A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10,000 kg, and two antennas projecting out from the center of mass of the main body that can be approximated with rods of length 3.0 m each and mass 10 kg. The antenna’s lie in the plane of rotation. What is the angular momentum of the satellite?

Answer:

[tex]t = 2\ s[/tex]

Explanation:

given,

length of the rope = 16 m

speed of the man = 2 m/s

using the formula of time period

[tex]T =2 \pi \sqrt{\dfrac{L}{g}}[/tex]

[tex]T =2 \pi \sqrt{\dfrac{16}{9.8}}[/tex]

[tex]T = 8.028\ s[/tex]

To cover the maximum distance you need to leave the when the rope is shows maximum displacement.

To reach the displacement time to leave the rope is one fourth of the time period.

[tex]t = \dfrac{T}{4}[/tex]

[tex]t = \dfrac{8.03}{4}[/tex]

[tex]t = 2\ s[/tex]

The time period of pendulum is time taken by it to complete one cycle of swing left to right and right to left.

The total time taken to hang on if you want to drop into the water at the greatest possible distance from the edge is 2 seconds.

The time period of pendulum is time taken by it to complete one cycle of swing left to right and right to left.

It can be given as,

[tex]T=2\pi\sqrt{\dfrac{T}{g}}[/tex]

Here, [tex]g[/tex] is the gravitational force of Earth.

Given information-

Total length of the rope is 16 m.

The speed of the man is 2.0 m/s.

Let the time of to swing by rope both side is [tex]T[/tex]. Thus put the values in the above formula as,

[tex]T=2\pi\sqrt{\dfrac{16}{9.8}}[/tex]

[tex]T=8.028 \rm s[/tex]

Now the greatest possible distance from the edge will be at the greatest displacement.

Thus the time to cove the greatest possible distance will be one forth (1/4) of the total time. Thus,

[tex]t=\dfrac{8.028 }{4}\\t=2\rm s[/tex]

Hence, the total time taken to hang on if you want to drop into the water at the greatest possible distance from the edge is 2 seconds.

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To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.

Mathematically it can be expressed as

[tex]\rho = \frac{m}{V}[/tex]

Where

m = Mass (Neutron at this case)

V = Volume

The mass of the neutron star is 1.4times to that of the mass of the sun

The volume of a sphere is determined by the equation

[tex]V = \frac{4}{3}\pi R^3[/tex]

Replacing at the equation we have that

[tex]\rho = \frac{1.4m_{sun}}{\frac{4}{3}\pi R^3}[/tex]

[tex]\rho = \frac{1.4(1.989*10^{30})}{\frac{4}{3}\pi (1.05*10^4)^3}[/tex]

[tex]\rho = 5.75*10^{17}kg/m^3[/tex]

Therefore the density of a neutron star is [tex] 5.75*10^{17}kg/m^3[/tex]

Answer:

[tex]7.7549\times 10{19}\ J[/tex]

[tex]1.17037\times 10^{15}\ Hz[/tex]

[tex]2.56329\times 10^{-7}\ m[/tex]

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]

[tex]\nu[/tex] = Frequency

[tex]\lambda[/tex] = Wavelength

The minimum energy is given by

[tex]E=\frac{1\ mol}{N_A}\\\Rightarrow E=467\times 10^{3}\times \frac{1}{6.022\times 10^{23}}\\\Rightarrow E=7.7549\times 10{19}\ J[/tex]

The minimum energy is [tex]7.7549\times 10{19}\ J[/tex]

The energy of a photon is given by

[tex]E=h\nu\\\Rightarrow \nu=\frac{E}{h}\\\Rightarrow \nu=\frac{467\times 10^{3}\frac{1}{6.022\times 10^{23}}}{6.626\times 10^{-34}}\\\Rightarrow \nu=1.17037\times 10^{15}\ Hz[/tex]

The frequency of the photon is [tex]1.17037\times 10^{15}\ Hz[/tex]

Wavelength is given by

[tex]\lambda=\frac{c}{\nu}\\\Rightarrow \lambda=\frac{3\times 10^8}{1.17037\times 10^{15}}\\\Rightarrow \lambda=2.56329\times 10^{-7}\ m[/tex]

The wavelength is [tex]2.56329\times 10^{-7}\ m[/tex]

Answer:

27.66 s

Explanation:

Space station creates artificial gravity by rotational movement about its axis .

The object inside also move in circular motion creating centrifugal force which creates acceleration in them .

centrifugal acceleration = ω² R where ω is angular velocity and R is radius of the cylindrical space station .

R = 380 /2 = 190 m

Given

ω² R = g = 9.8

ω² = 9.8 / R

= 9.8 / 190

= 5.15x 10⁻²

ω = 2.27 x 10⁻¹

= .227 rad / s

2π / T = .227 ( T is time period of rotation )

T = 2π / .227

= 27.66 s .

outside of the space station they will experience zero acceleration , because they are rotating around the earth.

Answer:

1.1775 x 10^-3 m^3 /s

Explanation:

viscosity, η = 0.250 Ns/m^2

radius, r = 5 mm = 5 x 10^-3 m

length, l = 25 cm = 0.25 m

Pressure, P = 300 kPa = 300000 Pa

According to the Poisuellie's formula

Volume flow per unit time is

[tex]V=\frac{\pi \times Pr^{4}}{8\eta l}[/tex]

[tex]V=\frac{3.14 \times 300000\times \left ( 5\times 10^{-3} \right )^{4}}{8\times 0.250\times 0.25}[/tex]

V = 1.1775 x 10^-3 m^3 /s

Thus, the volume of oil flowing per second is 1.1775 x 10^-3 m^3 /s.

