Answer:
(a) 0.115 m
(b) 2.08 x 10^-5 J
Explanation:
mass of bob, m = 81 g = 0.081 kg
The equation of oscillation is given by
θ = 0.068 Cos {9.2 t + Ф}
Now by comparison
The angular velocity
ω = 9.2 rad/s
(a) [tex]\omega^{2} =\frac{g}{L}[/tex]
where, L be the length of the pendulum
[tex]L =\frac{g}{\omega ^{2}}[/tex]
[tex]L =\frac{9.8}{9.2 \times 9.2}[/tex]
L = 0.115 m
(b) A = L Sinθ
A = 0.115 x Sin 0.068
A = 7.8 x 10^-3 m
Maximum kinetic energy
K = 0.5 x mω²A²
K = 0.5 x 0.081 x 9.2 x 9.2 x 7.8 x 7.8 x 10^-6
K = 2.08 x 10^-5 J
An 15-cm-long bicycle crank arm, with a pedal at one end, is attached to a 21-cm-diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 62 rpm to 95 rpm in 12 s . A) What is the tangential acceleration of the pedal?
B)What length of chain passes over the top of the sprocket during this interval?
The tangential acceleration of the pedal is 0.259 m/s^2. The length of chain passing over the top of the sprocket depends on the number of teeth on the sprocket, which is not provided in the question.
Explanation:A) The tangential acceleration of the pedal can be determined using the equation:
at = r × α
where at is the tangential acceleration, r is the radius of the pedal (half the length of the crank arm), and α is the angular acceleration. To find the angular acceleration, we can use the equation:
α = Δω / Δt
where Δω is the change in angular velocity and Δt is the change in time. Plugging in the given values:
α = (95 rpm - 62 rpm) * 2π / 60 s = (33 * 2π) / 60 s ≈ 3.459 rad/s2
Substituting this value and the radius into the first equation, we get:
at = 0.075 m * 3.459 rad/s2 = 0.259 m/s2
Therefore, the tangential acceleration of the pedal is 0.259 m/s2.
B) The length of chain passing over the top of the sprocket during the given time interval can be calculated using the formula:
d = 2π * r * (ωf - ωi) * t / N
where d is the length of chain, r is the radius of the sprocket, ωi and ωf are the initial and final angular velocities respectively, t is the time interval, and N is the number of teeth on the sprocket. Plugging in the given values:
d = 2π * (21 cm / 2) * [(95 rpm - 62 rpm) * 2π / 60 s] * 12 s / N
The length of chain depends on the number of teeth on the sprocket. Since the number of teeth is not provided in the question, we cannot calculate the exact length of chain. However, we have all the necessary equation components to calculate it once the number of teeth is known.
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A glass flask whose volume is 1000 cm3 at a temperature of 0.300 ∘C is completely filled with mercury at the same temperature. When the flask and mercury are warmed together to a temperature of 52.0 ∘C , a volume of 8.25 cm3 of mercury overflows the flask.
Final answer:
When the glass flask and mercury are warmed together, the mercury expands and overflows the flask. To calculate the initial volume of the mercury, use the equation: Volume of mercury = Volume of flask + Volume overflowed.
Explanation:
When the glass flask and mercury are warmed together, both substances expand due to the increase in temperature. As a result, some of the mercury overflows the flask. To calculate the initial volume of the mercury, we can use the equation:
Volume of mercury = Volume of flask + Volume overflowed
So, the initial volume of mercury is 1000 cm³ + 8.25 cm³ = 1008.25 cm³.
If the pendulum is taken into the orbiting space station what will happen to the bob?
a. It will continue to oscillate in a vertical plane with the same period.
b. It will no longer oscillate because there is no gravity in space.
c. It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.
d. It will oscillate much faster with a period that approaches zero.
Answer:
Option C
Explanation:
The bob of the pendulum, when the pendulum is taken into the orbiting space station will be in free fall as there is no gravity in space and also the point to which it is attached.
The movement of the bob of the pendulum about its mean position is because of its weight due to which it oscillates and in the free fall, no force acts on it due to its weight since the body is not under the attraction force due to gravity thus it experiences weightlessness and does not oscillate.
On an orbiting space station with microgravity, a pendulum will not oscillate because the pendulum and its point of attachment are in free fall, absent of the normal gravitational forces.
