4082 N⋅m. Since both torques are acting to rotate the system in the same direction (counterclockwise), we can sum them to find the net torque.
The magnitude of the net torque is 11.225 N⋅m + 10.4082 N⋅m = 21.6332 N⋅m, or to three decimal places, 21.633 N⋅m.
You have just completed the first part of this lab and have five time values for a particular height: 1.8, 1.7, 1.9, 0.8, and 1.9 seconds. a) Give one quantitative reason why you think that 0.8 sec is or is not consistent with the other measurements.
Answer: because 1.8 is an outlier.
Explanation: it can been seen that from the data given to us, 1.7, 1.8, 1.9, 1.9 are all close to each other with a difference of 0.2 or 0.1.
By considering 0.8, this value (0.8) is at a very far distance away from other observational data with a difference of at least 0.9.
This henceforth makes 0.8 an outlier ( a data which is at a far distance away from other observational data)
The 0.8-second measurement is likely inconsistent with the others due to its significant deviation from the clustered times around 1.8 seconds, pointing to it as an outlier potentially caused by an experimental error.
One quantitative reason to suspect that the time value of 0.8 seconds is inconsistent with the other measurements of 1.8, 1.7, 1.9, and 1.9 seconds is the concept of measurement uncertainties and standard deviation. When comparing the given times, we see that the majority of the values are clustered around 1.8 seconds, suggesting a certain range of natural variability in repeated measurements. However, 0.8 seconds is significantly lower than the others, indicating that it is an outlier and might have been affected by experimental error or a faulty measurement. The use of mean and standard deviation can give us a formal way to assess consistency among data. If we calculated the mean and standard deviation, we would likely find 0.8 seconds to fall outside an acceptable range based on the precision of our measurements, reinforcing the idea that this result is inconsistent with the others.
Three identical resistors, when connected in series, transform electrical energy into thermal energy at a rate of 12 W (4.0 W per resistor). Part A Determine the power consumed by the resistors when connected in parallel to the same potential difference. Express your answer with the appropriate units.
Answer:
108 Watts
Explanation:
The total circuit resistance when the resistors are connected in series is
R + R + R = 3R
When he resistors are connected in parallel, the resistance reduces from 3R in the series circuit to become;
[tex]\frac{1}{R} + \frac{1}{R} + \frac{1}{R}[/tex]
= [tex]\frac{R}{3}[/tex] Ω
[tex]Power = \frac{V^{2}}{R}[/tex]
The voltage supply was given to be constant for both the series and parallel circuits. This implies that V² is constant and power is inversely proportional to resistance.
Therefore;
Power for the parallel connected circuit = [tex]\frac{3R}{\frac{R}{3} } * 12 W[/tex]
= 9 × 12 W = 108 Watts
Answer:
Explanation:
Potential difference, V and let each resistance, R
Resistors are in series, total resistance, Rₓ = R1 + R2 + R3
= R + R + R
= 3R
Power, P = V²/Rₓ
12 = V²/3R
V²/R = 36
Resistors are in parallel, total resistance, 1/Rₓ = 1/R1 + 1/R2 + 1/R3
Rₓ = R/3
P = V²/Rₓ
P = V²/(R/3)
P = 3(V²/R)
= 3(36)
= 108 W.
ou place the spring vertically with one end on the floor. You then drop a book of mass 1.40 kgkg onto it from a height of 0.800 mm above the top of the spring. Find the maximum distance the spring will be compressed.
Complete question:
A spring of negligible mass has force constant k = 1600 N/m. (a) How far must the spring be compressed for 3.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a 1.40-kg book onto it from a height of 0.800 m above the top of the spring. Find the maximum distance the spring will be compressed.
Answer:
(a) 0.063 m
(b) 0.126 m
Explanation:
Given;
force constant, K = 1600 N/m
Part (a)
Elastic potential energy is given as;
U = ¹/₂Kx²
where;
x is the extension in the spring
[tex]x = \sqrt{\frac{2U}{K} } = \sqrt{\frac{2*3.2}{1600} } = 0.063 \ m[/tex]
Part (b)
given;
mass of the book, m = 1.4 kg
height above the spring from which the book was dropped, h = 0.8 m
From the principle of conservation of energy;
Gravitational potential energy = Elastic potential energy
mgH = ¹/₂Kx²
H is the total vertical distance from floor to 0.8 m = maximum distance the spring will be compressed + h
let the maximum distance = A
mg(A+h) = ¹/₂KA²
1.4 x 9.8(A + 0.8) = ¹/₂ x 1600A²
13.72 (A + 0.8) = 800A²
13.72A + 10.976 = 800A²
800A² - 13.72A - 10.976 = 0
This is a quadratic equation, and we solve using formula method, where a = 800, b = - 13.72 and c = - 10.976
A = 0.126 m
Final answer:
The question involves determining the maximum compression of a spring by using energy conservation to equate the gravitational potential energy of a falling book to the elastic potential energy of the spring. The calculation needs the spring constant, which is not given in the question.
