Answer:
0.0085 Wb
Explanation:
a = Side of square
r = Radius of circle
[tex]\phi_s[/tex] = Magnetic flux through square loop = [tex]6.68\times 10^{-3}\ Wb[/tex]
Magnetic flux is given by
[tex]\phi=BA[/tex]
For square
[tex]\phi_{s}=Ba^2[/tex]
The length of the square will be equal to the circumference of the circle
[tex]4a=2\pi r\\\Rightarrow r=\frac{2a}{\pi}[/tex]
For circle
[tex]\phi_{c}=B\pi r^2\\\Rightarrow \phi_{c}=B\pi \left(\frac{2a}{\pi}\right)^2\\\Rightarrow \phi_c=Ba^2\frac{4}{\pi}\\\Rightarrow \phi_c=\phi_s\frac{4}{\pi}\\\Rightarrow \phi_c=6.68\times 10^{-3}\frac{4}{\pi}\\\Rightarrow \phi_c=0.0085\ Wb[/tex]
The flux that passes through the circular loop is 0.0085 Wb
The magnetic flux that passes through the circular loop is 0.00848 Wb.
What is the magnetic flux?The magnetic flux is defined as the measurement of the total magnetic field which passes through a given area.
Given that the magnetic flux that passes through the square loop is 6.68 × 10-3 Wb.
Let's consider that r is the radius of the circle and a is the side of the square.
The magnetic flux through the square loop is given below.
[tex]\phi_s = Ba^2[/tex]
Where B is the magnetic field.
The length of the square will be equal to the circumference of the circle
[tex]4a = 2\pi r[/tex]
[tex]r = \dfrac {2a}{\pi}[/tex]
The magnetic flux through the circular loop is given below.
[tex]\phi = BA[/tex]
Where A is is the cross-sectional area of the circular loop.
[tex]\phi = B \pi r^2[/tex]
[tex]\phi = B \pi (\dfrac {2a}{\pi})^2[/tex]
[tex]\phi = B\pi \times \dfrac {4a^2}{\pi^2}[/tex]
[tex]\phi = Ba^2 \dfrac {4}{\pi}[/tex]
[tex]\phi = \phi_s \dfrac {4}{3.14}[/tex]
[tex]\phi = 6.68 \times 10^{-3} \times 1.27[/tex]
[tex]\phi = 0.00848 \;\rm Wb[/tex]
Hence we can conclude that the magnetic flux that passes through the circular loop is 0.00848 Wb.
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There is an almost isotropic radiation coming to Earth from outer space from all direction, the so-called background radiation. This radiation corresponds to a temperature of 3 K. To which wavelength should we tune our antenna in order to detect this radiation?A. 966 mmB. 8.69 mmC. 8.69 μmD. 966 μm
Answer:
D. 966 μm
Explanation:
b = Wien's displacement constant = [tex]2.89\times 10^{-3}\ mK[/tex]
T = Temperature
[tex]\lambda_m[/tex] = Peak wavelength
Here the Wien's displacement law is used
[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.89\times 10^{-3}}{3}\\\Rightarrow \lambda_m=0.000963\\\Rightarrow \lambda_m=963\times 10^{-6}\ m=963\ \mu m\approx 966\ \mu m[/tex]
The wavelength we should tune our antenna in order to detect this radiation is 966 μm
Starting with an initial value of P(0) = 75, the population of a prairie dog community grows at a rate of
P(t) =50 − t/5 (in units of prairie dogs/month), for 0 <_t <_250.
a. What is the population 99 months later?
b. Find the population P(t) for 0 <_t <_250.
A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straight down the slope. If the slope were frictionless, the cylinder would slide down the slope without rotating. What minimum coefficient of static friction is needed for the cylinder to roll down without slipping?
Answer:
[tex]\mu_s=\frac{1}{3}\tan \theta[/tex]
Explanation:
Let the minimum coefficient of static friction be [tex]\mu_s[/tex].
Given:
Mass of the cylinder = [tex]M[/tex]
Radius of the cylinder = [tex]R[/tex]
Length of the cylinder = [tex]L[/tex]
Angle of inclination = [tex]\theta[/tex]
Initial velocity of the cylinder (Released from rest) = 0
Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.
We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,
[tex]\tau =I\alpha[/tex]
Now, angular acceleration is given as:
[tex]\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}[/tex]
Also, moment of inertia for a cylinder is given as:
[tex]I=\frac{MR^2}{2}[/tex]
Therefore, the torque acting on the cylinder can be rewritten as:
[tex]\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1[/tex]
Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are [tex]mg\sin \theta\ and\ f[/tex]. The net force acting along the incline is given as:
[tex]F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2[/tex]
Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.
So, [tex]N=Mg\cos \theta[/tex]
Plugging in [tex]N=Mg\cos \theta[/tex] in equation (2), we get
[tex]F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3[/tex]
Now, as per Newton's second law,
[tex]F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4[/tex]
Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.
[tex]\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)[/tex]
Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:
[tex]\mu_s=\frac{1}{3}\tan \theta[/tex]
The minimum coefficient of static friction that needed for cylinder to roll down without slipping is [tex]\mu_s= \frac{tan\theta}{3}[/tex]
Explanation:A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straight down the slope. If the slope were frictionless, the cylinder would slide down the slope without rotating. What minimum coefficient of static friction is needed for the cylinder to roll down without slipping?