Explanation:If a pendulum is taken into an orbiting space station, its behavior will be altered due to the absence of an external gravitational force. This situation is best explained by option c. The pendulum will no longer oscillate because the pendulum and the point to which it is attached are in free fall, also known as microgravity. In an environment of microgravity, the forces that would ordinarily cause the pendulum bob to swing are absent, thus the pendulum will essentially float instead of oscillating back and forth.
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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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Determine the ratio of the flow rate through capillary tubes A and B (that is, Qa/Qb).
The length of A is twice that of B, and the radius of A is one-half that of B.
The pressure across both tubes is the same.
To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.
It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.
The mathematical equation that expresses this concept is
[tex]\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}[/tex]
Where
P = Pressure at each point
r = Radius
[tex]\eta =[/tex] Viscosity
l = Length
Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]
From the problem two terms are given
[tex]R_A = \frac{R_B}{2}[/tex]
[tex]L_A = 2L_B[/tex]
Replacing we have to
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}[/tex]
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}[/tex]
[tex]\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}[/tex]
Therefore the ratio of the flow rate through capillary tubes A and B is 1/32
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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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)
In a downtown office building, you notice each of the four sections of a rotating door has a mass of 75 kg. What is the width, in meters, of each section of the door if a force of 56 N applied to the outer edge of a section produces an angular acceleration of 0.420 rad/s2?
Answer:
each door has a width of 2.666 meters
Explanation:
from Newton's second law applied to rotational motion:
ζ = I α
where ζ= torque , I = moment of inertia , α = angular acceleration
the moment of inertia for a flat plate around its central axis is
I = 1/12 m a² , where m= mass, a= total width = 2L
therefore the moment of inercia for a flat plate with length 2L ( 2 doors, one in each side of the central axis) is
I1 = 1/12 m (2L)² = 1/3 m L²
if we have 4 doors , that is 2 flat plates with length of 2L perpendicular to each other:
I = Ix + Iy = 2*I1 = 2/3 m L²
thus
ζ = I α
4* F * L = 2/3 * (4*m) L² * α
L = 3/2* F/ ( m*α) = 3/2* 56 N / ( 75 Kg * 0.420 rad/seg²) = 2.666 m
Suppose there are two identical gas cylinders. One contains the monatomic gas krypton (Kr), and the other contains an equal mass of the monatomic gas neon (Ne). The pressures in the cylinders are the same, but the temperatures are different. Determine the ratio KEKr/KENe of the average kinetic energy of an atom of krypton to the average kinetic energy of an atom of neon.
Answer:
Explanation:
Let equal mass of Ne and Kr be m gm
no of moles of Ne and Kr will be m / 20 and m / 84 ( atomic weight of Ne and Kr is 20 and 84 )
Let the pressure and volume of both the gases be P and V respectively .
The temperature of Ne be T₁ and temperature of Kr be T₂.
For Ne
PV = (m / 20) x R T₁
For Kr
PV = (m / 84) x R T₂
T₁ / T₂ = 84 / 20
We know that
average KE of an atom of mono atomic gas = 3 / 2 x k T
k is boltzmann constant and T is temperature .
KEKr/KENe = T₂ / T₁
= 20 / 84
The ratio of the average kinetic energy of an atom of krypton to that of neon is 1.
Explanation:The average kinetic energy of an atom is directly proportional to the temperature. In this case, the temperatures of the krypton (Kr) and neon (Ne) gases are different. Since the temperatures are different, the ratio of their average kinetic energies will be equal to the ratio of their temperatures. Therefore, the ratio KEKr/KENe is equal to Tkrypton/Tneon.
Using the given information, we can find the ratio of the temperatures: Tkrypton/Tneon = 300K/300K = 1. Therefore, the ratio of the average kinetic energy of an atom of krypton to that of neon is 1.
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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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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/
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:
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°
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
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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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.
A toy cannon uses a spring to project a 5.35-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force constant of 8.07 N/m. When the cannon is fired, the ball moves 16.0 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.033 0 N on the ball.