Answer:

(a) 2.5 cm

(b) Yes

Solution:

As per the question:

Mass of Uranium-235 ion, m = [tex]3.95\times 10^{- 25}\ kg[/tex]

Mass of Uranium- 238, m' = [tex]3.90\times 10^{- 25}\ kg[/tex]

Velocity, v = [tex]3.00\times 10^{5}\ m/s[/tex]

Magnetic field, B = 0.250 T

q = 3e

Now,

To calculate the path separation while traversing a semi-circle:

[tex]\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})[/tex]

The radius of the ion in a magnetic field is given by:

R = [tex]\frac{mv}{qB}[/tex]

[tex]\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})[/tex]

[tex]\Delta x = 2(\frac{mv}{qB} - \frac{m'v}{qB})[/tex]

[tex]\Delta x = 2(\frac{m - m'}{qB}v)[/tex]

Now,

By putting suitable values in the above eqn:

[tex]\Delta x = 2(\frac{3.95\times 10^{- 25} - 3.90\times 10^{- 25}}{3\times 1.6\times 10^{- 19}\times 0.250}\times 3.00\times 10^{5}) = 2.5\ cm[/tex]

[tex]\Delta x = 1.25\ cm[/tex]

(b) Since the order of the distance is in cm, thus clearly this distance is sufficiently large enough in practical for the separation of the two uranium isotopes.

In a mass spectrometer, uranium-235 and uranium-238 ions can be separated based on their masses and velocities. The separation distance between their paths can be determined using the equation: d = mv/(qB). The distance between their paths is practical for the separation of uranium-235 from uranium-238.

In a mass spectrometer, triply charged uranium-235 and uranium-238 ions are separated based on their masses and velocities.

The separation of their paths when they hit a target after traversing a semicircle can be calculated using the equation:

d = mv/(qB)

where d is the separation, m is the mass of the ion, v is the velocity, q is the charge, and B is the magnetic field.

The distance between their paths seems to be big enough to be practical in the separation of uranium-235 from uranium-238, as even a small separation can result in significant enrichment over multiple passes through the mass spectrometer.

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Answer:

6.36 105 J

Explanation:

In calorimetry all the energy given to a system is converted to heat and the equation for heat is

     Q = m  [tex]c_{e}[/tex] ΔT

The temperature can be in degrees Celsius or Kelvin since the interval between them is the same, substitute and calculate

Q = 1.9 4186 (373-293)  

Q = 6.36 105 J  

This heat equals the energy supplied

Li+  [He]

N³-  [Ne]

In³+  [Kr] 4d10

Tl+  [Xe] 4f14 5d10 6S2

The ground-state electron configuration of the ions Li+, N3−, In3+, and Tl+ are [He], [Ne], [Kr]4d10, and [Xe]4f145d106s26p1 respectively. This notation suggests these ions have similar electronic structures to the noble gases and additional electrons in certain cases.

The ground-state electron configurations of the ions Li+, N3−, In3+, and Tl+ can be described using the noble gas core electron configuration. The noble gas core is essentially the electron configuration of the closest noble gas with less atomic number than the atom we are considering.

(a) Li+ has lost an electron compared to neutral Lithium. Its electron configuration becomes [He] - it resembles helium, a noble gas.

(b) N3− has gained three electrons compared to neutral Nitrogen and its electron configuration becomes [Ne] - it resembles neon, a noble gas.

(c) In3+ has lost three electrons compared to neutral Indium. Its electron configuration becomes [Kr]4d10 - core is like Kr (krypton), a noble gas, plus 10 electrons added in the d orbital.

(d) Tl+ has lost one electron compared to neutral Thallium and its electron configuration is [Xe]4f145d106s26p1 - core is like Xe (xenon), a noble gas, plus additional electrons.

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Answer:

A)  W = 1885 J , B) [tex]K_{f}[/tex] = 1885 J , C)  w = 8.68 rad / s , D)  t = 8,687 s , E)  P = 109 W  F) P = 2 [tex]P_{rms}[/tex]

Explanation:

Part A    The work in the rotational movement is

       W = τ θ

Let's look at the rotated angle

      θ = 12.0 rot (2pi rad / 1rot) = 75.398 rad

     W = 25.0 75.40

     W = 1885 J

Part B   Let's use the relationship between work and kinetic energy

      W = ΔK = Kf - Ko

As the body leaves the rest w₀ = 0     ⇒ K₀ = 0

      W = [tex]K_{f}[/tex] -0

      [tex]K_{f}[/tex] = 1885 J

Part C     The formula for kinetic energy is

      K = ½ I w²

     w² = 2k / I

     w = √ (2 1885/50)

     w = 8.68 rad / s

Part D     The power in the rotational movement

     P = τ w

     P = 25 8.68

     P = 217 W

     P = W / t

     t = W / P

     t = 1885/217

     t = 8,687 s

Part E   At average power is

     P = τ ([tex]w_{f}[/tex] -w₀)/ 2

We look for angular velocity with kinematics

    [tex]w_{f}[/tex = w₀ + α t

     τ = I α

      α = τ / I

      α = 25/50

      α = 0.5 rad / s²

calculate

      P = 25 (0.5 8.687)

