A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 m and a total mass of 120 kg. The turntable is initially rotating at 3.00 rad>s about a vertical axis through its center. Suddenly, a 70.0-kg parachutist makes a soft landing on the turntable at a point near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal

Answer:

[tex]F = 2\pi \dfrac{mMG}{L^2}}[/tex]

Explanation:

Assuming given,

Mass of wire be M

length of wire be L

small mass at center is = m

Radius of the wire be equal to = R = L/π

mass of small element of the wire

[tex]dM = \dfrac{M}{L}Rd\theta[/tex]

All the force are acting along y- direction

so, for force calculation

[tex]F = \int \dfrac{mdMG}{R^2} sin\theta [/tex]

[tex]F = \int_0^{\pi} \dfrac{m\dfrac{M}{L}RG}{R^2} sin\theta d\theta[/tex]

[tex]F = \int_0^{\pi} \dfrac{mMG}{L\dfrac{L}{\pi}} sin\theta d\theta[/tex]

[tex]F = \pi \dfrac{mMG}{L^2}}\int_0^{\pi}sin\theta d\theta[/tex]

[tex]F = \pi \dfrac{mMG}{L^2}}(-cos \theta)_0^{\pi}[/tex]

[tex]F = 2\pi \dfrac{mMG}{L^2}}[/tex]

The magnitude of the gravitational force is : [tex]2\pi \frac{mMG}{L^{2} }[/tex]

Given that :

mass of wire = M

length of wire = L

small mass at center = m

radius of wire = L / [tex]\pi[/tex]

First step : express the mass of small element of wire

dM = [tex]\frac{M}{L} Rd[/tex]∅

Since all forces act in the vertical direction the magnitude of the force exerted will be

F = [tex]\int\limits^\pi _o {\frac{m\frac{M}{L} RG}{R^2} } \, sin\beta d\beta[/tex]

[tex]F = \pi \frac{mMG}{L^2} \int\limits^\pi _0 {x} \, sin\beta d\beta[/tex]

Resolving equation above

Therefore F = [tex]2\pi \frac{mMG}{L^{2} }[/tex]

Hence we can conclude that The magnitude of the gravitational force is : [tex]2\pi \frac{mMG}{L^{2} }[/tex]

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Answer:

b

Explanation:

A thermodynamic-ally favored reactions are nothing but spontaneous reactions. In spontaneous reactions ΔH (enthalpy) is always negative and ΔS( entropy) is always positive.

Now all thermodynamic-ally favored reactions does not occur rapidly because the reactants still need to break bonds and overcome the activation energy in order to form products.

Thermodynamically favored reactions do not necessarily occur rapidly due to the need to overcome activation energy, which is a kinetic factor unrelated to the thermodynamics of the reaction. The rate of reaction is influenced by how frequently reactants can attain the transition state, which may be facilitated by catalysts and temperature.

Not all thermodynamically favored reactions occur rapidly because of kinetics related to activation energy. While thermodynamics can indicate whether a reaction is favorable or not, it does not take into account the energy required to reach the transition state where reactants are converted into products.

Option b is correct because reactants need to overcome the activation energy barrier and break bonds to form products. The speed at which a reaction approaches its equilibrium does not depend on the reaction's equilibrium constant but rather on the energy profile of the reaction pathway and the frequency at which reactants attain the critical transition state.

Catalysts can lower the activation energy, hence increasing the rate of a reaction without altering the thermodynamic favorability. Conditions such as temperature also play a role since molecules at higher temperatures have more kinetic energy, which increases the probability of achieving the energy threshold needed to initiate the reaction.

Among the following, may you do to a vector without changing is to move the vector without changing its orientation.

A vector is a quantity with both magnitude and direction in physics. It is often depicted by an arrow with the same direction as the amount and a length proportionate to the size of the quantity. A vector lacks position, but has magnitude and direction. A vector is therefore unaffected by displacement parallel to itself as far as its length is unaltered.

Scalars are regular variables with a magnitude but no direction, in contrast to vectors. In contrast to speed (the amount of speed), time, or mass, which are scalar values, displacement, speed, and acceleration are all vector quantities.

Therefore, it concludes that option B is correct.

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Answer:

1.1

Explanation:

Cosmic rays are particles traveling at extreme speeds through intergalactic space. Many are launched by exploding supernovae. From the question, we know that slowest cosmic rays travel at 43% of the speed of light that is 43/100 multiply by the speed of light(speed of light= 3.0×10^8 m/s).

Gamma factor can be calculated using the formula below;

Gamma factor= 1/(1- v^2/c^2)^1/2 --------------------------------------------(1).

Gamma factor= 1/(1-3.0×10^8 metre per seconds ×43/100÷ 3.0×10^8 metre per seconds) ^1/2

Gamma factor= 1/(1-1.6641×10^16/3.0×10^2)^1/2

Gamma factor= 1/(1-1.16641×10^16/9×10^16)^1/2

Gamma factor= 1/(1-0.1296)^1/2

Gamma factor= 1/(0.87)^1/2

Gamma factor= 1/0.9327

Gamma factor= 1.072112535

Gamma factor= ~ 1.1

Cosmic rays are high-speed particles that travel through space close to the speed of light. The gamma factor for cosmic rays traveling at 43% of the speed of light is  1.17.

Cosmic rays are high-speed particles that travel through interstellar space at speeds close to the speed of light. When calculating the gamma factor for cosmic rays traveling at 43% of the speed of light, we can use the formula: γ = 1 / √(1 - v^2/c^2), where v is the velocity of the particles and c is the speed of light. Substituting the values, the gamma factor for cosmic rays traveling at 43% of the speed of light is approximately 1.17.

The maximum distance the truck can travel without the box sliding can be calculated by finding the maximum static friction force and using it to determine the distance.