Explanation:
The student's question involves finding the maximum compression of the spring after dropping a 1.40 kg book on it from a certain height. This is a classic energy conservation problem in physics, where the gravitational potential energy of the book is converted into the elastic potential energy of the spring upon impact.
Firstly, the initial gravitational potential energy (Ug) of the book can be calculated using Ug = mgh, where m is the mass of the book, g is the acceleration due to gravity (approximately 9.81 m/s2), and h is the height from which the book is dropped.
The elastic potential energy stored in the spring (Ue) at maximum compression can be expressed as Ue = (1/2)kx2, where k is the spring constant and x is the compression of the spring.
Assuming no energy is lost due to friction or air resistance, energy conservation dictates that the initial gravitational potential energy will equal the elastic potential energy at the point of maximum compression. Therefore, mgh = (1/2)kx2. To find the maximum compression, x, you would rearrange this equation to solve for x and plug in the values for m, g, h, and k.
The actual computation requires the value of the spring constant k, which was not provided in the question. If it were provided, one would simply calculate the maximum compression using the specified formula.
A cylinder with moment of inertia 41.8 kg*m^2 rotates with angular velocity 2.27 rad/s on a frictionless vertical axle. A second cylinder, with moment of inertia 38.0 kg*m^2, initially not rotating, drops onto the first cylinder and remains in contact. Since the surfaces are rough, the two eventually reach the same angular velocity. Calculate the final angular velocity.
Explanation:
When any body rotates about its axis , the angular momentum of the body remains constant .
Thus the product of moment of inertia and its angular velocity remains constant .
In first case
The moment of inertia of cylinder = 41.8 kg m²
and Angular velocity = 2.27 rad/s
Thus angular momentum L₁ = 41.8 x 2.27 N-m s
In the second case
The moment of inertia = 41.8 + 38.0 = 79.8 kg m²
Suppose the angular velocity = ω
Thus angular momentum L₂ = 79.8 x ω
But according to principle of conservation of momentum
L₁ = L₂
41.8 x 2.27 = 79.8 x ω
Thus ω = [tex]\frac{41.8x2.27}{79.8}[/tex]
ω = 1.19 rad/s
Answer:
Final Angular Velocity=[tex]\omega[/tex]=1.189 rad/s
Explanation:
Given Data:
Moment of Inertia of 1st cyclinder=[tex]I_1=41.8 kg/m^{2}[/tex]
Angular Velocity of 1st cyclinder=[tex]\omega_1[/tex]=2.27 rad/s
Moment of Inertia of After contact=[tex]I_2=(41.8+38) kg/m^{2}=79.8 kg/m^{2}[/tex]
Required:
Final Angular velocity =[tex]\omega[/tex]=?
Formula:
Angula Momentum=L=[tex]I\omega[/tex]
Solution:
According to the conservation of angular momentum:
[tex]L_1=L_2[/tex]
[tex]I_1 \omega_1=I_2\omega\\41.8*2.27=(41.8+38)*\omega\\\omega=\frac{41.8*2.27}{79.8}\\\omega= 1.189\ rad/s[/tex]
Final Angular Velocity=[tex]\omega[/tex]=1.189 rad/s
A bullet is fired with a muzzle velocity of 1446 ft/sec from a gun aimed at an angle of 5 degrees above the horizontal. Find the vertical component of the velocity.
Answer:
126.03 ft/sec
Explanation:
From the question above,
V₁ = VsinФ............. Equation 1
Where V₁ = vertical component of the velocity, V = Velocity acting on the x-y plane, Ф = angle to the horizontal.
Given: V = 1446 ft/sec, Ф = 5°
Substitute into equation 1
V₁ = 1446sin(5)
V₁ = 1446(0.0872)
V₁ = 126.03 ft/sec.
Hence the vertical component of the velocity = 126.03 ft/sec
Why do you think it would be more practical to use an electromagnet to move scrap metal than to use a permanent magnet?
Using an electromagnet to lift scrap metal is advantageous because it can be turned off to release the metal, and its strength can be adjusted to handle different loads.