(a) With what speed does the projectile leave the barrel of the cannon? m/s
(b) At what point does the ball have maximum speed? cm (from its original position)
(c) What is this maximum speed? m/s
Answer:
a) the velocity is v=1.385 m/s
b) the ball has its maximum speed at 4.68 cm away from its compressed position
c) the maximum speed is 1.78 m/s
Explanation:
if we do an energy balance over the ball, the potencial energy given by the compressed spring is converted into kinetic energy and loss of energy due to friction, therefore
we can formulate this considering that the work of the friction force is equal to to the energy loss of the ball
W fr = - ΔE = - ΔU - ΔK = Ui - Uf + Kf - Ki
therefore
Ui + Ki = Uf + Kf + W fr
where U represents potencial energy of the compressed spring , K is the kinetic energy W fr is the work done by the friction force. i represents inicial state, and f final state.
since
U= 1/2 k x² , K= 1/2 m v² , W fr = F*L
X= compression length , L= horizontal distance covered
therefore
Ui + Ki = Uf + Kf + W fr
1/2 k xi² + 1/2 m vi² = 1/2 k x² + 1/2 m vf² + F*L
a) choosing our inicial state as the compressed state , the initial kinetic energy is Ki=0 and in the final state the ball is no longer pushed by the spring thus Uf=0
1/2 k X² + 0 = 0 + 1/2 m v² + F*L
1/2 m v² = 1/2 k X² - F*L
v = √[(k/m)x² -(2F/m)*L] = √[(8.07N/m/5.35*10^-3 Kg)*(-0.0508m)² -(2*0.033N/5.35*10^-3 Kg)*(0.16 m)] = 1.385 m/s
b) in any point x , and since L= d-(X-x) , d = distance where is no pushed by the spring.
1/2 k X² + 0 = 1/2 k x² + 1/2 m v² + F*[d-(X+x)]
1/2 m v² =1/2 k X²-1/2 k x² - F*[d-(X-x)] = (1/2 k X²+ F*X) - 1/2k x² - F*x + F*d
taking the derivative
dKf/dx = -kx - F = 0 → x = -F/k = -0.033N/8.07 N/m = -4.089*10^-3 m = -0.4cm
at x m = -0.4 cm the velocity is maximum
therefore is 5.08 cm-0.4 cm=4.68 cm away from the compressed position
c) the maximum speed is
1/2 m v max² = (1/2 k X²+ F*X) - 1/2k x m² - F*(x m) + 0
v =√[ (k/m) (X²-xm²) + (2F/m)(X-xm) ] = √[(8.07N/m/5.35*10^-3 Kg)*[(-0.0508m)² - (-0.004m)²] + (2*0.033N/5.35*10^-3 Kg)*(-0.0508m-(-0.004m)] = 1.78 m/s
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
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 motorcycle and a police car are moving toward one another. The police car emits sound with a frequency of 523 Hz and has a speed of 32.2 m > s. The motorcycle has a speed of 14.8 m > s. What frequency does the motorcyclist hear?
To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.
It can be described as
[tex]f = \frac{c\pm v_r}{c\pm v_s}f_0[/tex]
c = Propagation speed of waves in the medium
[tex]v_r[/tex]= Speed of the receiver relative to the medium
[tex]v_s[/tex]= Speed of the source relative to the medium
[tex]f_0 =[/tex]Frequency emited by the source
The sign depends on whether the receiver or the source approach or move away from each other.
Our values are given by,
[tex]v_s = 32.2m/s \rightarrow[/tex] Velocity of car
[tex]v_r = 14.8 m/s \rightarrow[/tex] velocity of motor
[tex]c = 343m/s \rightarrow[/tex] Velocity of sound
[tex]f_0 = 523Hz \rightarrow[/tex]Frequency emited by the source
Replacing we have that
[tex]f = \frac{c + v_r}{c - v_s}f_0[/tex]
[tex]f = \frac{343 + 14.8}{343 - 32}(523)[/tex]
[tex]f = 601.7Hz[/tex]
Therefore the frequency that hear the motorcyclist is 601.7Hz
Final answer:
To determine the frequency heard by the motorcyclist, we apply the Doppler shift formula using the given speeds of the motorcycle and police car and the emitted frequency of the police car's siren. The calculated frequency is the observed frequency by the motorcyclist due to the relative motion of the two vehicles. The final answer is about 602.1 Hz
Explanation:
The subject of the question is the Doppler Effect, which is related to the change in frequency of sound waves due to the relative motion of the source and the observer. To calculate the frequency the motorcyclist hears, we use the following Doppler shift formula for sound:
f' = f(v + vo) / (v - vs)
Where:
f' is the observed frequency by the motorcyclist,
f is the emitted frequency by the police car (523 Hz),
v is the speed of sound in air, which can be assumed to be approximately 343 m/s at room temperature,
vo is the speed of the observer (motorcyclist) towards the source (14.8 m/s),
vs is the speed of the source (police car) towards the observer (32.2 m/s).