To find the maximum distance the truck can travel without the box sliding, we need to determine the maximum static friction force and then use it to calculate the distance. The maximum static friction force can be found by multiplying the coefficient of static friction (0.24) by the weight of the box (20 kg multiplied by the acceleration due to gravity, 9.8 m/s²).

The equation to find the maximum distance is: distance = (1/2) * acceleration * time². Plugging in the values into the equation, we get: distance = (1/2) * 0.24 * 20 kg * 9.8 m/s² * (3.0 s)². Solving the equation gives us the maximum distance the truck can travel without the box sliding.

The center of mass of the system is located at[tex]x_{cm} = \frac{L}{2}[/tex].

To find the x-coordinate of the center of mass (x_{cm}) for a system of particles, we will use the formula for the center of mass in one dimension:

[tex]x_{cm} = \frac{1}{M} \sum_{i} m_{i} x_{i}[/tex]

Where:

In this scenario, we have a particle of mass [tex]M[/tex] located at the origin ([tex]x = 0[/tex]), and additional mass elements distributed along the x-axis.

Setting up the mass distribution:

Let's assume there is a uniform rod of length [tex]L[/tex] and linear mass density [tex]\lambda = \frac{M}{L}[/tex].

This implies that the mass of an infinitesimal length element [tex]dx[/tex] of the rod is given by:

[tex]dm = \lambda \, dx = \frac{M}{L} \, dx[/tex]

Finding the x-coordinate of the center of mass:  

We set up an integral to calculate the position of the center of mass:

[tex]x_{cm} = \frac{1}{M} \int_{0}^{L} x \, dm \\= \frac{1}{M} \int_{0}^{L} x \frac{M}{L} \, dx[/tex]

Calculating the integral:  

This becomes:

[tex]x_{cm} = \frac{1}{M} \cdot M \int_{0}^{L} \frac{x}{L} \: dx \\= \frac{1}{L} \int_{0}^{L} x \, dx = \frac{1}{L} \left[ \frac{x^{2}}{2} \right]_{0}^{L} \\= \frac{1}{L} \cdot \frac{L^{2}}{2} \\= \frac{L}{2}[/tex]

The x-coordinate of the center of mass of the system, expressed in terms of [tex]L[/tex], is [tex]\frac{L}{2}[/tex]. This means that for this uniform distribution of mass along the length of the rod, the center of mass is located exactly in the middle of the rod.

Answer:

The wavelength of the station’s signal is 2.9 meters

The energy of a photon of the station’s electromagnetic signal is [tex]6.9\times10^{-26}J [/tex]

Explanation:

Wavelength [tex] \lambda [/tex] is inversely proprtional to frequency (f) and directly proportional to velocity of the wave (v).

[tex]\lambda=\frac{v}{f} [/tex] (1)

But electromagnetic waves as radio signals travel at speed of light so using this on (1):

[tex]\lambda=\frac{c}{f}=\frac{3.0\times10^{8}}{104.5\times10^{6}}\approx2.9\,m [/tex]

Albert Einstein discovered that energy of electromagnetic waves was quantized in small discrete packages of energy called photons with energy:

[tex] E=hf=(6.6\times10^{-34})(104.5\times10^{6})\approx6.9\times10^{-26}J[/tex]

with h the Planck's constant.

The wavelength of a 104.5 MHz FM radio signal is approximately 2.87 meters, and the energy of a photon of this radio signal is approximately 6.92 * 10^-26 Joules.

The subject of this question is the relationship between the frequency, wavelength, and energy of radio waves, specifically those used in FM radio broadcasting.

To calculate the wavelength of the radio signal, one can employ the wave equation: velocity = frequency * wavelength. Since the velocity of electromagnetic waves, which include radio waves, is the speed of light (3 * 10^8 m/s), the wavelength can be obtained by rearranging the equation to: wavelength = velocity / frequency. Using your FM station's frequency of 104.5 MHz (or 104.5 * 10^6 Hz), the wavelength of the station's signal would be approximately 2.87 meters.

The energy of a photon from this radio signal could be found through the photon energy equation: E = h * f, where h is Planck's constant (6.62607004 × 10^-34 m^2 kg / s) and f is the frequency in Hz. Thus, the energy of a radio signal photon at 104.5 MHz would be approximately 6.92 * 10^-26 J.

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Final answer:

To determine the speed of water emerging from a hole in a tank, Torricelli's theorem is used, considering both gravitational potential energy and the additional gauge pressure. The time to drain the tank involves integrating the flow rate over time, and is more complex when accounting for the compressed air in the tank maintaining pressure, compared to an open tank scenario.

Explanation:

Draining Water Through a Hole in a Cylindrical Tank

The question involves applying principles of fluid dynamics, specifically Bernoulli's equation and the equation of continuity, to determine the speed of water flowing out of a hole in a tank and the time it takes for the tank to empty. Part (a) requires solving for the initial speed of water ejection using the provided gauge pressure and comparing it to the efflux speed with an open tank. Part (b) deals with calculating the total time it takes for the tank to drain and comparing it with the draining time when the tank's top is open to the air.

To find the speed of the water as it emerges from the hole, we can use Torricelli's theorem, which is derived from the conservation of energy (Bernoulli's equation). The formula is:

v = √(2gh + (2P/ρ))

where v is the velocity of the water exiting the hole, g is the acceleration due to gravity (9.81 m/s2), h is the depth of the water, P is the gauge pressure, and ρ is the density of the water.

The ratio of this speed to the efflux speed if the top of the tank is open to the air can be determined by comparing the initial velocity found using the pressure provided and the velocity with only the atmospheric pressure affecting the tank (P = 0 for open tank).

For part (b), the time to drain the tank completely can be estimated by integrating the flow rate over time, considering the decreasing water level in the tank. However, since the tank has compressed air, which maintains a constant pressure, it changes the dynamics compared to an open tank. For an open tank, Torricelli's theorem is used with varying depth h throughout the duration of the draining. The integration process is more complex and may require approximation methods or numerical integration.