Using an electromagnet to move scrap metal is more practical than using a permanent magnet for several reasons. Firstly, an electromagnet can be turned on and off, useful when you want to release the metal after lifting it. This is not possible with a permanent magnet, which would require a physical effort to detach the metal pieces. Secondly, the strength of an electromagnet can be adjusted by controlling the electric current. This allows for strong magnetic effects that can be finely tuned for the weight and type of scrap being lifted. Lastly, industrial electromagnets can be designed to lift thousands of pounds of metallic waste, which might not be feasible with the size and strength of a permanent magnet.
However, there are limits to how strong electromagnets can be made, mainly due to coil resistance leading to overheating. In cases where extremely strong magnetic fields are necessary, such as in particle accelerators, superconducting magnets may be employed, although these also have their limits, since superconducting properties can be destroyed by excessively strong magnetic fields.
A projectile is fired with initial speed vo at an angle of 45o above the horizontal. Assume no air resistance.
i: During the flight, the x-component of the projectile's momentum remains constant.
ii: During the flight, the y-component of the projectile's momentum remains constant.
a) i is True and ii is False
b) i is True and ii is True
c) i is False and ii is True
d) i is False and ii is False
Answer:
The correct answer is a
Explanation:
At projectile launch speeds are
X axis vₓ = v₀ = cte
Y axis [tex]v_{y}[/tex] = v_{oy} –gt
The moment is defined as
p = mv
For the x axis
pₓ = mvₓ = m v₀ₓ
As the speed is constant the moment is constant
For the y axis
p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)
Speed changes over time, so the moment also changes over time
Let's examine the answer
i True
ii False. The moment changes with time
The correct answer is a
In projectile motion without air resistance, the horizontal component of momentum remains constant while the vertical component changes due to gravity. A projectile with initial velocity 2i + j will have a final velocity of 2i - 2j before striking the ground. A projectile with velocity 2i + 3j m/s is ascending towards its maximum height.
Explanation:The student's question regards the momentum components of a projectile fired at a 45-degree angle assuming no air resistance. The correct answer to the student's multiple-choice question is that statement (i) is True and (ii) is False. This is because, in projectile motion, the horizontal component of momentum, which is dependent on the horizontal component of velocity (Vx), remains constant due to the absence of horizontal forces acting on the projectile (assuming no air resistance). On the other hand, the vertical component of momentum changes because gravity acts in the vertical direction, affecting the vertical component of velocity (Vy) over time.
Considering the examples provided, the velocity of a projectile with an initial velocity of 2i + j (in terms of unit vectors i and j) before striking the ground is 2i - 2j, considering that gravity (g = 10 m/s²) acts downward, affecting only the vertical component of velocity. For the second example, a projectile with a velocity of 2i + 3j m/s could be either ascending or descending, but since the vertical component of velocity is still positive, it suggests that the projectile is ascending to the maximum height. So, the answer would be option (d).
Tarik winds a small paper tube uniformly with 189 turns 189 turns of thin wire to form a solenoid. The tube's diameter is 7.99 mm 7.99 mm and its length is 2.19 cm 2.19 cm . What is the inductance, in microhenrys, of Tarik's solenoid?
Answer:
102.8 μH
Explanation:
The (self) inductance of a coil based on its own geometry is given as
L = (μ₀N²A)/l
where
μ₀ = magnetic constant = (4π × 10⁻⁷) H/m
N = number of turns = 189
A = Cross sectional Area = (πD²/4) = (π×0.00799²/4) = 0.00005014 m²
l = length of the solenoid = 2.19 cm = 0.0219 m
L = (4π × 10⁻⁷ × 189² × 0.00005014)/0.0219
L = 0.0001028131 H = (1.028 × 10⁻⁴) H = (102.8 × 10⁻⁶) H = 102.8 μH
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A 84.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 36.0 m/s. If both are initially at rest and if the ice is frictionless, how far (in m) does the player recoil in the time it takes the puck to reach the goal 24.0 m away? (Enter the magnitude.)