Given: radius R, length L, angle θ, and mass M
We need to calcuate the minimum static friction coefficient. It is useful so the cylinder will roll without slipping down the incline. The cylinder is also released from rest. As the cylinder is rolling, we have to consider the moment of inertia. Rolling of cylinder is happened due to the friction force
By applying Newton law of motion
[tex]F = M a\\\tau = I \alpha\\\tau = I \frac{a}{R} \\\tau = \frac{1}{2} M R^2 \frac{a}{R}[/tex]
From diagram
[tex]Mg sin\theta - f_{fr} = Ma\\f_{fr} = \mu_s N\\f_{fr} = \mu_s Mg cos \theta\\a = g sin \theta - \mu_s cos \theta[/tex]
Then also
[tex]\tau = f_{fr} R\\f_{fr} = \frac{Ma}{2} \\\mu_s Mg cos \theta = \frac{Mg (sin\theta - \mu_s cos \theta)}{2} \\\frac{3}{2} \mu_s cos\theta = \frac{sin\theta}{2}\\ \mu_s = \frac{tan\theta}{3}[/tex]
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Bonnie sits on the outer rim of a merry-go-round, and Jill sits midway between the center and the rim. The merrygo-round makes one complete revolution every 2 seconds. Jill's linear velocity is:
a. four times Bonnie's.
b. one-quarter of Bonnie's.
c. the same as Bonnie's.
d. twice Bonnie's.
e. half of Bonnie's.
Answer:
e. half of Bonnie's.
Explanation:
Jill and Bonnie move in a circular path with the same angular speed of the merry-go-round.
The tangential velocity of the body is calculated as follows:
v = ω*R
where:
v is the tangential velocity or linear velocity (m /s)
ω is the angular speed (rad/s)
R is radius where the body is located from the center of the circular path
Data
1 rev = 2π rad
ω = 1 rev/2s = 2π rad/2s = π rad/s
R : radio of the merry-go-round
Bonnie's linear velocity (vB)
vB = ω*R = π*R (m/s)
Jill's linear velocity (vJ)
vJ = ω*(R /2) = (1/2 )(π*R) (m/s)
Jill's linear velocity is half of Bonnie's because linear velocity depends on the distance from the center of the merry-go-round, and Jill is situated at half the radius from the center compared to Bonnie.
Explanation:The question pertains to understanding the relationship between linear and angular velocities of points located at different radii on a rotating platform, such as a merry-go-round. Given that the merry-go-round makes one complete revolution every 2 seconds, the angular velocity is the same for all points on the merry-go-round. However, linear velocity depends on the radius. Bonnie, who sits on the outer rim, would have the highest linear velocity because the further you are from the center, the greater the distance you cover per rotation. Jill, sitting midway, would cover half the distance in the same time period as Bonnie, resulting in Jill having a linear velocity that is half of Bonnie's.
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The energy in an oscillating LC circuit containing a 1.57 H inductor is 5.76 μJ. The maximum charge on the capacitor is 201 μC. For a mechanical system with the same period, find the
(a) mass
(b) spring constant
(c) maximum displacement
(d) maximum speed.
Answer:
(a)1.57 kg
(b) 281.17 N/m
(c) 201 micrometer
(d) [tex]2.69\times 10^{-3}m/sec[/tex]
Explanation:
We have given that value of inductor L = 1.57 Henry
Inductive energy [tex]E=5.76\mu j=5.76\times 10^{-6}J[/tex]
Maximum charge [tex]Q=201\mu C=201\times 10^{-6}C[/tex]
(A) In electrical mechanical system mass corresponds to inductance
So mass will be m = 1.57 kg
(B) We have given energy [tex]E=\frac{Q^2}{2C}[/tex]
[tex]C=\frac{Q^2}{2E}=\frac{(201\times 10^{-6})^2}{2\times 5.7\times 10^{-6}}=3543.94\times 10^{-6}[/tex]
In electrical mechanical system spring constant is equivalent to [tex]\frac{1}{C}[/tex]
So spring constant [tex]k=\frac{1}{C}=\frac{1}{3543.94\times 10^{-6}}=282.17N/m[/tex]
(c) Displacement is equivalent to maximum charge
So displacement will be [tex]x=201\mu m[/tex]
(d) Maximum speed is correspond to maximum current
As maximum current [tex]i_m=\frac{Q}{\sqrt{LC}}=\frac{201\times 10^{-6}}{\sqrt{1.57\times 3543.94\times 10^{-6}}}=2.69\times 10^{-3}A=2.69\times 10^{-3}m/sec[/tex]
Niobium metal becomes a superconductor when cooled below 9 K. Its superconductivity is destroyed when the surface magnetic field exceeds 0.100 T. In the absence of any external magnetic field, determine the maximum current a 4.50-mm-diameter niobium wire can carry and remain superconducting.
To develop the problem it is necessary to apply the concepts related to Magnetic Field.