      P = 108.6 W

      P = 109 W

Part F    

The average power is

      [tex]P_{rms}[/tex] = τ ([tex]w_{f}[/tex] -w₀) /

The instant power is

      P = τ w

The difference is that in one case the angular velocity is instantaneous and between averages

P / [tex]P_{rms}[/tex] = τ w / (τ ([tex]w_{f}[/tex]-w₀) / 2)

P / [tex]P_{rms}[/tex]= 2 w / Δw

For this case w₀ = o

p / [tex]P_{rms}[/tex] = 2

The motor exerts rotational motion to do 1885 Joules of work on the platform. The final angular velocity of the platform is 8.68 rad/s. The time it takes to do this work is 17.4 seconds. The average power delivered by the motor is 109 Watts and the instantaneous power is twice the average power.

A motor exerts a constant torque on a horizontal platform and we need to determine the work done, the rotational kinetic energy, the angular velocity, the time it takes, the average power, and compare the instantaneous power to the average power.

Firstly, the work done by the motor is calculated using the formula W = Torque x angular displacement. The angular displacement for 12.0 rotations will equals to 12.0 x 2π radians. So, W = 25.0 N.m x 12 x 2π rad = 1885 J. Hence, the motor does 1885 Joules of work on the platform.

For the rotational kinetic energy, since there is no friction, all of the work done on the platform is converted into kinetic energy, so, K_rot,f = 1885 J.

The final angular velocity ωf can be found from the rotational kinetic energy and the moment of inertia by the relation K_rot,f = 1/2 I ωf^2. From this, we can find ωf = √(2K_rot,f / I)= 8.68 rad/s.

The time it takes Δt to do the work can be calculated using Δt = angular displacement / average angular velocity. Hence, Δt= (12 x 2π) / ((0 + ωf) / 2)= 17.4 s.

The average power P_avg is given by the total work done divided by the total time, which gives P_avg= W / Δt = 109 Watts.

On the final part, the instantaneous power Pf is proportional to the final angular velocity. As Pf = Torque x ωf, we get Pf = 2 x P_avg. So, the instantaneous power being delivered by the motor at the end is twice the average power calculated before.

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Explanation:

Red dwarf and brown dwarf masses are less than a typical white dwarf mass measuring around 1.2 solar masses. But it's only a few kilometers of the radius. This is precisely because there is no force to overcome the contraction due to gravity. There is a constant  battle between the external force of fusion (who wants to expand the star) and inward pressure because of gravity (who wants to compact the star) of regular stars on the main sequence. There remains a balance between these two forces as long as the star remains on the celestial equator.

Red dwarfs are helped by the nuclear fusion force, but brown dwarfs were not large enough to cause the fusion of hydrogen, they are massive enough to generate sufficient energy in the core by fusing deuterium to sustain their volume.  However as soon as the star runs out of hydrogen to burn it weakens the force of the external fusion and gravity starts to compact the center of the star. The contraction heats up the core into more massive stars and helium fusion begins, rendering the star once again stable. However this helium fusion does not occur in stars with masses below 1.44Mo. Tightness persists for such stars until the star's gasses degenerate.

To develop this problem it is necessary to apply the related concepts at relative speed.

When an observer perceives the relative speed of a second observer, the function is described,

[tex]v' = \frac{v_1-v_2}{1-\frac{v_1v_2}{c^2}}[/tex]

Where,

[tex]v_1[/tex] = The velocity of the first escape pod

[tex]v_2[/tex] = The velocity of the second escape pod

c = Speed of light

v' = Speed of the first escape pod relative to the second escape pod.

Our values are given as,

[tex]v_1[/tex]= 0.7c

[tex]v_2[/tex]= -0.76c

Replacing we have,

[tex]v' = \frac{v_1-v_2}{1-\frac{v_1v_2}{c^2}}[/tex]

[tex]v' = \frac{0.7c-(-0.76c)}{1-\frac{(0.7c)(-0.76c)}{(3*10^8)^2}}[/tex]

[tex]v' = \frac{0.7c-(-0.76c)}{1-\frac{(0.7c)(-0.76c)}{(3*10^8)^2}}[/tex]

[tex]v' = 2.85*10^8m/s[/tex]

Therefore the speed of the first escape pod measure for the second escape pod is [tex]v' = 2.85*10^8m/s[/tex]

The relative speed of the second escape pod as measured by the first escape pod, using relativistic velocity addition, is approximately -0.04c (or just 0.04c considering the magnitude), where c is the speed of light.

The question involves calculating the relative speed of one escape pod as observed by the other escape pod in a scenario where they are moving in opposite directions. To do this, we must use the formula for relativistic velocity addition. The formula is as follows:

V = (v1 + v2) / (1 + v1*v2/c²),

where V is the relative velocity as measured by one escape pod, v1 and v2 are the velocities of the escape pods, and c is the speed of light. For this particular problem:

We substitute the values into the relativistic velocity addition formula to find the relative speed:

V = (0.70c - 0.76c) / (1 - 0.70*(-0.76)c²/c²)

Doing the calculations:

V = -0.06c / (1 + 0.532)c²/c²)

V = -0.06c / 1.532

V = -0.039c ≈ -0.04c

The negative sign indicates that the second escape pod is moving in the direction opposite to the first pod as measured by the first pod. It's important to keep in mind that this result is only an approximation, rounded to two significant figures as per the question's request.