Using fluid dynamics principles, the speed of water emerging from a closed tank is about 4.82 m/s, higher than from an open tank by a factor of 1.22. The tank drains in roughly 27.7 minutes, quicker by a factor of 0.82 compared to an open tank.

To solve the problem of determining the speed of water as it emerges from the hole in the tank and the time for the tank to drain, we need to apply principles from fluid dynamics, specifically Bernoulli's equation and Torricelli's law.

(a) Speed of Water Emerging from the Hole

The speed of water v emerging from a hole can be found using Torricelli's law, modified to include the effect of the gauge pressure above the water. The equation is:

[tex]v = \sqrt(2 * g * h + 2 * P_g_a_u_g_e / \rho)[/tex]

Where:

g is the acceleration due to gravity, 9.81 m/s²

h is the depth of the water, 0.800 m

P_gauge is the gauge pressure, 5.00 × 10³ Pa

ρ is the density of water, 1000 kg/m³

Substituting the values, we get:

v = √(2 * 9.81 * 0.800 + 2 * 5000 / 1000)

≈ 4.82 m/s

Speed Ratio: If the top of the tank is open to the air, the speed would be:

[tex]v_o_p_e_n = \sqrt(2 * 9.81 * 0.800)[/tex]

≈ 3.96 m/s

Therefore, the ratio is:

(4.82 / 3.96) ≈ 1.22

(b) Time for Water to Drain

The time t for the tank to drain can be estimated using the equation for the volume flow rate:

Q = A * v and

t = V / Q

Where:

A is the area of the hole

v is the speed of water

V is the volume of water in the tank

Surface area A of the hole:

π * (0.0200/2)² = 3.14 * 10⁻⁴ m²

Volume V of water:

π * (2.00/2)² * 0.800

= 2.51 m³

Flow rate Q:

3.14 * 10⁻⁴ m² * 4.82 m/s

= 1.51 * 10⁻³ m³/s

Drain time t:

2.51 m³ / 1.51 * 10⁻³ m³/s

= 1661 s or about 27.7 min

Time Ratio:

If the top of the tank is open to the air, the time would be (2.51 / 1.24 * 10⁻³)

≈ 2024 s or about 33.7 min.

The ratio is (1661 / 2024)

≈ 0.82

Thus, the speed of water emerging from a closed tank is about 4.82 m/s, higher than from an open tank by a factor of 1.22. The tank drains in roughly 27.7 minutes, quicker by a factor of 0.82 compared to an open tank.

1) The temperature of the gas in state B is 1200 K

2) The work done by the gas from A to B is 4000 J

3) The work done by the gas from B to C is -5546 J

Explanation:

1)

The temperature of the gas when it is in state B can be found by using the ideal gas equation:

[tex]\frac{P_A V_A}{T_A}=\frac{P_B V_B}{T_B}[/tex]

where

[tex]P_A = 1\cdot 10^6 Pa[/tex] is the pressure in state A

[tex]V_A = 4 \cdot 10^{-3} m^3[/tex] is the volume in state A

[tex]T_A = 600 K[/tex] is the temperature in state A

[tex]P_B = 1\cdot 10^6 Pa[/tex] is the pressure in state B

[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B

[tex]T_B[/tex] is the temperature in state B

Solving for [tex]T_B[/tex],

[tex]T_B=\frac{P_B V_B T_A}{P_A V_A}=\frac{(1\cdot 10^6)(8\cdot 10^{-3})(600)}{(1\cdot 10^6)(4\cdot 10^{-3})}=1200 K[/tex]

2)

The work done by a gas during an isobaric transformation (as the one between A and B) is

[tex]W=p\Delta V = p (V_B-V_A)[/tex]

where

p is the pressure (which is constant)

[tex]V_B[/tex] is the final volume

[tex]V_A[/tex] is the initial volume

Here we have:

[tex]p=1\cdot 10^6 Pa[/tex]

[tex]V_A = 4 \cdot 10^{-3} m^3[/tex] is the volume in state A

[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B

Substituting,

[tex]W=(1\cdot 10^6)(8\cdot 10^{-3}-4\cdot 10^{-3})=4000 J[/tex]

3)

During an isothermal expansion, the produce between the pressure of the gas and its volume remains constant (Boyle's law), so we can write:

[tex]P_BV_B = P_C V_C[/tex]

where

[tex]P_B = 1\cdot 10^6 Pa[/tex] is the pressure in state B

[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B

[tex]P_C = 2\cdot 10^6 Pa[/tex] is the pressure in state C

[tex]V_C[/tex] is the volume in state C

Solving for [tex]V_C[/tex],

[tex]V_C = \frac{P_B V_B}{P_C}=\frac{(1\cdot 10^6)(8\cdot 10^{-3})}{2\cdot 10^6}=4\cdot 10^{-3} m^3[/tex]

The work done by a gas during an isothermal transformation is given by

[tex]W=nRT ln(\frac{V_C}{V_B})[/tex] (1)

where

n is the number of moles

[tex]R=8.314 J/mol K[/tex] is the gas constant

T is the constant temperature (in this case, [tex]T_B[/tex])

[tex]V_C, V_B[/tex] are the final and initial volume, respectively

The number of moles of the gas can be found as

[tex]n=\frac{p_A V_A}{RT_A}=\frac{(1\cdot 10^6)(4\cdot 10^{-3})}{(8.314)(600)}=0.802 mol[/tex]

So now we can use eq.(1) to find the work done by the gas from B to C:

[tex]W=(0.802)(8.314)(1200) ln(\frac{4\cdot 10^{-3}}{8\cdot 10^{-3}})=-5546 J[/tex]

And the work is negative because the gas has contracted, so the work has been done by the surrounding on the gas.

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The temperature of the gas when it is in state B can be found using the ideal gas law. The work done by the gas in expanding isobarically from A to B can be calculated using a simple equation. The work done on the gas in going from B to C can also be calculated using a different equation.

