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:

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

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.

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.

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.

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 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)

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°

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.

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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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

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

Answer:yes

Explanation:

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]

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

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

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

Answer:

René Descartes

Explanation:

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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.

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.

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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:

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]

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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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.

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

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.

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.

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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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)).

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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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.

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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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

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

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

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.

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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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.

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.

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