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To solve the problem it is necessary to apply the concepts related to Newton's second law, as well as to the sum of forces in this type of bodies.

According to the description I make a diagram that allows a better understanding of the problem.

Performing sum of forces in the angular direction in which it is inclined we have to

[tex]\sum F = ma_{\theta}[/tex]

[tex]N - W cos\theta = ma_{\theta}[/tex]

[tex]N = ma_{\theta}+Wcos\theta[/tex]

Tangential acceleration can be expressed as

[tex]a_{\theta} = (r\ddot{\theta}+2\dot{r}\dot{\theta})[/tex]

Our values are given by,

[tex]\dot{\theta} = 2.9rad/s[/tex]

[tex]m = 0.15 lb[/tex]

[tex]\theta = 36\°[/tex]

[tex]v = 4.8ft/s[/tex]

Substituting [tex]\ddot{\theta}=0rad/s^2 ,  \dot{r}=-4.8ft/s, \dot{\theta}=2.9 rad/s[/tex]

[tex]a_{\theta} = r*0+2*(4.8*2.9)\\a_{\theta}=27.84ft/s^2[/tex]

At the same time we acan calculate the mass of the particle, then

W = mg

Where,

W = Weight of the particle

m = mass

g = acceleration due to gravity

[tex]0.15lb = m(32.2ft/s^2)[/tex]

[tex]m = 4.66*10^{-3}Lb[/tex]

Now using our first equation we have that

[tex]N = ma_{\theta}+Wcos\theta[/tex]

[tex]N = (4.66*10^{-3})(27.84)+0.2cos36[/tex]

[tex]N = 0.2914Lb[/tex]

Therefore the normal force exerted by the wall of the tube on the particle at this instant is 0.2914Lb

The magnitude of the normal force [tex]\( N \)[/tex] exerted by the wall of the tube on the particle at the instant when the angle [tex]\( \theta = 36^\circ \)[/tex] is approximately 0.249 lb

To determine the normal force [tex]\( N \)[/tex]exerted by the wall of the tube on the particle, we analyze the forces acting on the particle in a rotating reference frame. Here’s the step-by-step process:

1. Identify the given data:

  - Angular velocity [tex]\( \omega = 2.9 \)[/tex] rad/sec

  - Particle weight [tex]\( W = 0.15 \)[/tex] lb

  - Relative velocity towards O [tex]\( v_r = 4.8 \)[/tex] ft/sec

  - Angle [tex]\( \theta = 36^\circ \)[/tex]

2. **Convert the weight to mass:**

  -[tex]\( W = mg \)[/tex]

  -[tex]\( m = \frac{W}{g} \)[/tex]

  - Using[tex]\( g = 32.2 \text{ ft/sec}^2 \)[/tex]:

  - [tex]\( m = \frac{0.15 \text{ lb}}{32.2 \text{ ft/sec}^2} = 0.00466 \text{ slugs} \)[/tex]

3. Calculate the distance ( r ):

  - Since [tex]\( v_r = 4.8 \)[/tex] ft/sec is towards O and the tube is rotating with [tex]\( \omega \),[/tex] the distance [tex]\( r \)[/tex] can be found using [tex]\( r = \frac{v_r}{\omega} \)[/tex]:

  - [tex]\( r = \frac{4.8 \text{ ft/sec}}{2.9 \text{ rad/sec}} = 1.655 \text{ ft} \)[/tex]

4. **Determine the forces:**

  - Centrifugal force [tex]\( F_c = m \omega^2 r \)[/tex]:

   [tex]\[ F_c = 0.00466 \text{ slugs} \times (2.9 \text{ rad/sec})^2 \times 1.655 \text{ ft} = 0.0638 \text{ lb} \][/tex]

  - Component of gravitational force in the radial direction [tex]\( F_{g, r} = W \cos \theta \)[/tex]:

 [tex]\[ F_{g, r} = 0.15 \text{ lb} \times \cos 36^\circ = 0.1214 \text{ lb} \][/tex]

5. Calculate the normal force [tex]\( N \)[/tex]:

  - Normal force [tex]\( N \)[/tex] must balance the radial forces:

[tex]\[ N = F_c + F_{g, r} + m \cdot (\omega^2 r) \][/tex]

[tex]\[ N = 0.0638 \text{ lb} + 0.1214 \text{ lb} + 0.0638 \text{ lb} \][/tex]

 [tex]\[ N = 0.249 \text{ lb} \][/tex]

Answer: entropy

Explanation: entropy is the degree of disorderliness or randomness in a system

The displacement of a 15.5 kg mass undergoing simple harmonic motion with a frequency of 9.73 Hz, at a point in time 1.25 s after it was at its maximum displacement of 14.6 cm, is found to be -14.1 cm.

The displacement of a mass undergoing simple harmonic motion at any given point of time can be found using the formula x(t) = A cos(wt + φ), where 'A' is the amplitude (maximum displacement), 'w' is the angular frequency, and 'φ' is the phase constant. Given that the maximum displacement or amplitude 'A' is 14.6 cm (or 0.146 m), the frequency 'f' is 9.73 Hz, and the phase constant φ = 0 (as the displacement is maximum at t = 0), the angular frequency 'w' can be calculated as 2πf, which equals approximately 61.1 rad/s. Substituting all these values into the formula, we find that the displacement at time t = 1.25 s is x(t) = 0.146 cos(61.1*1.25 + 0) = -0.141 m, or -14.1 cm. Note that the negative sign indicates that the displacement is in the opposite direction of the initial maximum displacement.

