A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the oscillating system is 6.74 J.

A. Determine the following maximum speed of the object?

B. The force constant?

C. The amplitude of the Motion?

Answer:

672.29 W/m²

Explanation:

[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]

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

I = Intensity of light = 1200 W/m²

[tex]E_m[/tex] = Maximum value electric field

Intensity of light is given by

[tex]I=\frac{1}{2}\epsilon_0cE_m^2\\\Rightarrow E_m=\sqrt{\frac{2I}{\epsilon_0c}}\\\Rightarrow E_m=\sqrt{\frac{2\times 1200}{8.85\times 10^{-12}\times 3\times 10^8}}\\\Rightarrow E_m=950.765\ N/C[/tex]

RMS value

[tex]E_r=\frac{E_m}{\sqrt2}\\\Rightarrow E_r=\frac{950.765}{\sqrt2}\\\Rightarrow E_r=672.29\ W/m^2[/tex]

The approximate magnitude of the electric field in the sunlight is 672.29 W/m²

To solve this problem it is necessary to apply the concepts related to the kinematic equations of motion.

By definition we know that speed can be expressed as

[tex]v= \frac{d}{t}[/tex]

Where,

t = time

d = distance

For this specific case we know that the speed traveled corresponds to the round trip, therefore

[tex]v = \frac{2d}{t}[/tex]

Re-arrange to find t (The time taken to here the echo)

[tex]t = \frac{2d}{v}[/tex]

Replacing with our values we have that the distance is equal to 65m

[tex]t = \frac{2*65}{1530}[/tex]

[tex]t = 0.085s[/tex]

Therefore the time before the dolphin hears the echoes of the clicks is 0.085s

To solve the problem, it is necessary to apply the concepts related to the diffraction given in circular spaces. By definition it is expressed as

[tex]\Delta \theta = 1.22\frac{\lambda}{d}[/tex]

Where,

\lambda = Wavelength

d = Optical Diameter

[tex]\theta =[/tex] Angular resolution

In turn you can calculate the angle through the diameter and the arc length, that is,

[tex]\Delta \theta = \frac{x}{D}[/tex]

Where,

x = The length of the arc

D = Distance

From known data we know that Jupiter's diameter is,

[tex]x_J = 1.43*10^8m[/tex]

[tex]D = 20*9.4608*10^{15}[/tex]

[tex]\lambda = 600*10^{-9}m[/tex]

Replacing we have that,

[tex]\frac{x}{D} = 1.22\frac{\lambda}{d}[/tex]

[tex]\frac{1.43*10^8}{20*9.4608*10^{15} } = 1.22\frac{600*10^{-9}}{d}[/tex]

Re-arrange to find d,

[tex]d = 968.5m = 0.968Km[/tex]

Therefore the minimum diameter of an optical telescope to resolve a Jupiter-sized planet is 0.968Km.

Answer:

A. increase

Explanation:

Stirling cycle having four processes

1.Two processes are constant temperature processes

2.Two processes are constant volume processes

The efficiency of Stirling cycle is same as the efficiency of Carnot cycle.

The efficiency of Stirling cycle given as

[tex]\eta=1-\dfrac{T_L}{T_H}[/tex]

[tex]T_L[/tex]=Lower temperature

[tex]T_H[/tex]=Upper temperature

If we increase then upper temperature while the lower temperature is constant then the efficiency of Stirling cycle will increase because the ratio of lower and upper temperature will decreases.

Therefore answer is

A. increase

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, [tex]V_{Hall} = 20\ mV = 0.02\ V[/tex]

Magnetic Field, B = 0.10 T

Hall emf, [tex]V'_{Hall} = 69\ mV = 0.069\ V[/tex]

Now,

Drift velocity, [tex]v_{d} = \frac{V_{Hall}}{B}[/tex]

[tex]v_{d} = \frac{0.02}{0.10} = 0.2\ m/s[/tex]

Now, the expression for the electric field is given by:

[tex]E_{Hall} = Bv_{d}sin\theta[/tex]                            (1)

And

[tex]E_{Hall} = V_{Hall}d[/tex]

Thus eqn (1) becomes

[tex]V_{Hall}d = dBv_{d}sin\theta[/tex]

where

d = distance

[tex]B = \frac{V_{Hall}}{v_{d}sin\theta}[/tex]                      (2)

(a) When [tex]\theta = 90^{\circ}[/tex]

[tex]B = \frac{0.069}{0.2\times sin90} = 0.345\ T[/tex]

(b) When [tex]\theta = 60^{\circ}[/tex]

[tex]B = \frac{0.069}{0.2\times sin60} = 0.398\ T[/tex]

Answer:A

Explanation:

Given

Two ponies are at Extreme end of turntable with mass m

suppose turntable is moving with angular velocity [tex]\omega [/tex]

Moment of inertia of Turntable and two ponies

[tex]I=I_o+2mr_0^2[/tex]

let say at any time t they are at a distance of r from center such [tex]r<r_0[/tex]

Moment of inertia at that instant

[tex]I'=I_0+2mr'^2[/tex]

[tex]I'<I_0[/tex]

conserving angular momentum as net Torque is zero

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

[tex]\omega '=\frac{I}{I'}\times \omega [/tex]

[tex]\omega '>\omega [/tex]

Thus we can say that angular velocity increases as they move towards each other.        