The magnetic field is defined as
[tex]B = \frac{\mu_0 I}{2\pi r}[/tex]
Where,
[tex]\mu_0 =[/tex] Permeability constant in free space
r = Radius
I = Current
Our values are given as,
B = 0.1T
d = 4.5mm
r = 2.25mm
If the maximum current that the wire can carry is I, then
[tex]B = \frac{\mu_0 2I}{4\pi r}[/tex]
[tex]I = \frac{Br}{2\frac{\mu_0}{4\pi}}[/tex]
[tex]I = \frac{(0.1T)(2.25*10^{-3}m)}{2(1*10^{-7}N/A^2)}}[/tex]
[tex]I = 1125A[/tex]
Therefore the maximum current is 1125A
A small loop of area A is inside of, and has its axis in the same direction as, a long solenoid of n turns per unit length and current i. If i = i0 sin(ωt), find the emf in the loop. (Use the following as necessary: A, n, i0, ω, t, and μ0.)
The EMF is then found using Faraday's Law, resulting in ε = - μ0 n i0 A ω cos(ωt).
To find the EMF induced in a small loop of area A placed inside a long solenoid with n turns per unit length and a current i(t) = i0 sin(ωt), we use the concept of changing magnetic flux.
First, determine the magnetic field inside the solenoid. For a solenoid, the magnetic field B inside is given by:
B = μ0nI
Since the current I is changing with time, so does the magnetic field:
B(t) = μ0n i0 sin(ωt)
The flux Φ through the small loop of area A is:
Φ = B(t) * A = μ0n i0 sin(ωt) * A
The induced EMF (ε) can be found using Faraday's Law of Induction:
ε = -dΦ/dt
Taking the derivative of Φ with respect to time t:
ε = -d/dt (μ0n i0 sin(ωt) * A)
Using the chain rule, we get:
ε = - μ0n i0 A d/dt (sin(ωt))
The derivative of sin(ωt) is ω cos(ωt):
ε = - μ0n i0 A ω cos(ωt)
Therefore, the EMF induced in the loop is:
ε = - μ0n i0 A ω cos(ωt)
A crate is sliding down a ramp that is inclined at an angle 30.6 ° above the horizontal. The coefficient of kinetic friction between the crate and the ramp is 0.360. Find the acceleration of the moving crate.
Answer:[tex]1.95 m/s^2[/tex]
Explanation:
Given
inclination [tex]\theta =30.6^{\circ}[/tex]
coefficient of kinetic friction [tex]\mu =0.36 [/tex]
As crate is moving Down therefore friction will oppose the motion
using FBD
[tex]mg\sin \theta -f_r=ma [/tex]
[tex]f_r=\mu N[/tex]
[tex]f_r=\mu mg\cos \theta [/tex]
[tex]mg\sin \theta -\mu mg\cos \theta =ma[/tex]
[tex]a=g\sin \theta -\mu g\cos \theta [/tex]
[tex]a=g(\sin (30.6)-0.36\cdot \cos (30.6))[/tex]
[tex]a=9.8\times 0.199[/tex]
[tex]a=1.95 m/s^2[/tex]
To determine the crate's acceleration down the ramp, we calculate and subtract the kinetic friction from the component of the gravitational force acting down the slope. We then divide the resulting net force by the mass of the crate.
Explanation:To find the acceleration of the crate, we need to note that there are two forces acting on the crate as it moves down the ramp: the force due to gravity and the force due to friction. Since these two forces act in opposite directions, the crate's acceleration will be less than if there were no friction.
First, we need to calculate the component of the gravitational force acting down the slope, which can be found using the formula mg sin θ, where m is the mass of the crate, g is the acceleration due to gravity, and θ is the angle of the incline.
Next, we calculate the kinetic friction using the formula μk mg cos θ, where μk is the coefficient of kinetic friction.
The net force acting on the crate is the difference between these two forces and since F = ma, where F is the net force and m is the mass of the crate, we can solve for acceleration (a) by dividing the net force by the mass of the crate.
This calculation will give us the acceleration of the crate down the incline, taking into account the presence of friction.
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A window in a house has a rectangular shape of 2.0 m by 1.0 m. The glass in the window is 0.5 cm thick, with a thermal conductivity of 0.84 W/(m oC). The temperature in the house is 20.5 °C, and the temperature outside is 4 oC. How much heat is lost through this window in one minute? Don't use scientific notations
Answer:
P = 5280 W
Explanation:
The conductivity of the materials determines that heat flows from the hot part to the cold part, the equation for thermal conductivity transfer is
P = Q / t = k A ([tex]T_{h}[/tex] -[tex]T_{c}[/tex]) / L
Where k is the thermal conductivity of the glass 0.8 W / ºC, A the area of the window, T the temperature and L is glass thickness
Let's calculate the window area
A = l * a
A = 2.0 1.0
A = 2.0 m²
Let's replace
L = 0.5 cm (1 m / 100 cm) = 0.005 m
P = 0.8 2 (20.5 - 4) / 0.005
P = 5280 W
The production of sound during speech or singing is a complicated process. Let's concentrate on the mouth. A typical depth for the human mouth is about 8.00 cm, although this number can vary. (Check it against your own mouth.) We can model the mouth as an organ pipe that is open at the back of the throat. A) What are the wavelengths of the first four harmonics you can produce if your mouth is open? (in cm/4 answers) B)What are the frequencies of the first four harmonics you can produce if your mouth is open? Use 354 (in Hz/4 answers) C)What are the wavelengths of the first four harmonics you can produce if your mouth is closed? (in cm/4 answers) D)What are the frequencies of the first four harmonics you can produce if your mouth is closed?