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The displacement from equilibrium at time t = 1.25s is -0.042m.

The displacement from equilibrium of the mass at time t = 1.25 s can be calculated using the formula for simple harmonic motion. The displacement at any given time t is given by the equation x = A * cos(2πft), where A is the amplitude and f is the frequency. In this case, the amplitude is 0.146 m and the frequency is 9.73 Hz. Plugging in the values, we get x = 0.146 * cos(2π * 9.73 * 1.25), which gives us x = -0.042 m.

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Answer:

W = 0.384 J

Explanation:

Work and energy in the rotational movement are related

    W = ΔK = [tex]K_{f}[/tex] - K₀

    W = ½ I [tex]w_{f}[/tex]² - 1 /2 I w₀²

where W isthe work, I is the moment of inertia and w angular velocity

With the system part of the rest the initial angular speed is zero (w₀ = 0)

The angular and linear quantities are related

    v = w r

    w = v / r

Let's replace

    W = ½ I (v / r)²

Let's calculate

    W = ½  1.2 10⁻⁴ 1.2² / (1.5 10⁻²)²

    W = 0.384 J

    W = 38.4 J

Answer:

v = 3.7 m/s

Explanation:

As the swing starts from rest, if we choose the lowest point of the trajectory to be the zero reference level for gravitational potential energy, and if we neglect air resistance, we can apply energy conservation as follows:

m. g. h = 1/2 m v²

The only unknown (let alone the speed) in the equation , is the height from which the swing is released.

At this point, the ropes make a 30⁰ angle with the vertical, so we can obtain the vertical length at this point as L cos 30⁰, appying simply cos definition.

As the height we are looking for is the difference respect from the vertical length L, we can simply write as follows:

h = L - Lcos 30⁰ = 5m -5m. 0.866 = 4.3 m

Replacing in the energy conservation equation, and solving for v, we get:

v = √2.g.(L-Lcos30⁰) = √2.9.8 m/s². 4.3 m =3.7 m/s

Explanation:

The angular momentum is given by the moment of inertia, multiplied by the angular speed of the rotating body:

[tex]L=I\omega[/tex]

The angular speed is given by:

[tex]\omega=2\pi f\\\omega=2\pi 5.2\frac{rev}{s}\\\omega=32.67\frac{rad}{s}[/tex]

Now, we calculate the angular momentum:

[tex]L=0.32kg\cdot m^2(32.67\frac{rad}{s})\\L=10.45\frac{kg\cdot m^2}{s}[/tex]

The average torque is defined as:

[tex]\tau=I\alpha[/tex]

[tex]\alpha[/tex] is the angular acceleration, which is defined as:

[tex]\alpha=\frac{\omega_f-\omega_0}{t}[/tex]

We have to calculate [tex]\omega_f[/tex]:

[tex]\omega_f=2\pi (2.75\frac{rad}{s})\\\omega_f=17.28\frac{rad}{s}[/tex]

Now, we calculate the angular acceleration:

[tex]\alpha=\frac{17.28\frac{rad}{s}-32.67\frac{rad}{s}}{12s}\\\alpha=-1.28\frac{rad}{s^2}[/tex]

Finally, we can know the average torque:

[tex]\tau=0.32kg\cdot m^2(-1.28\frac{rad}{s^2})\\\tau=-0.41N\cdot m[/tex]

(a) The angular momentum of the skater is 10.45 kgm²/s

(b) The magnitude of the average torque that was exerted, is 0.41 Nm.

The angular momentum of the skater is calculated as follows;

L = Iω

where;

ω = 5.2 rev/s x 2π rad = 32.67 rad/s

L = 0.32 x 32.67

L = 10.45 kgm²/s

The angular acceleration is calculated as follows;

[tex]\alpha= \frac{\omega _f - \omega _i}{t}[/tex]

[tex]\alpha = \frac{17.28 -32.67 }{12} \\\\\alpha = -1.28 \ rad/s^2[/tex]

The magnitude of the average torque that was exerted, is calculated as;

τ = Iα

τ = 0.32 x (1.28)

τ = 0.41 Nm.

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Answer:

15.57 g/cm³

Explanation:

[tex]\rho_g[/tex] = Density of gold = 19.3 g/cm³

[tex]\rho_s[/tex] = Density of silver = 10.5 g/cm³

[tex]\rho_c[/tex] = Density of copper = 8.9 g/cm³

Assuming total mass as 1000 g

Volume of gold

[tex]V_g=\frac{0.75\times 1000}{19.3}\\\Rightarrow V_g=38.86\ cm^3[/tex]

Volume of silver

[tex]V_g=\frac{0.16\times 1000}{10.5}\\\Rightarrow V_g=15.238\ cm^3[/tex]

Volume of copper

[tex]V_c=\frac{0.09\times 1000}{8.9}\\\Rightarrow V_c=10.11\ cm^3[/tex]

Density of the alloy would be

[tex]\rho=\frac{M}{V_g+V_s+V_c}\\\Rightarrow \rho=\frac{1000}{38.86+15.238+10.11}\\\Rightarrow \rho=15.57\ g/cm^3[/tex]

The overall density of this alloy is 15.57 g/cm³

The overall density of an 18-karat gold alloy, which consists of 75% gold, 16% silver, and 9% copper, is approximately 15.2 grams per cubic centimeter given the known densities of the pure metals.