The angular speed of the turntable increases as the ponies walk towards each other due to conservation of angular momentum. The angular momentum remains constant as there are no external forces acting upon the system.

As the two ponies of equal mass walk toward each other across the turntable, the angular speed of the system increases. This is because the moment of inertia of the system decreases as the ponies move closer to the center, and according to the principle of conservation of angular momentum, a decrease in the moment of inertia must be accompanied by an increase in angular speed to keep the system's angular momentum constant.

In this scenario, yes, the angular momentum of the system is conserved. There are no external forces acting on the system, thus the sum of the angular momenta of the ponies and the turntable stays the same.

This principle can be exemplified by the behavior of figure skaters as they increase their rotation speed by bringing their arms in closer to the body, decreasing their moment of inertia and consequently increasing their angular velocity, to conserve their angular momentum.

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Answer:

d= 14 sin 31.41 t

Explanation:

Lets take the general equation of the motion

d= D sinωt

d=Displacement

ω=Natural angular frequency

t=Time

D=Amplitude

Given that D= 14 cm

We know that time period T

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

Given that T= 0.2 s

[tex]0.2=\dfrac{2\pi}{\omega}[/tex]

ω=31.41 rad/s

D= 14 cm

Therefore

d= D sinωt

d= 14 sin 31.41 t

Final answer:

The equation modeling the displacement d of an object in simple harmonic motion with an amplitude of 14 cm and a period of 0.2 seconds, starting at 0 cm and moving in a positive direction at t = 0, is d(t) = 14cos(10πt + π/2) cm.

Explanation:

The displacement d of an object in simple harmonic motion can be modeled by the function d(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase constant. Since the amplitude A is 14 cm and the period T is 0.2 seconds, and given that the initial displacement at time t = 0 is 0 cm and it initially moves in a positive direction, we can determine ω using the formula ω = 2π/T and φ using the fact that the cosine function must start at 0.

Therefore, the angular frequency ω = 2π/0.2 s = 10π rad/s, and since the function starts at 0 and moves in a positive direction, φ = +π/2. Hence, the equation that models the displacement d as a function of time t is d(t) = 14cos(10πt + π/2) cm.

To solve this problem it is necessary to use the concepts of Force of a spring through Hooke's law, therefore,

[tex]F = kx[/tex]

Where,

k = Spring constant

x = Displacement

Initially our values are given,

[tex]F = 685N[/tex]

[tex]x = 0.88 cm[/tex]

PART A ) With this values we can calculate the spring constant rearranging the previous equation,

[tex]k = \frac{F}{x}[/tex]

[tex]k = \frac{685}{0.88*10^-2}[/tex]

[tex]k = 77840.9N/m[/tex]

PART B) Since the constant is unique to the spring, we can now calculate the force through the new elongation (0.38cm), that is

[tex]F = kx[/tex]

[tex]F = (77840.9)(0.0038)[/tex]

[tex]F = 295.79N[/tex]

Therefore the weight of another person is 265.79N.

Final answer:

To determine the spring constant (k) and the weight of another person, Hooke's law is applied. The spring constant is calculated using the weight of the first person and the distance the spring was compressed. Then, with the found spring constant, the weight for another compression distance is determined.

Explanation:

To solve both parts of the question, we make use of Hooke's law, which states that the force (F) needed to compress or extend a spring by some distance (x) is directly proportional to that distance. The law is usually formulated as F = kx, where k is the spring constant. To find the spring constant, we rearrange the formula to k = F/x.

Part A:We are given that a person weighing 685 N compresses the spring by 0.88 cm (which is 0.0088 m). Using the formula k = F/x, we plug in the values to get k = 685 N / 0.0088 m, which gives us the spring constant.

Part B: With the spring constant from part A, we can then determine the weight of another person by using the same Hooke's law. If the spring is compressed by 0.38 cm (which is 0.0038 m), we use the formula F = kx to calculate the new weight (force).

(A) A device that converts heat into work with 100% efficiency

It clearly violates the second law of thermodynamics because it warns that while all work can be turned into heat, not all heat can be turned into work. Therefore, despite the innumerable efforts, the efficiencies of the bodies have only been able to reach 60% at present.