Answer:A) The wavelengths are:8cm for first harmonic
5.3cm for second harmonic
4cm for third harmonic
3.2cm for fourth harmonic
B) The frequencies are:
44.25Hz for first harmonic
66.38Hz for second harmonic
88.5Hz for third harmonic
110.63Hz for fourth harmonic
C) The wavelengths are:
10.66Cm for first harmonic
6.4 cm for second harmonic
4.57cm for third harmonic
3.5cm for fourth harmonic
D) The frequencies are:
33.19Hz for first harmonic
55.31 Hz for second harmonic
77.44Hz for third harmonic
99.56Hz for fourth harmonic
Explanation: since the mouth is being modelled as an organ pipe, then we can say that an open mouth is a model of an organ pipe open at both ends.
For an open pipe all harmonics are possible but only odd harmonic's are possible in closed pipes. A closed mouth is a model of a closed pipe, since closed pipes are open at one end.
Chloe read an essay that claims the body and mind, which are made up of different substances, interact both harmoniously and competitively in a living person. Which philosopher’s work is she most likely reading?
Answer:
René Descartes
Explanation:
https://quizlet.com/172472862/psychologys-early-history-flash-cards/
A centrifugal pump discharges 300 gpm against a head of 55 ft when the speed is 1500 rpm. The diameter of the impeller is 15.5 in. and the brake horsepower is 6.0. (a) Determine the efficiency of the pump. (b) What should be the speed of a geometrically similar pump of 15 in. diameter running at a capacity of 400 gpm?
Answer: Question 1: Efficiency is 0.6944
Question 2: speed of similar pump is 2067rpm
Explanation:
Question 1:
Flow rate of pump 1 (Q1) = 300gpm
Flow rate of pump 2 (Q2) = 400gpm
Head of pump (H)= 55ft
Speed of pump1 (v1)= 1500rpm
Speed of pump2(v2) = ?
Diameter of impeller in pump 1= 15.5in = 0.3937m
Diameter of impeller in pump 2= 15in = 0.381
B.H.P= 6.0
Assuming cold water, S.G = 1.0
eff= (H x Q x S.G)/ 3960 x B.H.P
= (55x 300x 1)/3960x 6
= 0.6944
Question 2:
Q = A x V. (1)
A1 x v1 = A2 x V2. (2)
Since A1 = A2 = A ( since they are geometrically similar
A = Q1/V1 = Q2/V2. (3)
V1(m/s) = r x 2π x N(rpm)/60
= (0.3937x 2 x π x 1500)/2x 60
= 30.925m/s
Using equation (3)
V2 = (400 x 30.925)/300
= 41.2335m/s
To rpm:
N(rpm) = (60 x V(m/s))/2 x π x r
= (60 x 41.2335)/ 2× π × 0.1905
= 2067rpm.
A typical meteor that hits the earth’s upper atmosphere has a mass of only 2.5 g, about the same as a penny, but it is moving at an impressive 40 km/s. As the meteor slows, the resulting thermal energy makes a glowing streak across the sky, a shooting star. The small mass packs a surprising punch. At what speed would a 900 kg compact car need to move to have the same kinetic energy
Answer:u=66.67 m/s
Explanation:
Given
mass of meteor [tex]m=2.5 gm\approx 2.5\times 10^{-3} kg[/tex]
velocity of meteor [tex]v=40km/s \approx 40000 m/s[/tex]
Kinetic Energy of Meteor
[tex]K.E.=\frac{mv^2}{2}[/tex]
[tex]K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}[/tex]
[tex]K.E.=2\times 10^6 J[/tex]
Kinetic Energy of Car
[tex]=\frac{1}{2}\times Mu^2[/tex]
[tex]=\frac{1}{2}\times 900\times u^2[/tex]
[tex]\frac{1}{2}\times 900\times u^2=2\times 10^6 [/tex]
[tex]900\times u^2=4\times 10^6[/tex]
[tex]u^2=\frac{4}{9}\times 10^4[/tex]
[tex]u=\frac{2}{3}\times 10^2[/tex]
[tex]u=66.67 m/s[/tex]
Answer:
v = 67 m/s
Explanation:
The meteor has a mass (m) of 2.5 g and a speed (v) of 40 km/s. In SI units:
2.5 g × (1 kg / 10³ g) = 2.5 × 10⁻³ kg
40 km/s × (10³ m / 1 km) = 4.0 × 10⁴ m/s
The kinetic energy (KE) is:
KE = 1/2 × m × v² = 1/2 × (2.5 × 10⁻³ kg) × (4.0 × 10⁴ m/s)² = 2.0 × 10⁶ J
A 900 kg compact car, with the same kinetic energy, must have the following speed.
KE = 1/2 × m × v²
2.0 × 10⁶ J = 1/2 × 900 kg × v²
v = 67 m/s
A 2.00 kg, frictionless block s attached to an ideal spring with force constant 550 N/m. At t = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 10.0 m/s. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Angular frequency, frequency, and period in shm.find:-a) the amplitude.b) the maximum acceleration of the block.c) the maximum force the spring exerts on the block.