The density of an alloy (a mixture of metals) is calculated by finding the volume-weighted average of the densities of the constituent metals. In the case of 18-karat gold, we have a mixture of 75% gold, 16% silver, and 9% copper. Using the given densities of these metals, we can calculate the overall density of the alloy as follows:

rho18kt = 0.75*rhogold+0.16*rhosilver+0.09*rhocopper = 0.75*19.3g/cm³+0.16*10.5g/cm³+0.09*8.90g/cm³ = 15.2g/cm³.

Therefore, the overall density of the 18-karat gold alloy, rho18kt, is approximately 15.2g/cm³.

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Answer:

doubled the initial value

Explanation:

Let the area of plates be A and the separation between them is d.

Let V be the potential difference of the battery.

The energy stored in the capacitor is given by

U = Q^2/2C   ...(1)

Now the battery is disconnected, it means the charge is constant.

the separation between the plates is doubled.

The capacitance of the parallel plate capacitor is inversely proportional to the distance between the plates.

C' = C/2

the new energy stored

U' = Q^2 /  2C'

U' = Q^2/C = 2 U

The energy stored in the capacitor is doubled the initial amount.

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

[tex]\omega =\sqrt{\dfrac{K}{m}}[/tex]

The maximum speed v given as

[tex]v=\omega A[/tex]

A=Amplitude

[tex]v=\sqrt{\dfrac{K}{m}}\times A[/tex]

[tex]0.32=\sqrt{\dfrac{13.5}{0.75}}\times A[/tex]

A=0.075 m

A= 0.75 cm

The speed at distance x

[tex]v=\omega \sqrt{A^2-x^2}[/tex]

[tex]v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}[/tex]

[tex]v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}[/tex]

v= 0.21 m/s

The amplitude of the subsequent oscillations can be calculated using the formula for potential energy and the initial kinetic energy. The block's speed at a given point can be found using the conservation of mechanical energy and the equation for kinetic energy.

To determine the amplitude of the subsequent oscillations, we need to first calculate the potential energy stored in the spring when the block is hit. Since the block is moving at a speed of 32.0 cm/s, we can calculate its kinetic energy. We can then equate this kinetic energy to the potential energy of the spring, using the formula potential energy = (1/2) * k * A^2, where k is the spring constant and A is the amplitude. Rearranging the formula, we find that the amplitude is equal to the square root of (2 * kinetic energy / k).

To find the block's speed when x = 0.750 A, we can use the conservation of mechanical energy. At the maximum displacement, all the energy is converted between potential and kinetic energy. At this point, the potential energy is zero, so the kinetic energy is equal to the initial kinetic energy. Using the formula for kinetic energy, we can find the speed when x = 0.750 A using the equation kinetic energy = (1/2) * k * (x^2 - A^2) and solving for the speed.

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Answer:

Explanation:

Inductance L = 1.4 x 10⁻³ H

Capacitance C = 1 x 10⁻⁶ F

a )

current I = 14 .0 t

dI / dt  = 14

voltage across inductor

= L dI / dt

= 1.4 x 10⁻³ x 14

= 19.6 x 10⁻³ V

= 19.6 mV

It does not depend upon time because it is constant at 19.6 mV.

b )

Voltage across capacitor

V = ∫ dq / C

= 1 / C ∫ I dt  

= 1 / C ∫ 14 t dt

1 / C x 14 t² / 2

= 7 t² / C

= 7 t² / 1 x 10⁻⁶

c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance

energy stored in inductor

= 1/2 L I²

energy stored in capacitor

= 1/2 CV²

After time t

1/2 L I² = 1/2 CV²

L I² =  CV²

L x ( 14 t )² = C x  ( 7 t² / C )²

L x 196 t² = 49 t⁴ / C

t² = CL x 196 / 49

t = 74.8 μ s

After 74.8 μ s energy stored in capacitor exceeds that of inductor.

Answer:

Explanation:

Let the velocity of lander with respect to earth be v .

In relativistic mechanism the expression for relative velocity is

v_r = [tex]\frac{v -u}{1-\frac{uv}{c^2}}[/tex]

Given u = .6c , v_r = .8 c

Substituting the values

.8c = [tex]\frac{v -0.6c}{1-\frac{.6c\times v}{c^2} }[/tex]

.8c-.48v = v - .6c

v = .946c

b )

Distance in terms of time = .2 ly

In relativistic mechanism , expression for relativistic time is given by the following relation

t = [tex]\frac{t_0}{\sqrt{1-\frac{v^2}{c^2} }}[/tex]

Substituting v = .946c

t₀ = .2

t = [tex]\frac{.2}{\sqrt{1-\frac{0.946\times .946c^2}{c^2}}}[/tex]

.2 / √.1050

= .62 ly

distance to the Earth at the time of lander launch, as observed by the aliens will be .62 ly.

Final answer:

The speed of the lander as observed on Earth is 0.865c. The distance to the Earth at the time of lander launch, as observed by the aliens, is 0 ly, which is an unreasonable result.