Answer:

a) 4 times larger. b) 16 times larger. c) 16 times smaller. d) Coulomb´s Law

Explanation:

Between any pair of charged particle, there exists a force, acting on the line that join the charges (assuming they can assimilated to point charges) directed from one to the other, which is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them.

F= k q1q2 / (r12)2

a) If the distance is reduced to the half of the original distance of separation, and we introduce this value in the force equation, we get:

F(r/2) = k q1q2 / (r12/2)2 = k q1q2 /((r12)2/4) = 4 F(r)

b) By the same token, if r= r/4, we will have F(r/4) = 16 F(r)

c) If the distance increases 4 times, as the force is inversely proportional to the square of the distance, the force will be the original divided by 16, i.e., 16 times smaller.

The empirical law that allows to find out easily these values, is the Coulomb´s Law.

Explanation:

From the definition of equilibrium, at the moment in which both children are sitting facing each other at a certain distance the torque within the seesaw will be zero.

However, if one of the children approaches the pivot, the center of mass will move towards the end of the other child, which will immediately cause the child to rise.

Another way of observing this problem is considering the distance between the two, the distance is proportional to the Torque,

[tex]\tau_1 = \tau_2[/tex]

[tex]F*d_1 = F*d_2[/tex]

Therefore by decreasing the distance - when walking towards the pivot - the torque of the child sitting at the other end will be greater because it keeps its distance.

Being said higher torque will cause the approaching child to rise.

The correct answer is B.

The child's side will rise because of shifting of weight of the child from its place.

Two children are balanced on opposite sides of a seesaw. If one child leans inward toward the pivot point, her side will rise because of shifting of weight of the child from its place. Due to movement from its place, the weight of other child increases which leads to lowering of its side.

So we can conclude that the child's side will rise because of shifting of weight of the child from its place.

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To produce a magnetic field of 90 pT at the center of an 8.0-cm-diameter loop, a current of approximately 0.045 Amps must circulate around the loop.

To calculate the current needed to produce a magnetic field of 90 pT at the center of an 8.0-cm-diameter loop, we can use the formula for the magnetic field strength at the center of a circular loop. The formula is given by B = (µ₀I)/(2R), where B is the magnetic field strength, µ₀ is the permeability of free space, I is the current, and R is the radius of the loop.

First, let's convert 90 pT to Tesla by dividing by 10^12. This gives us a value of 9 x 10^-11 T. We can now substitute this value into the formula along with the given radius of 4.0 cm (half of the diameter) to solve for the current.

B = (µ₀I)/(2R)

9 x 10^-11 T = (4π x 10^-7 T m/A)(I)/(2(0.04 m))

Simplifying the equation, we find:

I = (2π(0.04 m)(9 x 10^-11 T))/(4π x 10^-7 T m/A) = 0.045 A

Therefore, a current of approximately 0.045 Amps must circulate around the 8.0-cm-diameter loop to produce a magnetic field of 90 pT at its center.

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Answer

given,

mass of cockroach = 0.117 Kg

radius = 17.3 cm

rotational inertia = 5.20 x 10⁻³ Kg.m²

speed of cockroach = 1.91 m/s

angular velocity of Susan (ω₀)= 2.87 rad/s

final speed of cockroach = 0 m/s

Initial angular velocity of Susan

L_s = I ω₀

L_s = 5.20 x 10⁻³ x 2.87

L_s=0.015 kg.m²/s

initial angular momentum of the cockroach

L_c = - mvr

L_c = - 0.117 x 1.91 x 0.173

L_c = - 0.0387 kg.m²/s

total angular momentum of Both

L = 0.015 - 0.0387

L = - 0.0237 kg.m²/s

after cockroach stop inertia becomes

I_f = I + mr^2

I_f = 5.20 x 10⁻³+ 0.117 x 0.173^2

I_f = 8.7  x 10⁻³ kg.m²/s

final angular momentum of the disk

L_f = I_f ω_f

L_f = 8.7  x 10⁻³ x ω_f

using conservation of momentum

L_i = L_f

-0.0237 =8.7  x 10⁻³ x ω_f

[tex]\omega_f = \dfrac{-0.0237}{8.7 \times 10^{-3}}[/tex]

[tex]\omega_f = -2.72\ rad/s[/tex]

angular speed of Susan is [tex]\omega_f = -2.72\ rad/s[/tex]

The value is negative because it is in the opposite direction of cockroach.

[tex]|\omega_f |= 2.72\ rad/s[/tex]

b) the mechanical energy is not conserved because cockroach stopped in between.  

Answer:

Explanation:

Potential energy on the surface of the earth

= - GMm/ R

Potential at height h

=  - GMm/ (R+h)

Potential difference

= GMm/ R -  GMm/ (R+h)

= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed  to launch an object from the surface of Earth to a height h above the surface.