Plugging in the given values, the equation becomes:
f' = 523 Hz (343 m/s + 14.8 m/s) / (343 m/s - 32.2 m/s) = 602.089 Hz
After performing the calculations, the frequency heard by the motorcyclist can be determined. This is an application of the Doppler effect as studied in high school physics.
A horizontal vinyl record of mass 0.105 kg and radius 0.0757 m rotates freely about a vertical axis through its center with an angular speed of 5.80 rad/s and a rotational inertia of 5.18 x 10-4 kg·m2. Putty of mass 0.0213 kg drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately afterwards?
Answer:
4.6939 rad/s
Explanation:
You have to use the conservation of angular momentum for both objects as an object is spinning and a collision happens. To properly solve, you have to know that the putty and the vinyl have the same angular speed after the putty lands on it and that the putty acts as a point mass so the formula of it's rotational inertia is mr^2.
A hollow spherical shell with mass 2.00kg rolls without slipping down a slope that makes an angle of 40.0^\circ with the horizontal. Find the magnitude of the acceleration a_cm of the center of mass of the spherical shell. Find the magnitude of the frictional force acting on the spherical shell.
Answer:
[tex]a_{cm} = 9.64m/s^2[/tex]
[tex]Ff=6.42N[/tex]
Explanation:
The sum of torque on the sphere is:
[tex]m*g*sin\theta*R=I*\alpha[/tex]
[tex]m*g*sin\theta*R=2/3*m*R^2*\alpha[/tex]
[tex]m*g*sin\theta*R=2/3*m*R*a_{cm}[/tex]
Solving for a:
[tex]a_{cm}=9.64m/s^2[/tex]
Now, the sum of forces will be:
[tex]m*g*sin\theta-Ff=m*a_{cm}[/tex]
Solving for Ff:
[tex]Ff=m*g*sin\theta-m*a_{cm}[/tex]
Ff=-6.42N The negative sing tells us that it actually points downwards.
Answer:
a) a = 3.783 m/s^2
b) F_f = 5.045 N
Explanation:
Given:
- Mass of shell m = 2.0 kg
- Angle of slope Q = 40 degrees
- Moment of inertia of shell I = 2/3 *m*R^2
Find:
a) Find the magnitude of the acceleration a_cm of the center of mass of the spherical shell.
b) Find the magnitude of the frictional force acting on the spherical shell.
Solution:
- Draw a Free body diagram for the shell. We see that the gravitational force F_g acting parallel to the plane of the inclined surface makes the sphere to roll down. The frictional force F_f between the inclined surface and the sphere gives the necessary torque for the sphere to roll down with out slipping. Under this conditions a sphere will roll down without slipping with some acceleration and the acceleration can be calculated from the equation of motion of the sphere:
m*g*sin(Q) - F_f = m*a
- Where, The frictional force produces the torque and due to this torque the sphere gets an angular acceleration.
- Then we can write the equation for the rotational motion as:
F_f*R = I*α
F_f = I*α / R
- Using moment mass inertia of the shell we have:
F_f = (2/3)*m*R^2*α/R
- Where the angular acceleration α is related to linear acceleration a with:
α = a / R
- combing the two equations we will have friction force F_f as:
F_f = (2/3)*m*R^2*a/R^2
F_f = (2/3)*m*a
- Now evaluate the equation of motion:
m*g*sin(Q) - (2/3)*m*a= m*a
- Simplify:
(5/3)*a = g*sin(Q)
a = (3/5)*g*sin(Q)
- Plug the values in: a = (3/5)*9.81*sin(40)
a = 3.783 m/s^2
- Now compute the Frictional force F_f from the expression derived above:
F_f = (2/3)*m*a
- Plug values in: F_f = (2/3)*2*3.783
F_f = 5.045 N
a bucket of water of mass 14.7 is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.280m with mass 11.6 kg. the cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.3m to the water. You can ignore the weight of the rope.