Explanation:

(a) To find the speed of the lander as observed on Earth, we need to use the relativistic velocity addition formula. The formula is given by:

v' = (v1 + v2)/(1 + (v1*v2)/c^2)

Substituting in the values, where v1 = 0.800c and v2 = 0.600c, we get:

v' = (0.800c + 0.600c)/(1 + (0.800c*0.600c)/c^2) = 1.280c/1.480 = 0.865c

So, the speed of the lander as observed on Earth is 0.865c.

(b) To find the distance to the Earth at the time of lander launch, as observed by the aliens, we can use the time dilation formula. The formula is given by:

t' = t/sqrt(1 - (v^2/c^2))

Where t' is the time measured by the aliens, t is the time measured on Earth, v is the velocity of the spaceship relative to Earth, and c is the speed of light.

In this case, t' = 0 (since the lander is launched at the same time as observed on Earth), t = 0.200 ly (given in the question), v = 0.600c (velocity of the spaceship relative to Earth), and c is the speed of light. Substituting the values, we get:

0 = 0.200/sqrt(1 - (0.600c)^2/c^2)

0 = 0.200/sqrt(1 - 0.360)

0 = 0.200/sqrt(0.640)

0 = 0.200/0.8

0 = 0.25

Since 0 = 0.25 is not possible, this result is unreasonable. It suggests that there is a discrepancy in the calculations or assumptions made.

Answer:

Explanation:

Distance between plates d = 2 x 10⁻³m

Potential diff applied = 5 x 10³ V

Electric field = Potential diff applied /  d

= 5 x 10³  / 2 x 10⁻³

= 2.5 x 10⁶ V/m

This is less than  breakdown strength for air  3.0×10⁶ V/m

b ) Let the plates be at a separation of d .so

5 x 10³ / d = 3.0×10⁶ ( break down voltage )

d = 5 x 10³  / 3.0×10⁶

= 1.67 x 10⁻³ m

= 1.67 mm.

Answer:

b. 0.75 mm

Explanation:

The distance between antinodes d is half the wavelength [tex]\lambda[/tex]. We can obtain the wavelength with the formula [tex]v=\lambda f[/tex], where f is the frequency given ([tex]f=1MHz=1\times10^6Hz[/tex]) and v is the speed of sound in body tissues (v=1540m/s), so putting all together we have:

[tex]d=\frac{\lambda}{2}=\frac{v}{2f}=\frac{1540m/s}{2(1\times10^6Hz)}=0.00077m=0.77mm[/tex]

which is very close to the 0.75mm option.

Answer:

The wavelength is 742.7 nm.

Explanation:

Given that,

Grating = 4500 lines/cm

Angle = 42°

Order number =2

We need to calculate the distance

[tex]d=\dfrac{1\times10^{-2}}{4500}[/tex]

[tex]d=2.22\times10^{-6}\ m[/tex]

We need to calculate the wavelength

Using diffraction formula

[tex]d\sin\theta=m\times\lambda[/tex]

[tex]\lambda=\dfrac{d\sin\theta}{m}[/tex]

[tex]\lambda=\dfrac{2.22\times10^{-6}\times\sin42}{2}[/tex]

[tex]\lambda=7.427\times10^{-7}[/tex]

[tex]\lambda=742.7\ nm[/tex]

Hence, The wavelength is 742.7 nm.

The wavelength of this light was found to be 7.08 × 10⁻⁴ m.

To find the wavelength of light that results in a second-order maximum at a 42° angle when striking a diffraction grating with 4500 lines per centimeter, we use the formula for diffraction grating: d sin θ = mλ, where d is the distance between adjacent grating lines, θ is the angle of the maximum, m is the order of the maximum, and λ is the wavelength of the light. First, calculate the distance between grating lines (d) as the reciprocal of the grating's line density (4500 lines/cm or 4.5×105 lines/m), yielding d = 1/4.5×105 m. Then, substituting the given values into the formula with m = 2 and θ = 42°, solve for the wavelength λ.
λ = (1/4.5×105) * sin(42°) / 2
  = 7.08 × 10⁻⁴ m

Explanation:

Given that,

Force, [tex]F=((-8i)+6j)\ N[/tex]

Position of the particle, [tex]r=(3i+4j)\ m[/tex]

(a) The toque on a particle about the origin is given by :

[tex]\tau=F\times r[/tex]

[tex]\tau=((-8i)+6j) \times (3i+4j)[/tex]

Taking the cross product of above two vectors, we get the value of torque as :

[tex]\tau=(0+0-50k)\ N-m[/tex]

(b) Let [tex]\theta[/tex] is the angle between r and F. The angle between two vectors is given by :

[tex]cos\theta=\dfrac{r.F}{|r|.|F|}[/tex]

[tex]cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }[/tex]

[tex]cos\theta=\dfrac{0}{50}[/tex]

[tex]\theta=90^{\circ}[/tex]

Answer:

25 N

Explanation:

Given that

Weight ,mg = 50 N

m=Mass of the box

g=acceleration due to gravity

Horizontal force F= 10 N

Coefficient of friction  ,μ = 0.4

Lets take vertical force = R N

In vertical direction

R + N = mg

N= mg - R

The friction force Fr

Fr= μ N

Fr= μ ( mg - R)

To start the motion

F > Fr

10   > 0.4 ( 50 - R )

25  > 50 - R

R > 50 - 25  

R > 25 N

Therefore minimum force R= 25 N

In the given problem, the minimum vertical force required to set the box into motion is greater than 15N. This force reduces the normal force and consequently the frictional force such that it becomes less than the horizontally applied force.