Extra  energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / ( R + h)

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel [tex]I=25kgm^2[/tex]

Initial angular velocity of the wheel [tex]\omega _0=10rad/sec[/tex]

Angular acceleration [tex]\alpha =15rad/sec^2[/tex]

(a) We know that [tex]\omega =\omega _0+\alpha t[/tex]

We have given t = 2 sec

So [tex]\omega =10+15\times  2=40rad/sec[/tex]

Now [tex]\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad[/tex]

(b) After 3 sec [tex]\omega =10+15\times 3=55rad/sec[/tex]

We know that kinetic energy is given by [tex]Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J[/tex]

Answer:

ρ = 1469  kg/m³

Explanation:

given,

mass of statue = 0.4 Kg

density of statue = 8 x 10³ kg/m³

tension in the string = 3.2 N

density of the fluid = ?

Volume of the statue

[tex]V = \dfrac{0.4}{8\times 10^3}[/tex]

V = 5 x 10⁻⁵ m³

W = ρ g V

W = ρ x 9.8 x 5 x 10⁻⁵

now, tension on the string will be equal to

T = mg - W

3.2 = 0.4 x 9.8 - ρ x 9.8 x 5 x 10⁻⁵

ρ x 9.8 x 5 x 10⁻⁵ = 0.72

ρ = 1469  kg/m³

Answer:

The recoil velocity of the raft is 1.205 m/s.

Explanation:

given that,

Mass of the swimmer, [tex]m_1=55\ kg[/tex]

Mass of the raft, [tex]m_2=210\ kg[/tex]

Velocity of the swimmer, v = +4.6 m/s

It is mentioned that the swimmer then runs off the raft, the total linear momentum of the  swimmer/raft system is conserved. Let V is the recoil velocity of the raft.

[tex]m_1v+m_2V=0[/tex]

[tex]55\times 4.6+210V=0[/tex]

V = -1.205 m/s

So, the recoil velocity of the raft is 1.205 m/s. Hence, this is the required solution.

Answer:

The recoil velocity of the raft would be [tex]v_{r}\approx 1.2\frac{m}{s}[/tex] (pointing to the left if the swimmer runs to the right)

Explanation:

The problem states that the swimmer has a mass of m=55 kg, and the raft has a mass of M=210 kg. Then, it says that the swimmer runs off the raft with a (final) velocity of v=4.6 m/s relative to the shore.

To analyze it, we take a system of "two particles", wich means that we will consider the swimmer and the raft as a hole system, aisolated from the rest of the world.

Then, from the shore, we can put our reference system and take the initial moment when the swimmer and the raft are stationary. This means that the initial momentum is equal to zero:

[tex]p_{i}=0[/tex]

Besides, we can use magnitudes instead of vectors because the problem will develope in only one dimension after the initial stationary moment (x direction, positive to the side of the running swimmer, and negative to the side of the recoling raft), this means that we can write the final momentum as

[tex]p_{f}=mv-Mv_{r}=0[/tex]

The final momentum is equal to zero due to conservation of momentum (because there are no external forces in the problem, for the system "swimmer-raft"), so the momentum is constant.

Then, from that previous relation we can clear

[tex]v_{r}=\frac{m}{M}v=\frac{55}{210}*4.6\frac{m}{s}=\frac{253}{210}\frac{m}{s}\approx1.2\frac{m}{s}[/tex]

wich is the recoil velocity of the raft, and it is pointing to the left (we established this when we said that the raft was going to the negative side of the system of reference, and when we put a minus in the raft term inside the momentum equation).

The concepts required to solve this problem are the thermodynamic expressions of expansion and linear expansion.

In mathematical terms the dilation of a body can be expressed as

[tex]\Delta L = L_0 \alpha \Delta T[/tex]

Where,

[tex]L_0 =[/tex] Initial Length

[tex]\alpha =[/tex] Thermal coefficient of linear expansion

[tex]\Delta T =[/tex] Change in Length

Our values are given as

[tex]L_0 = 300m[/tex]

[tex]T_f = 25\°C[/tex]

[tex]T_i = 2\°C[/tex]

[tex]\alpha = 12*10^{-6}C^{-1} (Iron)[/tex]

Replacing at the equation we have,

[tex]\Delta L = L_0 \alpha \Delta T[/tex]

[tex]\Delta L = (300)(12*10^{{-6})(25-2)[/tex]

[tex]\Delta L = 0.0828m[/tex]

Therefore the change in the height is 8.28cm

The Eiffel Tower's height is subject to changes due to thermal expansion. In fact, it's around 8.3 cm taller in July than in January, with the temperature rise contributing to this change.

The height change of the Eiffel Tower due to thermal expansion can be estimated using the linear expansion formula: ΔL = α*L*ΔT. Here, ΔL is the change in length, α is the thermal expansion coefficient, L is the original length, and ΔT is the change in temperature. Wrought iron has an expansion coefficient of approximately 12*10^-6 °C^-1. The height of the Eiffel Tower is approximately 300 m. The difference in average temperatures is approximately 23°C (25°C-2°C).