The amplitude of the motion is 0.6 m. The maximum acceleration of the block is 165 m/s^2. The maximum force exerted by the spring on the block is 330 N.
Explanation:The problem you're asking about is related to Simple Harmonic Motion (SHM) associated with a spring-block system on a frictionless surface. An ideal spring obeys Hooke's law which states that the force it exerts is proportional to the displacement from its equilibrium position.
Amplitude of the motion can be determined using the principle of conservation of energy. In the absence of frictional forces, the kinetic energy of the block at maximum displacement converts entirely into potential energy of the spring. So, (1/2)mv^2 = (1/2)kA^2, where m is the mass of the block, v is the velocity of the block, k is the spring constant and A is the amplitude. Substituting the given values, we have A = sqrt((2*2*10^2)/550) = 0.6 m. The maximum acceleration a of the block happens at the maximum displacement, i.e., at the amplitude. According to Hooke's law, F = -kx, where F is the force exerted by the spring and x is the displacement. So, we have a = F/m = kA/m = 550*0.6/2 = 165 m/s^2. The maximum force the spring exerts on the block is when the block is at the amplitude. It equals F = kA = 550*0.6 = 330 N.Learn more about Simple Harmonic Motion here:
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A potter's wheel has the shape of a solid uniform disk of mass 13.0 kg and radius 1.25 m. It spins about an axis perpendicular to the disk at its center. A small 1.7 kg lump of very dense clay is dropped onto the wheel at a distance 0.63 m from the axis. What is the moment of inertia of the system about the axis of spin?
Answer:[tex]10.82 kg-m^2[/tex]
Explanation:
Given
Mass of solid uniform disk [tex]M=13 kg[/tex]
radius of disk [tex]r=1.25 m[/tex]
mass of lump [tex]m=1.7 kg[/tex]
distance of lump from axis [tex]r_0=0.63[/tex]
Moment of inertia is the distribution of mass from the axis of rotation
Initial moment of inertia of disk [tex]I_1=\frac{Mr^2}{2}[/tex]
[tex]I_1=\frac{13\times 1.25^2}{2}=10.15 kg-m^2[/tex]
Final moment of inertia [tex]I_f[/tex]=Moment of inertia of disk+moment of inertia of lump about axis
[tex]I_f=\frac{Mr^2}{2}+mr_0^2[/tex]
[tex]I_f=10.15+1.7\times 0.63^2[/tex]
[tex]I_f=10.15+0.674[/tex]
[tex]I_f=10.82 kg-m^2[/tex]
Find a glass jar with a screw-top metal lid. Close the lid snugly and put the jar into the refrigerator. Leave it there for about 10 minutes and then take the jar out and try to open the lid. (a) Did the lid become tighter or looser? Explain your observation.
Answer:
The lid becomes tighter
It becomes tighter because metals have a lower heat capacity than glass meaning their temperature drops (or increases) much faster than glass for the same energy change. So in this example, the metal will contract faster than the glass causing it to be more tighter around the glass.
19.A 20 kg sign is pulled by a horizontal force such that the single rope (originally vertical) holding the sign makes an angle of 21° with the vertical. Assuming the sign is motionless, findA) the magnitude of the tension in the rope andB) the magnitude of the horizontal force.
Answer:
A)[tex]T=209.94N[/tex]
B) [tex]F=75.24N[/tex]
Explanation:
Using the free body diagram and according to Newton's first law, we have:
[tex]\sum F_y=Tcos(21^\circ)-mg=0(1)\\\sum F_x=F-Tsin(21^\circ)=0(2)[/tex]
A) Solving (1) for T:
[tex]T=\frac{mg}{cos(21^\circ)}\\T=\frac{20kg(9.8\frac{m}{s^2})}{cos(21^\circ)}\\T=209.94N[/tex]
B) Solving (2) for F:
[tex]F=Tsin(21^\circ)\\F=(209.94N)sin(21^\circ)\\F=75.24N[/tex]
A ring made from brass has an inner radius of 2.50000 cm and an outer radius of 3.50000 cm, giving the ring a thickness of 1.00000 cm. The thermal expansion coefficient of brass is 0.000019/°C. If the temperature of the ring is increased from 20.0°C to 90.0°C, by how much does the thickness of the ring change? _______ cm
Answer:0.00133 cm
Explanation:
Given
change in temperature [tex]\Delta T=90-20=70^{\circ}C[/tex]
Thermal Expansion coefficient of brass [tex]\alpha =0.000019 /^{\circ}C[/tex]
thickness t=1 cm
change in thickness is given by
[tex]\Delta t=t\cdot \alpha \cdot \Delta T[/tex]
[tex]\Delta t=1\times 0.000019\times 70[/tex]
[tex]\Delta t=0.00133\ cm[/tex]
You are trying to determine the specific gravity of a solid object that floats in water. If m is the mass of your object, mS is the apparent mass of the combination of 2 masses with one (the sinker) submerged, and mOS is the apparent mass of the combination of 2 masses with both submerged, what is the formula for specific gravity
Answer:
Specific Gravity = m/[m(s)-m(os)]
Explanation:
Specific gravity, also called relative density, is the ratio of the density of a substance to the density of a reference substance. By this definition we need to find out the ratio of density of the object of mass m to the density of the surrounding liquid.