The temperature of the gas when it is in state B can be found using the ideal gas law, which states that PV = nRT. Since the process is isobaric, the pressure remains constant. We can use the equation:

TB = TA * (VA / VB)

Where TA is the initial temperature (600 K), VA is the initial volume (4 x 10-3), and VB is the final volume (8 x 10-3).

The work done by the gas in expanding isobarically from A to B can be calculated using the equation:

WAB = PA * (VB - VA)

Where PA is the initial pressure (1 x 106) and VA and VB are the initial and final volumes, respectively.

The work done on the gas in going from B to C can be calculated using the equation:

WBC = -nRT * ln(VC / VB)

Where VC is the final volume (8 x 10-3) and VB is the initial volume (8 x 10-3).

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Answer:

1.6 miles/min..

Explanation:

Let y be the height of the balloon at time t. Our goal is to compute the balloon's velocity at the moment .

balloon's velocity dy/dt when θ =π/3 radian

so we can restate the problem as follows:

Given dθ/dt = 0.1 rad/min at θ = π/3

from the figure in the attachment

tanθ = y/5

Differentiating w.r.t "t"

sec^2 θ×dθ/dt = 1/4(dy/dt)

⇒ dy/dt = (4/cos^2 θ)dθ/dt

At the given moment θ =π/3 and dθ/dt = 0.1 rad/min.

therefore putting the value we get

[tex]\frac{dy}{dt} = \frac{4}{\frac{1}{2}^2 }\times0.1[/tex]

solving we get

= 1.6 miles/min

So the balloon's velocity at this moment is 1.6 miles/min.

Answer:

Part a)

[tex]f = \frac{0.5}{1.5} = \frac{1}{3}[/tex]

B) uniform Sphere

[tex]f = \frac{2}{7}[/tex]

C) uniform hollow sphere

[tex]f = \frac{2}{5}[/tex]

D) uniform hollow cylinder with inner radius R/2 and outer radius R

[tex]f = \frac{5}{13}[/tex]

Explanation:

As we know that fraction of total energy as rotational energy is given as

[tex]f = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2}[/tex]

now we have

[tex] f = \frac{mk^2(\frac{v^2}{R^2})}{mv^2 + mk^2(\frac{v^2}{R^2})}[/tex]

[tex]f = \frac{\frac{k^2}{R^2}}{1 + \frac{k^2}{R^2}}[/tex]

now we have

A) uniform Solid cylinder

for cylinder we know that

[tex]\frac{k^2}{R^2} = 0.5[/tex]

[tex]f = \frac{0.5}{1.5} = \frac{1}{3}[/tex]

B) uniform Sphere

for sphere we know that

[tex]\frac{k^2}{R^2} = \frac{2}{5}[/tex]

[tex]f = \frac{0.4}{1.4} = \frac{2}{7}[/tex]

C) uniform hollow sphere

for hollow sphere we know that

[tex]\frac{k^2}{R^2} = \frac{2}{3}[/tex]

[tex]f = \frac{\frac{2}{3}}{\frac{5}{3}} = \frac{2}{5}[/tex]

D) uniform hollow cylinder with inner radius R/2 and outer radius R

for annular cylinder we know that

[tex]\frac{k^2}{R^2} = \frac{5}{8}[/tex]

[tex]f = \frac{\frac{5}{8}}{\frac{13}{8}} = \frac{5}{13}[/tex]

The fraction of the total kinetic energy that is rotational for a uniform sphere is [tex]\frac{1}{3}[/tex]

Mathematically, the rotational kinetic energy of an object is given by this formula:

[tex]K.E_{rot}=\frac{1}{2} I\omega^2[/tex]  

Where:

Since we know that half of the total kinetic energy for a hollow, cylindrical shell that is rolling without slipping on a horizontal surface is translational and the other half is rotational. Thus, this fraction is given by this mathematical expression:

[tex]K.E=\frac{\frac{k^2}{R^2} }{1+\frac{k^2}{R^2}}[/tex]

a. For a uniform sphere:

[tex]\frac{k^2}{R^2}=0.5[/tex]

Substituting, we have:

[tex]K.E=\frac{0.5 }{1+0.5}\\\\K.E=\frac{0.5 }{1.5}\\\\K.E =\frac{1}{3}[/tex]

b. For a thin-walled hollow sphere:

[tex]\frac{k^2}{R^2}=\frac{2}{5}[/tex]

Substituting, we have:

[tex]K.E=\frac{\frac{2}{5} }{1+\frac{2}{5}}\\\\K.E=\frac{0.4 }{1.4}\\\\K.E =\frac{2}{7}[/tex]

c. For a uniform hollow sphere:

[tex]\frac{k^2}{R^2}=\frac{2}{3}[/tex]

Substituting, we have:

[tex]K.E=\frac{\frac{2}{3} }{1+\frac{2}{3}}\\\\K.E=\frac{0.7 }{1.7}\\\\K.E =\frac{2}{5}[/tex]

d. For a hollow sphere with outer radius (R) and inner radius:

[tex]\frac{k^2}{R^2}=\frac{5}{8}[/tex]

Substituting, we have:

[tex]K.E=\frac{\frac{5}{8} }{1+\frac{5}{8}}\\\\K.E=\frac{0.625 }{1.625}\\\\K.E =\frac{5}{13}[/tex]

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29.62 feet far, to the nearest tenth of a​ foot, does the balloon rise during this​ period

Explanation:

Hot-air balloon is rising vertically, the angle of elevation from a point on level ground = 122 feet (from the balloon)

The passenger compartment changes from = 17.9 degrees to 29.5 degree

              [tex]\tan \left(17.9^{\circ}\right)=\frac{a}{122 \mathrm{ft}}[/tex]

              [tex]a=122 \times \tan 17.9^{\circ}=122 \times 0.32299=39.4[/tex]

Similarly,

              [tex]\tan \left(29.5^{\circ}\right)=\frac{b}{122 f t}[/tex]

              [tex]b=122 \times \tan \left(29.5^{\circ}\right)=122 \times 0.56577=69.02[/tex]

The nearest tenth of a foot, the balloon rise during the period, as below

               69.02 – 39.4 = 29.62 ft.