Let's plug these values into the formula: ΔL = 12*10^-6 °C^-1 * 300m * 23°C, which gives us a change in height of approximately 0.083m, or 8.3 cm. Therefore, the Eiffel Tower is an estimated 8.3 cm taller in July than it is in January, due to the effects of thermal expansion.

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Answer:

The potential energy in the water will be 0.0769 mJ

Explanation:

We know that energy stored in the capacitor due to charge is given by [tex]U=\frac{Q^2}{2C}[/tex]

From the relation we can see that potential energy is inversely proportional to the capacitance

And we also know that capacitance is directly proportional to the dielectric constant

So the new potential energy will be [tex]=\frac{6}{78}=0.0769mJ[/tex]

So the potential energy in the water will be 0.0769 mJ

Final answer:

When a capacitor filled with air and charged is filled with water without discharging its plates, the energy stored increases due to the higher dielectric constant of water. The formula used is U' = K_water * U, leading to the capacitor storing 468 mJ of energy after being filled with water.

Explanation:

The energy stored in a capacitor is given by the formula U = ½ CV^2, where U is the stored energy, C is the capacitance and V is the voltage across the capacitor. When a dielectric is introduced between the plates of a capacitor that is charged but disconnected from any voltage supply, the capacitance of the capacitor increases by a factor equal to the dielectric constant of the material inserted, K. Since the charge Q remains constant, and since C increases while U is directly proportional to C, the energy stored in the capacitor will increase as well.

For this particular question, the initial energy is 6 mJ and the dielectric constant for water is 78. After filling the capacitor with water, the new energy stored U' can be found using the dielectric constant. However, since dielectric constant for air is approximately 1, and capacitance is directly proportional to dielectric constant, we can simply multiply the original energy by the dielectric constant of water to find the energy stored after it is filled.

U' = K_water * U  

  = 78 * 6 mJ  

  = 468 mJ

Therefore, the energy the capacitor continues to store after it is filled with water is 468 mJ.

In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,

[tex]KE = \fract{1}{2}mv^2[/tex]

[tex]PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))[/tex]

Where,

M = Mass of Earth

m = Mass of Object

v = Velocity

r = Radius

G = Gravitational universal constant

Our values are given as,

[tex]m = 910 Kg[/tex]

[tex]r_1 = 1200 + 6371 km = 7571km[/tex]

[tex]r_2 = 6371 km,[/tex]

Replacing we have,

[tex]\frac{1}{2} mv^2 =  -GMm(\frac{1}{r_1}-\frac{1}{r_2})[/tex]

[tex]v^2 =  -2GM(\frac{1}{r_1}-\frac{1}{r_2})[/tex]

[tex]v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})[/tex]

[tex]v = 4456 m/s[/tex]

Therefore the speed of the object when striking the surface of earth is 4456 m/s

The speed of a 910-kg object striking the Earth after falling from 1200 km above the North Pole can be found using conservation of mechanical energy and the formula for gravitational potential energy.

To determine the speed at which a 910-kg object strikes the surface of the Earth when released from rest at an altitude of 1200 km above the North Pole, we can use the conservation of mechanical energy. The total mechanical energy (kinetic plus potential) at the initial and final points must be equal since no external work is done on the system (we ignore atmospheric friction).

Initially, the object has potential energy due to its altitude above Earth and no kinetic energy since it's at rest. When it strikes the Earth, it has kinetic energy and minimal potential energy. Setting the initial and final total energies equal gives:

Ui + Ki = Uf + Kf,

where U is potential energy and K is kinetic energy. The potential energy can be calculated using the formula U = -GMEarthm/r, where m is the object's mass, r is the distance from the object to the center of Earth, and G is the gravitational constant. The kinetic energy is given by K = (1/2)mv2, where v is the velocity of the object.

The initial distance ri to the center of Earth is 6357 km + 1200 km (release altitude), and when the object strikes Earth, rf = 6357 km (polar radius of Earth). We know Ki = 0 and Uf is small enough to be negligible at the Earth's surface.

By plugging in the values, solving the equation for v, and taking the square root, we find the speed of the object when it strikes the ground.