m = mass of the object
Weight in air
W (air) = mg, where g is the gravitational acceleration
Weight with submerged with only one mass
m(s)g + Fb = mg + m(b)g, consider this to be equation 1
where Fb is the buoyancy force
Weight with submerged with both masses
m(os)g + Fb’ = mg + m(b)g, consider this to be equation 2
equation 1 – equation 2 would give us
m(s)g – m(os)g = Fb’ – Fb
where Fb = D x V x g, where D is the density of the liquid the object is submerged in, g is the force of gravity and V is the submerged volume of the object
m(s)g – m(os)g = D(l) x V x g
m(s) – m(os) = D(l) x V
we know that Mass = Density x V, which in our case would be, D(b) x V, which also means
V = Mass/D(b), where D(b) is the density of the mass
Substituting V into the above equation we get
m(s) – m(os) = [D(l) x m)/ D(b)]
Rearranging to get the ratio of density of object to the density of liquid
D(b)/D(l) = m/[m(s)-m(os)], where D(b)/D(l) denotes the specific gravity
The New River Gorge Bridge in West Virginia is a steel arch bridge 518 m in length. How much does the total length of the roadway decking change between temperature extremes of –20.0°C and 35.0°C? The result indicates the size of the expansion joints that must be built into the structure.
Final answer:
The change in length of the New River Gorge Bridge due to thermal expansion can be calculated using the coefficient of linear expansion for steel and the temperature change noted.
Explanation:
The question is about calculating the change in length of a steel structure, the New River Gorge Bridge, due to thermal expansion between its temperature extremes. Using a standard linear expansion formula, ΔL = αLδT, where ΔL is the change in length, α is the coefficient of linear expansion for steel (typically around 11×10^-6 1/°C), L is the length of the steel at its coldest, and δT is the change in temperature. For the New River Gorge Bridge with a length of 518 m and a temperature change from -20.0°C to 35.0°C (a change of 55.0°C), the expansion can be calculated as follows: ΔL = (11×10^-6 1/°C)×518 m×55.0°C, which gives the total change in length of the roadway decking.
A beam of electons ( which have negative charge q ) is coming straight toward you. you put the north pole of a magnet directly above the beam. the magnetic field B from magnet points straight down. which way will the electron beam deflect?
Answer:
Towards left
Explanation:
Applying Fleming Right hand rule in which the middle finger denotes downward magnetic field. Thumb pointing towards you is the direction of motion of electron beam.The direction of force on the electron indicated by the index finger is towards left. Hence the electron beam is deflected towards left.
The electron beam will be deflected to the left-hand side. This is determined by the left-hand rule, due to the negative charge on the electron.
Explanation:This particular scenario of an electron beam and a magnet involves the concept of the magnetic force on a charged particle. According to the right-hand rule, when a charged particle moves in a magnetic field, the magnetic force on the particle is perpendicular to both the velocity of the particle and the magnetic field. In this case, since electrons have a negative charge, we'll be using your left hand for the rule.
Begin by extending your left hand flat, with your thumb pointing in the direction of the electron beam (towards you), and your fingers pointing in the direction of the magnetic field (downward). You'll notice that your palm would then be facing toward your left side. This indicates that the electron beam will be deflected to the left-hand side.
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The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelion) is 1.471 1011 m. The Earth's orbital speed at perihelion is 3.027 104 m/s. Ignore the effect of the Moon and other planets. (a) Determine the Earth's orbital speed at aphelion. m/s (b) Determine the kinetic and potential energies of the Earth–Sun system at perihelion. Kp = J Up = J (c) Determine the kinetic and potential energies at aphelion. Ka = J Ua = J (d)Is the total energy constant
Answer:
29274.93096 m/s
[tex]2.73966\times 10^{33}\ J[/tex]
[tex]-5.39323\times 10^{33}\ J[/tex]
[tex]2.56249\times 10^{33}\ J[/tex]
[tex]-5.21594\times 10^{33}[/tex]
Explanation:
[tex]r_p[/tex] = Distance at perihelion = [tex]1.471\times 10^{11}\ m[/tex]
[tex]r_a[/tex] = Distance at aphelion = [tex]1.521\times 10^{11}\ m[/tex]
[tex]v_p[/tex] = Velocity at perihelion = [tex]3.027\times 10^{4}\ m/s[/tex]
[tex]v_a[/tex] = Velocity at aphelion
m = Mass of the Earth = 5.98 × 10²⁴ kg
M = Mass of Sun = [tex]1.9889\times 10^{30}\ kg[/tex]
Here, the angular momentum is conserved
[tex]L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s[/tex]
Earth's orbital speed at aphelion is 29274.93096 m/s
Kinetic energy is given by
[tex]K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J[/tex]
Kinetic energy at perihelion is [tex]2.73966\times 10^{33}\ J[/tex]
Potential energy is given by
[tex]P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}[/tex]
Potential energy at perihelion is [tex]-5.39323\times 10^{33}\ J[/tex]
[tex]K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J[/tex]
Kinetic energy at aphelion is [tex]2.56249\times 10^{33}\ J[/tex]
Potential energy is given by
[tex]P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}[/tex]
Potential energy at aphelion is [tex]-5.21594\times 10^{33}\ J[/tex]
Answer:
:) *_* :3 ^-^ {.}{.}
Explanation:
3. If a car's wheels are replaced with wheels of greater diameter, will the reading of the speedometer change? Explain.
Answer:yes
Explanation:
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6640 N (1493 lbf). If the length of the rod is 370 mm (14.57 in.), what must be the diameter to allow an elongation of 0.53 mm (0.02087 in.)?