Part A
What is the tension in the rope while the bucket is falling
-Take the free fall acceleration to be g=9.80 m/s squared
Part B
with what speed does the bucket strike the water?
-Take the free fall acceleration to be g=9.80 m/s^2
Part C
What is the time of fall
-Take the free fall acceleration to be g=9.80m/s^2
Part D
While the bucket is falling, what is the force exerted on the cylinder by the axle?
-Take the free fall acceleration to be g=9.80 m/s^2
Answer:
Explanation:
Tension T in the rope will create torque in solid cylinder ( axle ). If α be angular acceleration
T R = 1/2 M R²α ( M is mass and R is radius of cylinder )
= 1/2 M R² x a / R ( a is linear acceleration )
T = Ma / 2
For downward motion of the bucket
mg - T = m a ( m is mass and a is linear acceleration of bucket downwards )
mg - Ma / 2 = ma
a = mg / ( M /2 + m )
Substituting the values
a = 14.7 x 9.8 / ( 5.8+ 14.7 )
= 7 m / s²
A )
T = Ma / 2
= 5.8 x 7
= 40.6 N
B ) v² = u² + 2 a h
= 2 x 7 x 10.3
v = 12 m /s
C )
v = u + a t
12 = 0 + 7 t
t = 1.7 s
A rigid cube (each side is 0.10 m) is filled with water and frozen solid. When water freezes its volume expands about 9%. How much pressure is exerted on the sides of the cube? Hint: Imagine trying to squeeze the block of ice back into the original cube.
Answer:
P = 1.89 10⁸ N / m²
Explanation:
To solve this problem we can use the definition of bulk modules
B = - P / (ΔV/V)
The negative sign is entered for the volume module to be positive, P is the pressure and ΔV/V is the volume change fraction
In this case the volume change is 9% this is
ΔV / V 100 = 9%
ΔV / V = 0.09
P = B ΔV / V
The bulk modulus value is that of water since it is in a liquid state and then freezes
B = 0.21 101¹⁰ N / m²
let's calculate
P = 0.21 10¹⁰ 0.09
P = 1.89 10⁸ N / m²
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]
3. If a car's wheels are replaced with wheels of greater diameter, will the reading of the speedometer change? Explain.
Answer:yes
Explanation:
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . Suppose that the quarterback takes 0.30 s to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?
Answer:
0.0241875 m
Explanation:
[tex]m_1[/tex] = Mass of quarterback = 80 kg
[tex]m_2[/tex] = Mass of football = 0.43 kg
[tex]v_1[/tex] = Velocity of quarterback
[tex]v_2[/tex] = Velocity of football = 15 m/s
Time taken = 0.3 seconds
In this system as the linear momentum is conserved
[tex]m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\frac{m_2v_2}{m_1}\\\Rightarrow v_1=-\frac{0.43\times 15}{80}\\\Rightarrow v_1=0.080625\ m/s[/tex]
Assuming this velocity is constant
[tex]Distance=Velocity\times Time\\\Rightarrow Distance=0.080625\times 0.3\\\Rightarrow Distance=0.0241875\ m[/tex]
The distance the quarterback will move in the horizontal direction is 0.0241875 m
Final answer:
The question is a physics problem regarding kinematics and requires calculating the horizontal distance an 80-kg quarterback moves while in the air after throwing a football. However, since the quarterback jumps vertically with zero initial horizontal speed and no horizontal force is acting after the throw, the resulting horizontal distance is zero.
Explanation:
The student is asking a question related to kinematics and the conservation of momentum in physics. Specifically, it concerns an 80-kg quarterback who jumps vertically and then throws a football horizontally. The core concept here is determining how far the quarterback will move horizontally during the time he is in the air, assuming a constant horizontal speed.
To calculate the distance (d) the quarterback moves horizontally, we can apply the equation for constant velocity motion: d = v * t. However, the information given does not directly provide the quarterback's horizontal speed after the throw; thus, we must assume that the horizontal speed of the quarterback is zero as the ball is thrown horizontally and no horizontal force on the quarterback has been mentioned.