Answer:

Mass in kg will be 60 kg

Explanation:

We have given initial speed of the man u = 100 m/sec

And final speed v = 0 m/sec

Time is given, t = 10 sec

So acceleration [tex]a=\frac{v-u}{t}=\frac{0-100}{10}=-10m/sec^2[/tex]

Force is given , F= 600 N

From Newton's law we know that F = ma

So [tex]600=m\times 10[/tex]

mass = 60 kg

So mass in kg will be 60 kg

Answer:

The power input is 0.102 kW

Solution:

As per the question:

Length of the loop, L = 40 m

Diameter of the loop, d = 1.2 cm

Velocity, v = 2 m/s

Loss coefficient of the threaded bends, [tex]K_{L, bend} = 0.9[/tex]

Loss coefficient of the valve, [tex]K_{L, valve} = 0.2[/tex]

Dynamic viscosity of water, [tex]\mu = 0.467\times 10^{- 3}\ kg/m.s[/tex]

Density of water, [tex]\rho = 983.3\ kg/m^{3}[/tex]

Roughness of the pipe of cast iron, [tex]\epsilon = 0.00026\ m[/tex]

Efficiency of the pump, [tex]\eta = 0.76[/tex]

Now,

We calculate the volume flow rate as:

[tex]\dot{V} = Av[/tex]

where

[tex]\dot{V}[/tex] = Volume rate flow

A = Area

v = velocity

[tex]\dot{V} = \frac{\pi}{4}d^{2}\times 2 = 2.262\times 10^{- 4}\ m^{3}/s[/tex]

For this, Reynold's N. is given by:

[tex]R_{e} = \frac{\rho vd}{\mu}[/tex]

[tex]R_{e} = \frac{983.3\times 2\times 0.012}{0.467\times 10^{- 3}} = 50533.62[/tex]

Since, [tex]R_{e}[/tex] > 4000, the flow is turbulent in nature.

Now,

With the help of the Colebrook eqn, we calculate the friction factor as:

[tex]\frac{1}{\sqrt{f}} = - 2log[\frac{\frac{\epsilon}{d}}{3.7} + \frac{2.51}{R_{e}\sqrt{f}}][/tex]

[tex]\frac{1}{\sqrt{f}} = - 2log[\frac{\frac{0.00026}{0.012}}{3.7} + \frac{2.51}{50533.62\sqrt{f}}][/tex]

f = 0.05075

Now,

To calculate the total head loss:

[tex]H_{loss} = (\frac{fL}{d} + 6K_{L, bend} + 2K_{L, valve})\farc{v^{2}}{2g}[/tex]

[tex]H_{loss} = (\frac{0.05075\times 40}{0.012} + 6\times 0.9 + 2\times 0.2)\farc{2^{2}}{2\times 9.8} = 35.71\ m[/tex]

Now,

The drop in the pressure can be calculated as:

[tex]\Delat P = \rho g H_{loss}[/tex]

[tex]\Delat P = 983.3\times 9.8\times 35.71 = 344.113\ kN/m^{2}[/tex]

Now,

to calculate the input power:

[tex]\dot{W} = \frac{\dot{W_{p}}}{\eta}[/tex]

[tex]\dot{W} = \frac{\dot{V}\Delta P}{0.76}[/tex]

[tex]\dot{W} = \frac{2.262\times 10^{- 4}\times 344.113\times 1000}{0.76} = 0.102\ kW[/tex]

Answer:[tex]18.85 rad/s^2[/tex]

Explanation:

Given

mass of turntable [tex]m=0.32 kg[/tex]

radius [tex]r=0.18 m[/tex]

[tex]N=24 rev/s[/tex]

[tex]\omega =2\pi N=2\pi 24[/tex]

[tex]\omega =48\pi rad/s[/tex]

time interval [tex]t=8 s[/tex]

using [tex]\omega =\omeag _0+\alpha t[/tex]

where [tex]\omega =final\ angular\ velocity[/tex]

[tex]\omega _0=initial\ angular\ velocity[/tex]

[tex]\alpha =angular\ acceleration[/tex]

[tex]t=time[/tex]

[tex]48\pi =0+\alpha \times 8[/tex]

[tex]\alpha =6\pi rad/s^2[/tex]

[tex]\alpha =18.85 rad/s^2[/tex]              

Answer:

d

Explanation:

Nuclear reactions unlike chemical reactions do not involve electrons or bonds.The main use for the reaction is neutrons. In nuclear there is only turning the element into another element and releasing a large amount of  energy. So catalyst are also redundant in nuclear reactions.  

Therefore, points III and 3 are true, and rest are all wrong. therefore option d is correct

In a nuclear reaction, elements are converted from one to another and the reaction rates are typically not affected by catalysts. Characteristics such as electrons in atomic orbitals being involved in breaking and forming bonds, and atoms being rearranged by breaking and forming of chemical bonds, pertain to chemical reactions, not nuclear ones.

The correct characteristics of a nuclear reaction are options II and III. In a nuclear reaction, elements are indeed converted from one to another (II). This happens as the structure of an atom's nucleus changes. Furthermore, nuclear reaction rates are typically not affected by catalysts (III). This contrasts with chemical reactions, where catalysts can significantly speed up reaction rates. As for options I and IV, they are characteristic of chemical reactions, not nuclear ones. In chemical reactions, electrons in atomic orbitals are involved in the breaking and forming of bonds (I), and atoms are rearranged by the breaking and forming of chemical bonds (IV). Therefore, the answer is (d) II and III.