Answer:
d= 7.32 mm
Explanation:
Given that
E= 110 GPa
σ = 240 MPa
P= 6640 N
L= 370 mm
ΔL = 0.53
Area A= πr²
We know that elongation due to load given as
[tex]\Delta L=\dfrac{PL}{AE}[/tex]
[tex]A=\dfrac{PL}{\Delta LE}[/tex]
[tex]A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}[/tex]
A= 42.14 mm²
πr² = 42.14 mm²
r=3.66 mm
diameter ,d= 2r
d= 7.32 mm
Answer:
[tex]d=7.32\ mm[/tex]
Explanation:
Given:
Young's modulus, [tex]E=110\times 10^3\ MPa[/tex]yield strength, [tex]\sigma_y=240\ MPa[/tex]load applied, [tex]F=6640\ N[/tex]initial length of rod, [tex]l=370\ mm[/tex]elongation allowed, [tex]\Delta l=0.53[/tex]We know,
Stress:
[tex]\sigma=\frac{F}{A}[/tex]
where: A = cross sectional area
Strain:
[tex]\epsilon = \frac{\Delta l}{l}[/tex]
& by Hooke's Law within the elastic limits:
[tex]E=\frac{\sigma}{\epsilon}[/tex]
[tex]\therefore 110\times 10^3=\frac{F}{A}\div \frac{\Delta l}{l}[/tex]
[tex]\therefore 110\times 10^3=\frac{6640\times 4}{\pi.d^2}\div \frac{0.53}{370}[/tex]
where: d = diameter of the copper rod
[tex]d=7.32\ mm[/tex]
If a rectangular area is rotated in a uniform electric field from the position where the maximum electric flux goes through it to an orientation where only half the flux goes through it, what has been the angle of rotation? A) 45° B) 26.6° C) 90° D) 30° E
The correct angle of rotation that results in half the electric flux going through a rectangular area in a uniform electric field is 60 degrees, which corresponds to option D) 30 degrees.
The angle of rotation that results in half the electric flux can be found by understanding the relationship between the angle of rotation and the flux through a surface in a uniform electric field. The electric flux through a surface is calculated using the formula [tex]\\( \Phi _{E} = EA \cos(\theta) \)[/tex], where E is the electric field strength, A is the area of the surface, and ([tex]\theta[/tex]is the angle between the field and the normal to the surface. The maximum flux is when [tex]\\(\theta)[/tex] is 0 degrees. To find the angle at which half of the maximum flux passes through, we set [tex](\cos(\theta))[/tex] equal to 0.5 (since [tex](\cos(\theta)\)[/tex] is the ratio of the current flux to the maximum flux), which corresponds to an angle [tex](\theta[/tex] = 60 degrees. Therefore, if the rectangle initially had maximum flux passing through it, the rotation angle that results in half the flux is 60 degrees, which corresponds to answer D) 30 degrees.
The angle of rotation at which the electric flux through the surface is halved is 60°, as it corresponds to the angle where the cosine of the angle equals 0.5. This corresponds to option D in the multiple-choice question.
The electric flux through a surface is given by the equation ∅ = E*A*cosФ where
E is the magnitude of the electric field, A is the area of the surface, and Ф is the angle between the field and the normal to the surface.For maximum flux, the angle is 0° since (cos(0°) = 1).
If the flux is halved, the equation becomes
Ф/2 = E*A* cosФ
To find the angle of rotation for which the flux is halved, we set (cos(Ф) = 1/2), which corresponds to ( Ф= 60°) or (Ф = 120°).
The angle of rotation from the position of maximum flux (0°) to the position where half the flux goes through it is therefore either 60° or 120°. However, since the question implies a single rotation direction, we would generally take the smaller angle, which is 60°, corresponding to option D.
A block of mass m attached to a horizontally mounted spring with spring constant k undergoes simple harmonic motion on a rictionless surface. How would the maximum speed of the block be affected i the spring constant was increased by a factor of 4 while holding the amplitude of oscillation constant?
1) It would remain unchanged.
2) It would increase by a factor of 2.
3) It would decrease by a factor of 1/4
4) It would increase by a factor of 4.
5) It would decrease by a factor of 1/2
To solve this problem it is necessary to apply the concepts related to simple harmonic movement.
The maximum speed from the simple harmonic motion is given as
[tex]V = A\sqrt{\frac{K}{m}}[/tex]
Where,
K = Spring constant
m = mass
At this case m is a constant then
[tex]V \propto \sqrt{K}[/tex]
then the ratio is given by
[tex]\frac{v_2}{v_1}=\sqrt{\frac{K_1}{K_2}}[/tex]
According the statement,
[tex]v_2 = \sqrt{\frac{K_1}{K_2}}v_1[/tex]
[tex]v_2 = 2v_1[/tex]
Therefore the maximum speed becomes double: 2) It would increase by a factor of 2.