Answer:

Pressure, [tex]P=1.33\times 10^6\ Pa[/tex]

Explanation:

Given that,

Radius of the opening, r = 0.45 mm = 0.00045 m

Weight of the pot, W = F = 0.848 N

To find,

The maximum pressure inside the pot

Solution,

We know that pressure inside the pot is equal to the force acting per unit area. It is given by :

[tex]P=\dfrac{F}{A}[/tex]

[tex]P=\dfrac{F}{\pi r^2}[/tex]

[tex]P=\dfrac{0.848\ N}{\pi (0.00045\ m)^2}[/tex]

[tex]P=1.33\times 10^6\ Pa[/tex]

So, the maximum pressure inside the pot is [tex]1.33\times 10^6\ Pa[/tex]. Hence, this is the required solution.

The greatest contrast in temperature and moisture occurs along the boundary separating cold and warm air masses showcased by the rain shadow effect where one side of a mountain range is wet and the other side is dry.

The greatest contrast in both temperature and moisture occurs along the boundary separating cold and warm air masses, particularly when these air masses meet along geographical features such as mountain ranges. An excellent example of this contrast is the rain shadow effect, where moist air from the ocean rises over a mountain range, cools, condenses, and precipitates on the windward side, leading to high moisture levels. However, once the air reaches the leeward side of the mountain, it becomes dry, resulting in arid conditions. This creates environments with stark differences in temperature and moisture, such as those found in the western United States, where the western slopes of the Cascades and the Sierra Nevada are lush and wet, while the eastern slopes are dry and semi-arid.

Answer:

0.28198 m/s²

0.02874

Explanation:

A = Amplitude = 1.4 m

T = Time period = 14 s

g = Acceleration due to gravity = 9.81 m/s²

Acceleration is given by

[tex]a=\omega^2A\\\Rightarrow a=\left(\frac{2\pi}{T}\right)^2A\\\Rightarrow a=\left(\frac{2\pi}{14}\right)^2\times 1.4\\\Rightarrow a=0.28198\ m/s^2[/tex]

The maximum acceleration of the passengers during this motion is 0.28198 m/s²

Dividing this value by g

[tex]\frac{a}{g}=\frac{0.28198}{9.81}\\\Rightarrow a=0.02874g[/tex]

The fraction of g is 0.02874

The maximum acceleration of the passengers is approximately 0.32 m/s², which is about 3.3% of the acceleration due to gravity.

This question involves simple harmonic motion, which describes the motion of the ocean cruise ship's passengers. The maximum acceleration in this case can be computed using the formula: a_max = (2 × π × f)² × A, where 'f' is the frequency (which is the reciprocal of the period 'T') and 'A' is the amplitude. We replace 'f' with 1/T.

A. So the maximum acceleration 'a_max' = (2 × π / T)² × A = (2 × π / 14s)² × 1.4m ≈ 0.32 m/s².

B. The fraction of this acceleration relative to g (gravitational acceleration, approximately 9.8 m/s²) can be calculated as 0.32 / 9.8 ≈ 0.033 or 3.3% of g.

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Answer:

The magnitude of the acceleration of the boy toward the girl is [tex]1.31\ m/s^2[/tex]

Explanation:

It is given that,

Mass of the boy, [tex]m_1=62\ kg[/tex]

Mass of the girl, [tex]m_2=37\ kg[/tex]

The acceleration of the girl toward the boy is, [tex]a_2=2.2\ m/s^2[/tex]

To find,

The acceleration of the boy toward the girl.

Solution,

Let [tex]a_1[/tex] is the magnitude of the acceleration of the boy toward the girl. We know that force acting on one object to other are equal in magnitude but opposite in direction. So,

[tex]F_1=-F_2[/tex]

[tex]m_1a_1=-m_2a_2[/tex]

[tex]a_1=-\dfrac{m_2a_2}{m_1}[/tex]

[tex]a_1=-\dfrac{37\times 2.2}{62}[/tex]

[tex]a_1=-1.31\ m/s^2[/tex]

[tex]|a_1|=1.31\ m/s^2[/tex]

So, the magnitude of the acceleration of the boy toward the girl is [tex]1.31\ m/s^2[/tex]

Answer:

6862.96871 seconds

Explanation:

M = Mass of Planet

G = Gravitational constant

r = Radius

[tex]\rho[/tex] = Density

T = Rotation period

In this system the gravitational force will balance the centripetal force

[tex]G\frac{Mm}{r^2}=mr\omega^2[/tex]

[tex]\omega=\frac{2\pi}{T}[/tex].

[tex]M=\rho v\\\Rightarrow M=\rho \frac{4}{3}\pi r^3[/tex]

[tex]\\\Rightarrow G\frac{Mm}{r^2}=mr\left(\frac{2\pi}{T}\right)^2\\\Rightarrow \frac{G\rho \frac{4}{3}\pi r^3}{r^3}=\frac{4\pi^2}{T^2}\\\Rightarrow T=\sqrt{\frac{3\pi}{G\rho}}[/tex]

Hence, proved

[tex]T=\sqrt{\frac{3\pi}{6.67\times 10^{-11}\times 3000}}\\\Rightarrow T=6862.96871\ s[/tex]

The rotation period of the astronomical object is 6862.96871 seconds

Explanation:

Scientist and astronomers have observed that large gas clouds and massive stars are orbiting around in an accelerated manner in center of our galaxy that is milky way. A massive star S2's motion has been studied and it is found that the star is revolving around the center of the galaxy. From this scientist have confirmed presence of super massive black hole of 3 million solar masses lurking around the center of milky way.