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To determine how long it takes for particles to settle at the bottom of a container, principles of sedimentation in physics are applied. The calculation considers gravitational forces, buoyancy, and drag forces. However, exact time estimation requires additional specific information on fluid characteristics.

The question revolves around the concept of sedimentation, which is a physical process where particulate matter settles down at the bottom of a container due to gravity. To calculate how long it takes for spherical particles with a diameter of 2.5 μm (micrometers) and a mass of 1.9 x 10⁻¹⁴ kg to settle at the bottom of a 15-cm-tall container, we need to understand the principles of particle sedimentation in fluids, a topic studied in physics. This involves the forces acting on the particles, including gravitational forces, buoyancy, and drag forces, along with the properties of the fluid and the particles themselves. However, without a specific formula or further details on the characteristics of the fluid (e.g., viscosity) and assuming we ignore any air resistance or buoyancy effects, it's difficult to provide an exact time for sedimentation.

Therefore, it takes approximately [tex]\(4657.76\)[/tex] seconds for all the particles to settle to the bottom of the container.

To find the time it takes for all the particles to settle to the bottom of the container, we can use Stokes' law, which relates the settling velocity of particles to their size and density.

Stokes' law states: [tex]\( v = \frac{{2gr^2(\rho_p - \rho)}}{{9\eta}} \),[/tex] where:

-  v  is the settling velocity of the particle,

-  g  is the acceleration due to gravity [tex](\( 9.8 \, \text{m/s}^2 \)),[/tex]

-  r  is the radius of the particle [tex](\( 1.25 \times 10^{-6} \, \text{m} \)),[/tex]

- [tex]\( \rho_p \)[/tex]is the density of the particle[tex](\( \frac{m}{V} = \frac{m}{\frac{4}{3}\pi r^3} = \frac{1.9 \times 10^{-14}}{\frac{4}{3}\pi(1.25 \times 10^{-6})^3} \)),[/tex]

- [tex]\( \rho \)[/tex] is the density of the fluid [tex](\( \rho = 1.2 \, \text{kg/m}^3 \)),[/tex] and

- [tex]\( \eta \)[/tex] is the viscosity of the fluid[tex](\( \eta = 1.81 \times 10^{-5} \, \text{Pa s} \)).[/tex]

After substituting the values into Stokes' law and solving for the settling velocity, we found it to be approximately[tex]\(3.22 \times 10^{-5} \, \text{m/s}\).[/tex]

Then, using the formula [tex]\(t = \frac{h}{v}\), where \(h\)[/tex]is the height of the container (0.15 m), we calculated the time it takes for the particles to settle to the bottom:

[tex]\[ t \approx 4657.76 \, \text{s} \][/tex]

Answer:b. In your hand

Explanation: The objects from an outer perspective are moving at 90km/h all of them, even if there was a fly flying around in the train. Then if there is no change of speed of the train the moment the ball leaves the hand its relative speed with respect to the hand is zero (0), then it will land in the hand again. So the only way for it not to happen is either you move the hand or the train changes its speed, therefore the relative speed of the

ball with respect to the hand would be differente from zero and further information would be required, but since it is not stated that neither the train accelerates nor the hand is going to be moved, we can affirm the ball will land in the hand.

a) The time taken for the wave to travel a distance of 8.40 m is 0.24 s

b) The time taken for a point on the string to travel a distance of 8.40 m is 6.46 s

c) The time in part a) does not change; the time in part b) is halved (3.23 s)

Explanation:

a)

In order to solve this part, we have to find the speed of the wave, which is given by the wave equation:

[tex]v=f \lambda[/tex]

where

v is the speed

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the wave in this problem,

[tex]f = 62.0 Hz\\\lambda =0.560 m[/tex]

So its speed is

[tex]v=(62.0)(0.560)=34.7 m/s[/tex]

Now we want to know how long does it take for the wave to travel a distance of

d = 8.40 m

Since the wave has a constant velocity, we can use the equation for uniform motion:

[tex]t=\frac{d}{v}=\frac{8.40}{34.7}=0.24 s[/tex]

b)

In this case, we want to know how long does it take for a point on the string to travel a distance of

d = 8.40 m

A point on the string does not travel along the string, but instead oscillates up and down. The amplitude of the wave corresponds to the maximum displacement of a point on the string from the equilibrium position, and for this wave it is

A = 5.20 mm = 0.0052 m

The period of the wave is the time taken for a point on the string to complete one oscillation. It can be calculated as the reciprocal of the frequency:

[tex]T=\frac{1}{f}=\frac{1}{62.0}=0.016 s[/tex]

During one oscillation (so, in a time corresponding to one period), the distance covered by a point on the string is 4 times the amplitude:

[tex]d=4A=4(0.0052)=0.0208 m[/tex]

Since d is the distance covered by a point on the string in a time T, then we can find the time t needed to cover a distance of

d' = 8.40 m

By setting up the following proportion:

[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0208}=6.46 s[/tex]

c)

As we can see from the equation used in part a), the speed of the wave depends only on its frequency and wavelength, not on the amplitude: therefore, if the amplitude is changed, the speed of the wave is not affected, therefore the time calculated in part a) remains the same.