In this physical situation, you need to understand the role of static friction and normal force in setting the box into motion. The frictional force is calculated by multiplying the coefficient of static friction (0.4 in this case) and normal force. The normal force on the box is the weight of the box minus the upward force applied. Initially, the upward force is zero, so the frictional force is 0.4 * 50N = 20N. This is greater than the 10N horizontal force applied, so the box does not move.

To make the box move, the vertical force has to reduce the normal force such that the frictional force (which is now less due to the decreased normal force) becomes less than the applied horizontal force (10N). Let's consider the vertical upward force needed as F. Hence, the new frictional force will be 0.4 * (50N - F) and should be less than 10N for the box to move. Solving this inequality, the minimum F needed is >15N, anything above this will make the box move.

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The concept necessary to develop this exercise is that of a magnetic field on a surface. The magnetic field is a tool to describe how a magnetic force is distributed in the space around and within something magnetic.

It can be defined as,

[tex]B = \frac{\mu_0 I}{2\pi R}[/tex]

Where,

R= The distance from the point

I = Current

[tex]\mu_0 =[/tex] Permeability constant in free space

Our values are given as,

[tex]\mu_0 = 4\pi * 10^{-7}H/m[/tex]

[tex]I = 200 A[/tex]

[tex]R = 20m[/tex]

Replacing ,

[tex]B = \frac{(4\pi*10^{-7})(200)}{2\pi 20}[/tex]

[tex]B = 2*10^{-6}T[/tex]

Therefore the magnetic field strength on the ground directly under such a transmission line is [tex]2*10^{-6}T[/tex]

The magnetic field strength on the ground directly under a typical high-voltage transmission line carrying a 200 A current at a potential of 110 kV, which is 20 m above the ground, is 2 μT. This is calculated using Ampere's Law. Current evidence does not conclusively support any health hazards associated with such exposure.

To answer your question about the strength of the magnetic field on the ground directly under a typical high-voltage transmission line carrying a 200 A current at a potential of 110 kV, we should first understand the basic physics principle. The magnetic field created by an electric current in a long straight wire forms concentric circles around the wire, and the strength B of the magnetic field is given by Ampere's law.

Ampere's law states that B = μ₀I / 2πr, where B is the magnetic field strength, μ₀ is the permeability of free space (4π × 10^-7 T m/A), I is the current, and r is the distance from the wire. In this case, the transmission line is 20 m above the ground, so r = 20 m. The current I = 200 A. Substituting these values into the equation, we get B = (4π × 10^-7 T m/A * 200 A) / (2π * 20 m) = 2 × 10^-6 T or 2 μT.

So, the strength of the magnetic field on the ground directly under the transmission line is 2 μT, which is less than a tenth of Earth's admittedly weak magnetic field. While there's an ongoing controversy regarding potential health hazards associated with exposure to these electromagnetic fields (E-fields), current evidence does not conclusively support these claims.

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For the development of this problem it is necessary to apply the concepts related to the wavelength depending on the frequency and speed of light.

By definition we know that frequency can be expressed as

[tex]f = \frac{v}{\lambda}[/tex]

Where,

v = Velocity

[tex]\lambda =[/tex] Wavelength

Our values are

[tex]v=345m/sec[/tex]

[tex]f=850Hz[/tex]

Re-arrange to find Wavelength

[tex]f = \frac{v}{\lambda}[/tex]

[tex]\lambda = \frac{v}{f}[/tex]

[tex]\lambda = \frac{345}{850}[/tex]

[tex]\lambda = 0.4058m[/tex]

Converting to centimeters,

[tex]\lambda = 0.4058m(\frac{100cm}{1m})[/tex]

[tex]\lambda = 40.58cm[/tex]

Therefore the wavelength of the sound waves in the tube is 40.58cm

Answer:

The carriage speed B was 67.4 mph

Explanation:

This is an exercise for the moment, that as a vector we must look for the solution of each axis (x, y). We define a system formed by the two cars, for this system the forces during the crash are internal, so the moment is preserved.

The data they give is the car more A m = 1000kg and its speed is v1₁₀ = 30 mph i^ and the mass of the car B M = 1500 kg

Let's write the moment for each axis

X axis

     p₀ₓ = [tex]p_{fx}[/tex]

     m v₁ₓ + 0 = (m + M) vₓ

Y axis

     poy = [tex]p_{fy}[/tex]

      0 + M [tex]v_{2y}[/tex] = (m + M) [tex]v_{y}[/tex]

Let's look for the components of the final velocity with trigonometry

     sin 66 = [tex]v_{y}[/tex]  / v

     cos 66 = vₓ / v

     [tex]v_{y}[/tex]  = v sin 66

     vₓ = v cos 66

We substitute and write the system of equations

     m v₁ₓ = (m + M) v cos 66

     M [tex]v_{2y}[/tex]  = (m + M) v sin66

From the first equation

    v = m / (m + M) v₁ₓ / cos 66

    v = 1000 / (1000 + 1500) 30 / cos 66

    v = 29.5 mph

From the second equation

   [tex]v_{2y}[/tex]  = (m + M)/m   v sin 66

  [tex]v_{2y}[/tex]  = (1000 + 1500) /1000     29.5 sin 66

   [tex]v_{2y}[/tex]  = 67.4 mph

The carriage speed B was 67.4 mph