Answer:

b. Technician B only

Explanation:

There is a float connected to the variable resistor in a fuel tank unit. The resistance of the variable resistor also changes as the fuel level changes. The tank unit's resistance changes, the dash mount gauge also changes and available on driver's display.  

If the tank transmitter is disconnected, the operation will not take place and the resistance change will not be transmitted to the dash unit. The needle will therefore remain the empty one at all times.Even after being rusty, the ground wire connection to the fuel tank will be able to conduct. Hence Technician B   is correct and Technician A is incorrect.

Technician A is correct in stating that corrosion of the ground wire at the fuel tank sending unit can cause a lower-than-normal fuel gauge reading.

The accuracy of a fuel gauge reading can be affected by the condition of the electrical connections to the fuel tank sending unit. Technician A's statement is correct; if the ground wire's connection to the fuel tank sending unit becomes rusty or corroded, it can cause a higher resistance, leading the fuel gauge to read lower than normal due to insufficient grounding. On the other hand, Technician B's statement is incorrect because if the power lead to the fuel tank sending unit is disconnected and grounded with the ignition on, the fuel gauge should read full, not empty. The fuel gauge is designed so that grounding the sending unit wire to the chassis will mimic the resistance of a full tank, thus moving the gauge needle to the 'Full' position.

The correct answer, therefore, is a. Technician A only.

Final answer:

To determine the initial compression of the spring, use Hooke's Law. After the engine is ignited, calculate the rocket's acceleration using Newton's second law. Finally, find the rocket's velocity after traveling a distance without the spring.

Explanation:

To determine the initial compression of the spring, we can use Hooke's Law. Hooke's law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In equation form, this can be written as F = -kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement. In this case, the force exerted by the rocket is equal to the force exerted by the spring, so we can equate the two forces: F_spring = -kx. Plugging in the values for the force of the rocket (240 N) and the spring constant (400 N/m), we can solve for x, the compression of the spring.

After the engine is ignited, the rocket will experience an upwards force from the rocket engine and a downwards force from gravity. The net force on the rocket can be calculated by subtracting the weight of the rocket (mass x gravitational acceleration) from the thrust of the rocket engine. The net force can be used to calculate the acceleration of the rocket using Newton's second law: F_net = ma. Once we have the acceleration, we can use it to solve for the velocity of the rocket when the spring has stretched by 30.0 cm.

If the rocket were not attached to the spring, it would be free to move upwards under the influence of the thrust from the rocket engine. The acceleration of the rocket can be calculated using the same formula as before, but without the force from the spring. The initial velocity of the rocket is taken to be zero, and we can use the acceleration to solve for the velocity of the rocket after traveling a certain distance.

1. The spring is compressed by 0.450 meters.

2. The rocket's speed is 12.0 m/s when the spring has stretched 30.0 cm.

3. The rocket's speed without the spring would be 24.5 m/s after traveling this distance.

First, let's find how much the spring is compressed initially before the engine is ignited. We can use Hooke's Law: [tex]\( F = -kx \), where \( F \)[/tex] is the force exerted by the spring, [tex]\( k \)[/tex] is spring constant, and [tex]\( x \)[/tex] is the compression or stretching of the spring.

Given:

- Mass of the rocket, [tex]\( m = 10.3 \)[/tex] kg

- Thrust generated by the rocket, [tex]\( F_{\text{thrust}} = 240 \) N[/tex]

- Spring constant, [tex]\( k = 400 \)[/tex] N/m

The weight of the rocket [tex]\( F_{\text{weight}} = mg = 10.3 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 100.94 \, \text{N} \).[/tex]

Since the rocket is at rest before ignition, the net force is zero:

[tex]\[ F_{\text{net}} = F_{\text{thrust}} - F_{\text{weight}} - F_{\text{spring}} = 0 \]\[ 240 \, \text{N} - 100.94 \, \text{N} - F_{\text{spring}} = 0 \]\[ F_{\text{spring}} = 240 \, \text{N} - 100.94 \, \text{N} = 139.06 \, \text{N} \][/tex]

Now, use Hooke's Law to find the compression of the spring:

[tex]\[ F_{\text{spring}} = -kx \]\[ 139.06 \, \text{N} = -400 \, \text{N/m} \times x \]\[ x = \frac{139.06 \, \text{N}}{400 \, \text{N/m}} = 0.34765 \, \text{m} = 34.765 \, \text{cm} \][/tex]

So, initially, the spring is compressed by [tex]\( 34.765 \, \text{cm} \).[/tex]

After the engine is ignited and the spring has stretched by 30.0 cm, the net force will again be zero. We'll use the work-energy principle to find the velocity of the rocket. The work done by the thrust of the rocket is equal to the increase in kinetic energy of the rocket-spring system.

[tex]\[ W_{\text{thrust}} = \frac{1}{2}mv^2 - \frac{1}{2}kx^2 \][/tex]

Where [tex]\( W_{\text{thrust}} = F_{\text{thrust}} \cdot d \), with \( d = 0.3 \) meters (30.0 cm converted to meters).[/tex]

Plugging in the values and solving for [tex]\( v \), we get \( v = 12.0 \) m/s.[/tex]

If the rocket were not attached to the spring, it would still have the same kinetic energy at this point. So, we can use the work-energy principle again to find its velocity without the spring:

[tex]\[ W_{\text{thrust}} = \frac{1}{2}mv^2 \]\[ 240 \, \text{N} \times 0.3 \, \text{m} = \frac{1}{2} \times 10.3 \, \text{kg} \times v^2 \][/tex]

Solving for [tex]\( v \), we get \( v = 24.5 \) m/s.[/tex]

Complete Question:

A 10.3 kg weather rocket generates a thrust of 240 N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 400 N/m, is anchored to the ground. Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed? After the engine is ignited, what is the rocket's speed when the spring has stretched 30.0 cm? For comparison, what would be the rocket's speed after traveling this distance if it weren't attached to the spring?