It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinetic energy is converted into heat via friction. Find the minimum speed of a lead bullet (initial temperature = 43.0° C) for such an event to happen? (Use Lf = 2.32 104 J/kg and melting point = 327.3° C.)
Answer:
[tex]v=346.05\ m.s^{-1}[/tex]
Explanation:
Given:
initial temperature of the lead bullet, [tex]T_i=43^{\circ}C[/tex]
latent heat of fusion of lead, [tex]L_f=2.32\times 10^4\ J.kg^{-1}[/tex]
melting point of lead, [tex]T_m=327.3^{\circ}C[/tex]
We have:
specific heat capacity of lead, [tex]c=129\ J.kg^{-1}.K^{-1}[/tex]
According to question the whole kinetic energy gets converted into heat which establishes the relation:
[tex]\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)[/tex]
[tex]\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f[/tex]
[tex]\frac{1}{2} m.v^2=m(c.\Delta T+L_f)[/tex]
[tex]\frac{v^2}{2} =129\times(327.3-43)+23200[/tex]
[tex]v=346.05\ m.s^{-1}[/tex]
The minimum speed of a lead bullet that will completely melt upon stopping suddenly can be found by calculating the kinetic energy that is converted into heat via friction and comparing it with the heat required to elevate the bullet's temperature to its melting point. The formula for calculating the minimum speed of the bullet is the square root of twice the product of the latent heat of fusion and the change in temperature (√(2Lf∆T)).
Explanation:The question asked pertains to the concept of kinetic energy transformations and heat generation due to friction in Physics. The kinetic energy of the bullet is transformed into thermal energy due to the sudden change in speed, causing the lead bullet to heat up and potentially melt if it is moving fast enough.
First, we calculate the change in temperature which is the difference between the melting point of lead and the initial temperature, ∆T = 327.3°C - 43.0°C = 284.3°C. Then, we utilize the formula for heat transfer Q = mLf, where m is the mass of the bullet, and Lf is the latent heat of fusion of lead. We rearrange this to find the mass of the bullet, m = Q/Lf.
Next, we use the principle of conservation of energy. All of the kinetic energy of the bullet (1/2mv²) is converted into heat (Q), leading to the equation 1/2m v² = Q. Solving for v (the bullet's speed) we get that v = √(2Q/m). Combining equations yields v = √(2Lf∆T).
This v is the minimum speed at which the bullet will completely melt upon stopping suddenly, assuming that all of its kinetic energy is converted into heat via friction and is equal to the energy required to raise the bullet's temperature to its melting point.
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Two boats leave a dock together . Each travels in a straight line . The angle between their courses measures 54° 10¢ . One boat travels 36.2 km/hr and the other travels 45.6 km/hr. How far apart will they be after three hours
Answer:
218.93 km
Explanation:
[tex]\theta[/tex] = Angle between the paths of the two boats = 54.10°
Distance = Speed × Time
Distance traveled by one boat = [tex]36.2\times 3\ km[/tex]
Distance traveled by other boat = [tex]45.6\times 3\ km[/tex]
From the triangle law of vectors we have
[tex]d=\sqrt{(36.2\times 3)^2+(45.6\times 3)^2+2\times 36.2\times 3\times 45.6\times 3\times cos54.10}\\\Rightarrow d=218.93\ km[/tex]
The distance they will be apart after 3 hours is 218.93 km
Part complete Horizontally-polarized light passes through a polarizing sheet, and only 25 % of the intensity of the incident light is transmitted through the sheet. What angle does the polarization of the light make with the horizontal after passing through the polarizer?
Answer:
60°
Explanation:
I₀ = Intensity of unpolarized light
θ = Angle between the axis of the filter and polarization direction
Intensity of polarzied light
[tex]I=I_0cos\theta[/tex]
Here, the light that is transmitted is reduced by 25% that means
[tex]I=0.25I_0[/tex]
So,
[tex]0.25I_0=I_0cos^2\theta\\\Rightarrow cos^2\theta =0.25\\\Rightarrow cos\theta =5\\\Rightarrow \theta= cos^{-1}0.5\\\Rightarrow \theta=60^{\circ}[/tex]
∴ The angle between the axes of the polarizer and the analyzer is 60°
Two identical billiard balls traveling at the same speed have a head-on collision and rebound. If the balls had twice the mass, but maintained the same size and speed, how would the rebound be different?
a. No difference.
b. At a higher speed.
c. At slower speed.
Doubling the mass of billiard balls during an elastic head-on collision would result in no difference in the rebound speed due to the conservation of momentum. The speed and motion of the balls post-collision would be unaffected by the change in mass.
Explanation:The question asks how the rebound of two identical billiard balls would be different if their mass were doubled, but they maintained the same size and speed during a head-on collision. According to the principle of conservation of momentum, which dictates that the total momentum of an isolated system remains constant if no external forces act upon it, the rebound should be unaffected by the change in mass.
In a head-on elastic collision, if we assume an ideal scenario without energy loss due to factors like friction or air resistance, doubling the mass of the balls while keeping the speed the same would not change the speed at which they rebound. This is because the momentum before the collision must equal the momentum after the collision for each ball, and since momentum is the product of mass and velocity (p = mv), the velocities would remain unchanged post-collision. Therefore, the correct answer would be (a) No difference.
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