On the contrary, the answer to part b) is affected by the amplitude. In fact, if the amplitude is doubled:

A' = 2A = 2(0.0052) = 0.0104 m

Then the distance covered during one period is

[tex]d=4A' = 4(0.0104)=0.0416 m[/tex]

The time period does not change, so it is still

T = 0.016 s

Therefore, the time needed to cover a distance of

d' = 8.40 m

this time is

[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0416}=3.23 s[/tex]

So, the time taken has halved.

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The time taken for a wave to travel a distance of 8.40 m is approximately 0.120 seconds. The time taken for a point on the string to travel 8.40 m once the wave train has reached it is also approximately 0.120 seconds. Doubling the amplitude of the wave does not affect the time taken.

To determine how long it takes for the wave to travel a distance of 8.40 m, we can use the formula:

Time = Distance / Speed

The speed of a wave is equal to the product of its frequency and wavelength:

Speed = Frequency x Wavelength

Substituting the given values into the formulas, we find that it takes approximately 0.120 seconds for the wave to travel 8.40 m along the string.

To calculate the time it takes for a point on the string to travel a distance of 8.40 m once the wave train has reached the point and set it into motion, we need to divide the distance by the speed of the wave pulse:

Time = Distance / Speed

Since the speed of the wave pulse is the same as the speed of the wave, the time it takes for a point on the string to travel 8.40 m is also approximately 0.120 seconds.

If the amplitude of the wave is doubled, it does not affect the time it takes for the wave to travel a distance of 8.40 m or for a point on the string to travel 8.40 m. The only thing that changes is the maximum displacement of the particles in the wave, but it does not impact the time taken.

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Answer:

K= 290.28 N/m

Explanation:

Given: mass= 85 kg

the time taken to reach point two more times in 6.8 s.

2×t= 6.8 sec

t= 6.8/2= 3.4 sec

then, the time period for oscillation is

[tex]t= 2\pi\sqrt{\frac{m}{k} }[/tex]

Here K= spring constant

m= mass of jumper

⇒[tex]K= \frac{4\pi^2m}{t^2}[/tex]

now plugging the values we get

[tex]K= \frac{4\pi^2\times85}{3.4^2}[/tex]

K= 290.28 N/m

The net force on the pressure cooker lid depends on the temperature inside the cooker. At 120°C, the net force is found using the ideal gas law and the given parameters. After the cooker cools to 20°C, and pressure is equalized, the net force is zero.

Part A: The net force on the pressure cooker lid at 120°C is dependent on the difference in pressure inside and outside the cooker. Using the ideal gas law, we find pressure is proportional to temperature (p = NkBT/V). The force exerted on the cooker lid is the product of this pressure and the lid's area (F = pA). Given that the pressure outside the pot is pa (atmospheric pressure) at a temperature of 20°C and the pressure inside increased to a temperature of 120°C, the net force on the lid is F120 = A(NkB(120+273)/V - pa), where T is in Kelvins.

Part B: After the pressure equalizes and the cooker has cooled down to 20°C, the pressures inside and outside the cooker are equal, therefore the net force is zero. This is because the difference in pressures is what causes the force (F20 = A(pa - pa) = 0)

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Final answer:

The angular acceleration of the disk is calculated using the torque exerted by the applied force and the moment of inertia of a uniform disk. The calculated angular acceleration is 5.33 rad/s².

Explanation:

The question involves calculating the angular acceleration of a rotating disk when a force is applied tangentially at its rim. To find the angular acceleration, we first need to calculate the torque exerted by the force and then use the moment of inertia of the disk to find the angular acceleration.

Given:

The torque (τ) exerted by the force is the product of the force and the radius at which it is applied (torque = force × radius), so:

τ = F × r = 4.0 N × 0.15 m = 0.60 N·m

The moment of inertia (I) for a uniform disk is given by (1/2)mr². Thus:

I = (1/2) × 5.0 kg × (0.15 m)² = 0.1125 kg·m²

Using the formula for angular acceleration (α = torque / moment of inertia), we get:

α = τ / I = 0.60 N·m / 0.1125 kg·m² = 5.33 rad/s²

Therefore, the angular acceleration of the disk is 5.33 rad/s².