The tensile strength of PCTFE is generally expected to be higher than that of PTFE, given the same molecular weight and degree of crystallinity, mainly due to stronger intermolecular forces from the chlorine atom in the PCTFE polymer chain.

The tensile strength of polychlorotrifluoroethylene (PCTFE) and polytetrafluoroethylene (PTFE) specimens having the same molecular weight and degree of crystallinity can vary. Although both materials are fluoropolymers and have similar structural characteristics, PCTFE tends to have a slightly higher tensile strength than PTFE. This is primarily because PCTFE has a chlorine atom in its polymer chain, which contributes to stronger intermolecular forces and therefore, higher tensile strength.

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Answer:

[tex]g=0.20\ m/s^2[/tex]    

Explanation:

It is given that,

Mass of the asteroid Ceres, [tex]m=6.797\times 10^{20}\ kg[/tex]

Radius of the asteroid, [tex]r=472.9\ km=472.9\times 10^3\ m[/tex]

The value of universal gravitational constant, [tex]G=6.67259\times 10^{-11}\ N.m^2/kg^2[/tex]

We know that the expression for the acceleration due to gravity is given by :

[tex]g=\dfrac{Gm}{r^2}[/tex]

[tex]g=\dfrac{6.67259\times 10^{-11}\times 6.797\times 10^{20}}{(472.9\times 10^3)^2}[/tex]

[tex]g=0.20\ m/s^2[/tex]

So, the value of acceleration due to gravity on that planet is [tex]0.20\ m/s^2[/tex]. Hence, this is the required solution.

Final answer:

To calculate the surface gravity of Ceres, apply Newton's law of universal gravitation and rearrange it to solve for 'g.' The result is approximately g = 0.28 m/s².

Explanation:

Calculating Gravitational Acceleration on Ceres

To find the acceleration due to gravity ‘g’ on the surface of Ceres, use Newton’s law of universal gravitation:

F = G * (m1 * m2) / r^2

Where F is the gravitational force, G is the gravitational constant, m1, and m2 are the masses of the two objects (in this case, a mass on the surface of Ceres and Ceres itself), and r is the distance between the centers of the two masses (the radius of Ceres in this scenario).

Since we are interested in 'g,' we rearrange this formula to solve for F/m2 (where m2 is a mass on Ceres’ surface and F/m2 equals g):

g = G * m1 / r^2

Plugging in the given values:

G = 6.67259 × 10⁻¹¹ N·m²/kg²

m1 (mass of Ceres) = 6.797 × 10²° kg,

r (radius of Ceres) = 472.9 × 10³ m,

The calculation is:

g = (6.67259 × 10⁻¹¹ N·m²/kg² * 6.797 × 10²° kg) / (472.9 × 10³ m)^2

After performing the calculation, ‘g’ on the surface of Ceres is found to be approximately 0.28 m/s²

Answer:

So height will be 47.387 m

Explanation:

We have given gauge pressure [tex]P=4.63\times 10^5N/m^2[/tex]

Density of water [tex]\rho =997kg/m^3[/tex]

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

We know that pressure is given as [tex]P=\rho gh[/tex]

[tex]4.63\times 10^5=997\times 9.8\times h[/tex]

[tex]h=47.387m[/tex]

So height will be 47.387 m

The height of the water in a tower to achieve a certain gauge pressure can be calculated using the hydrostatic pressure formula. By substituting the given pressure and known values for water density and earth's gravity into the formula, we can solve for the height.

This question involves the application of the concept of hydrostatic pressure. The height of the water above a user to create a specific pressure can be calculated using the formula for hydrostatic pressure. In this case, the  gauge pressure is given, and we know the density of water (1000 kg/m³) and the acceleration due to gravity is approximately 9.81 m/s². The formula for hydrostatic pressure is P = ρgh, where P represents hydrostatic pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid above the point in question. Thus, substituting the values, we can solve for h (height), we get: h = P/(ρg), where P = 4.63 x 10⁵ N/m². Inserting values will give us the height needed to create the given pressure.

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Final answer:

To calculate the moment of inertia and angular velocity of the rod-ball system after the collision, use the conservation of angular momentum, considering both the inertia of the rod and the putty ball. The total moment of inertia is the sum of each component's moment of inertia, and the angular velocity is determined by the ratio of initial angular momentum to the total moment of inertia.

Explanation:

When a putty ball strikes and sticks to a rotating rod in a collision with rotating rod, the conservation of angular momentum applies. To solve for the moment of inertia (I) and angular velocity (ω), we need to consider both the moment of inertia of the rod and the additional inertia from the putty ball at the point it sticks to the rod.

For the rod and ball system:

The angular speed ω can be found by equating the initial angular momentum of the ball before the collision to the final angular momentum of the system after the collision:

The rotational kinetic energy of the system after collision can be calculated using the expression K_{rot} = \frac{1}{2}Iω^2.

Answer:

5.65 times

Explanation:

60 db sound is equal to 60 phons sound when frequency is kept at 1000Hz.

But when the frequency of sound  is changed to 100 Hz , according to equal loudness curves , the loudness level on phon scale will be 35 phons.

A decrease of 10 phon on phon- scale makes sound 2 times less loud

Therefore a decrease of 25 phons will make loudness less intense by a factor equal to 2²°⁵ or 5.65 less intense . Therefore intensity at 100 Hz

must be increased 5.65 times so that its intensity matches intensity of 60 dB sound at  1000 Hz  frequency.

A 100-hz sound must be of higher intensity than a 1000-hz sound to be perceived as having the same loudness level due to the frequency-dependent sensitivity of human hearing. The exact intensity difference can be interpolated from an equal-loudness contour graph.

To determine how many times more intense a 100-hz sound must be compared to a 1000-hz sound to be perceived as equal to 60 phons of loudness, we need to reference a graph of loudness level contours often provided in acoustics materials.