Therefore, under the assumption that the quarterback's horizontal speed remains zero, the distance he moves horizontally d would also be zero, since he moves vertically up and down with no horizontal velocity component. If the problem had given a horizontal speed for the quarterback post-throw, we would use that given speed to calculate the distance using the equation.
A 19 g bullet is fired into the bob of a ballistic pendulum of mass 1.3 kg. When the bob is at its maximum height, the strings make an angle of 60° with the vertical. The length of the pendulum is 2.3 m. Find the speed of the bullet.
Answer:
217.43298 m/s
Explanation:
[tex]m_1[/tex] = Mass of bullet = 19 g
[tex]m_2[/tex] = Mass of bob = 1.3 kg
L = Length of pendulum = 2.3 m
[tex]\theta[/tex] = Angle of deflection = 60°
u = Velocity of bullet
Combined velocity of bullet and bob is given by
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s[/tex]
As the momentum is conserved
[tex]m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s[/tex]
The speed of the bullet is 217.43298 m/s
The problem involves conservation of momentum and energy principles. Initially, bullet's momentum equals the final momentum of the system. The bullet's speed can be found by solving these equations, using the provided values.
Explanation:This problem can be solved using principles from both conservation of momentum and conservation of energy. To find the speed of the bullet, we need to consider two scenarios: the before and after the bullet is fired into the bob. Initial momentum is the mass of the bullet multiplied by its velocity and final momentum is the combined mass of the bullet and bob at their highest point. Assuming there's no external force acting, we can have:
m_bullet * v_bullet = (m_bullet + m_bob) * v_final.
The final velocity here is the vertical component of the velocity when the pendulum reach its highest point. This can be calculated by:
v_final = sqrt(2*gravity*height).
The height can be calculated using trigonometry:
height = length - length * cos(60).
Filling all the given values into the equations will give the speed of the bullet.
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The specifications for a ceiling fan you have just bought indicate that the total moment of inertia of its blades is 0.23 kg · m2 and they experience a net torque of 2.8 N · m.
(a) What is the angular acceleration of the blades in rad/s2? rad/s2
(b) When the blades rotate at 205 rpm, what is the rotational kinetic energy, in joules? J
Answer:(a)12.17 rad/s
Explanation:
Given
Moment of Inertia [tex]I=0.23 kg.m^2[/tex]
Torque [tex]T=2.8 N-m[/tex]
(a)Torque is given by Product of Moment of inertia and angular acceleration
[tex]T=I\cdot \alpha [/tex]
[tex]2.8=0.23\cdot \alpha [/tex]
[tex]\alpha =\frac{2.8}{0.23}=12.17 rad/s[/tex]
(b)RPM of blades [tex]N=205 rpm [/tex]
angular velocity [tex]\omega =\frac{2\pi N}{60}[/tex]
[tex]\omega =\frac{2\pi 205}{60}=21.47 rad/s[/tex]
Rotational Kinetic Energy [tex]=\frac{I\omega ^2}{2}[/tex]
[tex]=\frac{0.23\times (21.47)^2}{2}=53.01 J[/tex]
The angular acceleration of the blades is approximately 12.1739 rad/s^2. When the blades rotate at 205 rpm, the rotational kinetic energy is approximately 0.0948 J.
Explanation:(a) To find the angular acceleration of the blades, we can use the formula:
torque = moment of inertia × angular acceleration
Plugging in the given values:
torque = 2.8 N · m
moment of inertia = 0.23 kg · m2
Rearranging the formula, we get:
angular acceleration = torque / moment of inertia
Substituting the values:
angular acceleration = 2.8 N · m / 0.23 kg · m2
Solving for angular acceleration gives us:
angular acceleration ≈ 12.1739 rad/s2
(b) To find the rotational kinetic energy, we can use the formula:
rotational kinetic energy = ½ × moment of inertia × (angular velocity)2
Plugging in the given values:
moment of inertia = 0.23 kg · m2
angular velocity = 205 rpm = 205 revolutions / 60 seconds = 3.4167 rev/s
Rearranging the formula, we get:
rotational kinetic energy = ½ × moment of inertia × (angular velocity)2
Substituting the values:
rotational kinetic energy = ½ × 0.23 kg · m2 × (3.4167 rev/s)2
Solving for rotational kinetic energy gives us:
rotational kinetic energy ≈ 0.0948 J
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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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