Answer: 3333333222135790075
Explanation:Set term u equal to initial velocity for simplicity
Set V equal to final velocity for simplicity
2
To begin this problem, one must look at the system to have multiple stages. These being before and after hitting the puck. In these first few steps, we look at BEFORE the human hits the puck
3
This collision is elastic because the puck and the human do not join together after interaction
4
Because the initial velocity of both the puck and the human are both 0, the terms on the left of the equal sign become 0
5
Solving for the final velocity of the human gives this formula. This number should be negative as the negative indicates the direction he is going (left)
The final velocity of the puck is already given in the problem
6
Because the ice is frictionless, the final velocity before hitting the puck is equal to the initial velocity after hitting the puck
Now we begin to look at the system AFTER the puck has been hit
7
Using the formula for final position allows us to solve for time it takes the puck to travel the distance given
8
9
Solve for time
10
We can now use the formula for the final position of the human to solve for the final answer
11
12nuewnfunw
Plugging in formulas from steps 5 and 9 gives the final answer
Again, this number should be negative as the negative sign denotes the direction the human is going. Because the problem does not ask for snijndij hinu9nub hvtj c v7 yf jhmb tfgnb nb fyhgbv
The distance traveled by the player ( recoil ) in the time the puck reaches the goal is 0.043m.
What is law of conservation of linear momentum?According to the law of conservation of momentum, the sum of the momentum of the object before and after the collision must be equal.
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
where m₁ and m₂ is the mass of the objects, u₁ and u₂ are initial speed while v₁ & v₂ is final speed.
Given the initial velocity of the player is u₁ = 0 and the puck is u₂ = 0
The mass of the player m₁ = 84 Kg
The mass of the puck, m₂ = 0.150 Kg
The final velocity of the puck, v₂ = 36 m/s
From the law of conservation of momentum, find the velocity of the player:
m₁ u₁ + m₂ u₂ =m₁ v₁ + m₂v₂
84 × 0 + m ×0 = 84 × v + 0.150 × 36
v = - 0.064 m/s
A negative sign shows the player and puck moving in the opposite direction.
Now, we calculte the time taken for the puck to trach the goal:
Time = Distance/ Velocity
t = 24/36 = 0.667 sec
Next, we calculate the distance traveled by the player( recoil ) in the time of 0.667 seconds.
Distance = Velocity× time
S = 0.064 ×0.667
S = 0.043m
Therefore, the distance covered by the player in the time the puck reaches the goal is 0.043m.
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There is a spot of paint on the front wheel of the bicycle. Take the position of the spot at time t=0 to be at angle θ=0 radians with respect to an axis parallel to the ground (and perpendicular to the axis of rotation of the tire) and measure positive angles in the direction of the wheel's rotation. What angular displacement θ has the spot of paint undergone between time 0 and 2 seconds? Express your answer in radians using three significant figures.
Answer:
[tex]\int\limits^2_0 {w} \, dt[/tex]
Explanation:
If Angular velocity w (omega ) is given, which is defined as, w = (change in angle)/(change in unit time).
then simply taking integral from 0 to 2 gives us the answer.
Answer:
Change in theta = 0.793rad
Explanation:
Please see attachment below.
A wire is formed into a circle having a diameter of 10.3 cm and is placed in a uniform magnetic field of 2.98 mT. The wire carries a current of 5.00 A. Find the maximum torque on the wire.
Answer:
T(max) = 1.17 × 10⁻⁴Nm
= 117μNm
Explanation:
T = BIA sinθ
A = area enclosed
θ = angle between normal plane
for max. torque θ = 90, (sin90° =1)
T = BIA sin90°
T= BI (πd/4)
T = [tex]T_m_a_x = \frac{1}{4} (2.98 * 10^-^3)(5)(\pi )\\T_m_a_x = 1.17 * 10^-^4Nm\\T_m_a_x = 117UNm[/tex]
Two thermally insulated cylinders, A and B, of equal volume, both equipped with pistons, are connected by a valve. Initially A has its piston fully withdrawn and contains a perfect monatomic gas at temperature T, while B has its piston fully inserted, and the valve is closed. Calculate the final temperature of the gas after the following operations, which each start with the same initial arrangement. The thermal capacity of the cylinders is to be ignored.
(a) The valve is fully opened and the gas slowly drawn into B by pulling out the piston B; piston A remains stationary.
(b) Piston B is fully withdrawn and the valve is opened slightly; the gas is then driven as far as it will go into B by pushing home piston A at such a rate that the pressure in A remains constant: the cylinders are in thermal contact
In scenario (a), the temperature of the gas decreases since it is an adiabatic process. In scenario (b), the final temperature depends on the initial and final volumes and the presence of heat exchange.
Explanation:In scenario (a), the gas is slowly drawn into cylinder B by pulling out the piston B while cylinder A remains stationary. Since the cylinders are thermally insulated, there is no heat exchange with the surroundings, and the process is adiabatic. As a result, the temperature of the gas decreases.
In scenario (b), the gas is driven as far as it will go into cylinder B by pushing the piston A at a rate that maintains constant pressure in cylinder A. In this case, the process is isobaric, and the gas expands while exerting work. Since there is thermal contact between the cylinders, heat can be exchanged between the gas and the surroundings, leading to a change in temperature.