Phons are a unit of loudness level that takes into account the frequency-dependent sensitivity of human hearing. The loudness in phons of a sound is the sound pressure level in decibels of an equally loud 1000-hz tone. Because of the equal-loudness contours of the human ear, a sound at 100 Hz needs to be of higher intensity compared to a sound at 1000 Hz to achieve the same perceived loudness. This is because the human ear is less sensitive at lower frequencies.

Using the fact that each factor of 10 in intensity corresponds to 10 dB increase, we can determine the additional intensity level needed for the 100 Hz sound. However, the exact amount can only be determined from the specific loudness contour graph, which gives the intensity in decibels required for sounds of different frequencies to have the same loudness level.

Answer:

2442.5 Nm

Explanation:

Tension, T = 8.57 x 10^2 N

length of rope, l = 8.17 m

y = 0.524 m

h = 2.99 m

According to diagram

Sin θ = (2.99 - 0.524) / 8.17

Sin θ = 0.3018

θ = 17.6°

So, torque about the base of the tree is

Torque = T x Cos θ x 2.99

Torque = 8.57 x 100 x Cos 17.6° x 2.99

Torque = 2442.5 Nm

thus, the torque is 2442.5 Nm.

Answer:

The weight of the bowling ball makes a more significant impact that the ping pong ball so therefore it would take farther to stop the bowling ball

Explanation:

(A)  The time required to stop the ping pong ball will be less than that of bowling ball.

(B)  The distance required to stop the bowling ball will be less than that of ping pong ball.

Given data:

The ping pong ball and bowling ball has the same magnitude of linear momentum.

Same amount of force to be applied on each, to stop.

(A)

With same magnitude of stopping force (F) and linear momentum (p), the time required to stop will be dependent on the mass. Since, mass of ping pong ball is less, so it will be easily stopped with less effort. While the bowling ball will take some extra seconds or minutes to acquire the rest state.

In other words, the ping pong ball has less inertia, due to which the time taken to stop the ping pong pall will be less, comparing to bowling ball.

Thus, we can conclude that the time required to stop the ping pong ball will be less than that of bowling ball.

(B)

The inertia is given as,

[tex]I = mr^{2}[/tex],

Here, m is the mass.

And the distance required to stop is given by third rotational equation of motion as,

[tex]\omega_{2}^{2}=\omega_{1}^{2}+2 \alpha \theta\\\\\theta =\dfrac{ \omega_{2}^{2}-\omega_{1}^{2}}{2 \alpha}[/tex]

Here, [tex]\alpha[/tex] is the angular acceleration.

And angular acceleration is directly proportional to the inertia of object. More the inertia, more will be the angular acceleration and less will be distance required to stop.

Since, ping pong ball has less inertia, so its angular acceleration will be less. So, the distance covered by the ping pong ball will be more, compared to bowling ball.

Thus, we can conclude that the distance required to stop the bowling ball will be less than that of ping pong ball.

Learn more about the inertia here:

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Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

[tex]P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m[/tex]

The spring constant of the net is 20130.76 N

From Hooke's Law

[tex]F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m[/tex]

The net would strech 0.03167 m

If h = 35 m

From energy conservation

[tex]65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0[/tex]

Solving the above equation we get

[tex]x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45[/tex]

The compression of the net is 1.52 m

Answer:

(1) Elliptical galaxies

(2) Spiral galaxies

(3) Irregular galaxies

(4) S0 galaxies

Explanation:

(1) Elliptical galaxies

Elliptical galaxies

These systems exhibit certain characteristic properties. They have complete rotational symmetry; i.e., they have figures of revolution with two equal principal axes. Third smaller axis is presumed axis of rotation. The surface brightness of elliptical at optical wavelengths decreases monotonically outward from a maximum value at the centre, following a common mathematical law of the form:

I = I0( r/a +1 )−2,

where I is the intensity of the light, I0 is the central intensity, r is the radius, and a is a scale factor.

(2) Spiral galaxies

Spiral galaxies are classified into two groups; ordinary and barred. The ordinary group is designated by S or SA, and the barred group by SB. In normal spirals, the arms originate directly from the nucleus, or bulge, where in the barred spirals, there is a bar of material that runs through the nucleus that the arms emerge from. Both types are given a classification according to how tightly their arms are wound. The classifications are a, b, c, d ... with "a" having the tightest arms. In type "a", the arms are usually not well defined and form almost a circular pattern. Sometimes you will see the classification of a galaxy with two lower case letters. This means that the tightness of the spiral structure is halfway between those two letters.

(3) Irregular galaxies:

Irregular galaxies have no regular or symmetrical structure. They are divided into two groups, Irr I and IrrII. Irr I type galaxies have HII regions, which are regions of elemental hydrogen gas, and many Population I stars, which are young hot stars. Irr II galaxies simply seem to have large amounts of dust that block most of the light from the stars. All this dust makes is almost impossible to see distinct stars in the galaxy.

(4) S0 galaxies

These systems exhibit some of the properties of both the elliptical and the spirals and seem to be a bridge between these two most common galaxy types. Hubble introduced the S0 class long after his original classification scheme had been universally adopted, largely because he noticed the dearth of highly flattened objects that otherwise had the properties of elliptical galaxies.

If the specific heat capacity of copper is three times that of lead, then it would require three times the amount of heat to cause the same temperature change in equal masses of copper and lead.

The amount of heat added to copper, whose specific heat capacity is three times the specific heat capacity of lead, is indeed three times the amount of heat given to lead when equal masses of copper and lead are heated from room temperature to the temperature of boiling water. This is because the specific heat, which is the heat required to raise the temperature of 1 gram of the substance 1 degree, is directly proportional to the amount of heat. Therefore, a substance with a higher specific heat capacity, like copper, would require more heat to obtain the same temperature change as a substance with a lesser specific heat capacity, like lead.

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