To calculate the final temperatures in both scenarios, it is necessary to know the initial pressure and volume of the gas in cylinder A, as well as the final volume of the gas in cylinder B, in each case.
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For the adiabatic free expansion in scenario (a), the final temperature remains the same as the initial temperature. In scenario (b), the gas undergoes adiabatic compression followed by isothermal expansion, resulting in the final temperature being the same as the initial temperature, T.
Final Temperature Calculation in Two Scenarios
Let's explore the gas dynamics in two scenarios involving thermally insulated cylinders A and B containing a perfect monatomic gas at initial temperature T.
Scenario (a): Valve is Fully Opened and Gas is Drawn into B
Initial conditions:
Volume of cylinder A (Va): VVolume of cylinder B (Vb): 0Initial temperature (T): TSince cylinder A is insulated and its piston remains stationary, there is no work done on or by the gas in cylinder A. The gas expands into cylinder B, which is an adiabatic free expansion:
The final temperature (T') will be the same as the initial temperature (T). Because the process is adiabatic and involves no work, the internal energy (and thus temperature) of the gas remains unchanged.
Scenario (b): Adiabatic Compression of A and Isothermal Expansion into B
Initial conditions:
Volume of cylinder A (Va): VVolume of cylinder B (Vb): 0Initial temperature (T): TWe perform an adiabatic process on cylinder A and isothermal expansion into B. The adiabatic compression affects the temperature of the gas in A before the gas is allowed to expand isothermally:
1. Adiabatic compression in A:
For adiabatic processes, TVγ-1 = constant, where γ = 5/3 for a monatomic ideal gas. The final volume of gas in A is V/2 because the gas expands equally into B.
2. Isothermal expansion into B:
After the compression, we allow the gas to expand isothermally into B at constant temperature T. Hence, the final temperature in both cylinders will be T because the gas reaches thermal equilibrium with the environment.
Assume: The small objects are point particles. The system of seven small 2 kg objects is rotating at an angular speed of 9 rad/s. The objects are connected by light, flexible spokes that can be lengthened or shortened.
A) What is the initial angular momentum of the object?
B)What is the new angular speed if the spokes are shortened from 9 m to 6 m ?
C)What is the new angular momentum of the object?
Answer: a) [tex]L = 10206 kg \cdot \frac{m^{2}}{s}[/tex], b) [tex]\omega_{f} = 20.25 rad/s[/tex], c) [tex]L = 10206 kg \cdot \frac{m^{2}}{s}[/tex]
Explanation:
The angular momentum of a system of particles rotating around an axis is:
[tex]L = \sum_{i=1}^{7} r_{i}^{2} \cdot m_{i} \cdot \omega[/tex]
a) The previous expression can be simplified to find the initial angular momentum:
[tex]L = 7 \cdot r^{2} \cdot m \cdot \omega\\L = 7 \cdot (9 m)^2 \cdot (2kg) \cdot (9 \frac{rad}{s} )\\L = 10206 kg \cdot \frac{m^{2}}{s}[/tex]
b) The new angular speed is calculated from the Principle of Angular Momentum Conservation, since there are no external forces influencing over the system:
[tex]L = 7 \cdot r_{f}^2 \cdot (m) \cdot \omega_{f}[/tex]
[tex]\omega_{f}=\frac{L}{7 \cdot r_{f}^2\cdot m}[/tex]
[tex]\omega_{f} = 20.25 rad/s[/tex]
c) The angular momentum remains constant, since there is no information that indicates the presence of external forces influencing the system.
(A) The initial angular momentum of the object is 10206 kgm²/s
(B) The new angular speed is 20.25 rad/s
(C) The angular momentum does not change
Conservation of angular momentum:All 7 objects have a mass of m = 2kg. The length of the spokes is R = 9m.
(A) The angular momentum of a system is given by:
L = Iω
where I is the moment of inertia of the system
ω is the angular speed = 9 rad/s (given)
L = 7×mR²×ω
L = 7×(2×9²)×9 kgm²/s
L = 10206 kgm²/s
(B) According to the law of conservation of angular momentum, the angular momentum of the system must remain conserved.
So, the new angular momentum is:
L' = 7×(2×6²)×ω'
where ω' is the new angular momentum
10206 = 7×(2×6²)×ω'
ω' = 20.25 rad/s
(C) The new angular momentum of the object is same as the original angular momentum of the object since there is no dissipative force so the angular momentum must be conserved.
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