** You pull a rope oriented at a 37° angle above the horizontal. The other end of the rope is attached to the front of the first of two wagons that have the same 30-kg mass. The rope exerts a force of magnitude T1 on the first wagon. The wagons are connected by a second horizontal rope that exerts a force of magnitude T2 on the second wagon. Determine the magnitudes of T1 and T2 if the acceleration of the wagons is 2.0 m⁄s2.

Answer:

Temperature will be 305 K  

Explanation:

We have given The asteroid has a surface area [tex]A=7.70m^2[/tex]

Power absorbed P = 3800 watt

Boltzmann constant [tex]\sigma =5.67\times 10^{-8}Wm/K^4[/tex]

According to Boltzmann rule power radiated is given by

[tex]P=\sigma AT^4[/tex]

[tex]3800=5.67\times 10^{-8}\times 7.70\times T^4[/tex]

[tex]T^4=87.0381\times 10^8[/tex]

[tex]T=305K[/tex]

So temperature will be 305 K  

Answer:

The wavelength is 539 nm.

Explanation:

Given that,

Distance of screen L= 10 m

Width s= 5.4 cm

Grating separation d= 0.1 mm

Suppose the pattern is displayed on a screen a distance L from the grating and the spots are separated by s. we need to find the wavelength.

The angle between the central and first maximum is given as

[tex]\tan\theta=\dfrac{s}{L}[/tex]

[tex]\theta=\tan^{-1}\dfrac{s}{L}[/tex]

[tex]\theta=\tan^{-1}\dfrac{5.4\times10^{-2}}{10}[/tex]

[tex]\theta=0.309[/tex]

We need to calculate the wavelength

The condition for maximum of a diffraction grating is

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

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

Put the value into the formula

[tex]\lambda=\dfrac{0.1\times10^{-3}\times\sin0.309}{1}[/tex]

[tex]\lambda=539\times10^{-9}\ m[/tex]

[tex]\lambda=539\ nm[/tex]

Hence, The wavelength is 539 nm.

The wavelength of a laser can be determined using the diffraction grating equation. Calculate the θ value with θ = tan⁻¹(y/L), then substitute the known values into the equation λ = (d sin θ) / m. Convert the resultant wavelength to nanometers.

Your question relates to the physics of light and diffraction grating. We can find the wavelength of the laser by using the diffraction grating equation. The equation is: d sin θ = mλ

Where:

In your case, we need to calculate θ first: θ = tan⁻¹(y/L) where y is the distance between the spots (5.4cm) and L is the distance from the grating to the screen (10m).

Substitute the known values and solve for λ: λ = (d sin θ) / m

Remember, the resultant wavelength should be converted to nanometers (1m = 1 x 10⁹ nm).

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Answer:

Hydrogen bonding

Ionic bonding

Non- induced dipole forces

Ion dipole forces

Van-der wall forces

Explanation:

Intermolecular forces are the forces which operates at molecular level they may be attractive or repulsive in nature. But the given question ask only about attractive molecular forces. These are of basically  five types:-

1. Hydrogen bonding- It is the force of attraction between hydrogen atom and an electro-negative atom.

2 .Ionic bonding- It is the force of attraction between cationic and anionic pair.

3.Non - Induced dipole forces- It is the dipole- dipole interaction between permanent electromagnetic materials.

4.Ion-dipole forces- These bonds are stronger than hydrogen bond and also stronger than dipole-dipole bond.

Van-der walls forces- It is the force of attraction between uncharged molecules.

In this part, we discuss the expected decrease in angular acceleration when the child is on the merry-go-round due to the greater moment of inertia. We calculate the child's moment of inertia by approximating the child as a point mass at a specific distance from the axis.

b. We expect the angular acceleration for the system to be less in this part because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia I, we first find the child's moment of inertia Ic by approximating the child as a point mass at a distance of 1.25 m from the axis. Then

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Final answer:

The rotational kinetic energy of the merry-go-round plus child at 29.2 rpm is 394.5834 J. The father must push for approximately 0.167 revolutions to reach this velocity from rest. A force of -8.975 N (in the opposite direction of motion) must be exerted to stop the merry-go-round in seven revolutions.

Explanation:

The problem addresses the concepts of angular velocity, rotational kinetic energy, and the application of work-energy principles within a physics context, specifically relating to a merry-go-round system.

(a) To calculate the rotational kinetic energy (KErot) of the merry-go-round plus child, we use the formula KErot = (1/2)Iω^2 where I is the moment of inertia and ω is the angular velocity in radians per second. First, convert the angular velocity from rpm to rad/s using the conversion factor 1 rpm = (2π/60) rad/s.
KErot = (1/2)(84.4 kg · m^2)(29.2 rpm × 2π/60 rad/s)^2
= (1/2)(84.4)(3.058 rad/s)^2
= (1/2)(84.4)(9.351)
= (1/2)(789.1668) J
= 394.5834 J

(b) The number of revolutions needed to achieve this angular velocity from rest can be found using the work-energy principle, where work done (W) by the father's force equals the rotational kinetic energy of the merry-go-round system.
W = τr(2πN)
394.5834 J = 375 N · m × (2πN)
N = 394.5834 J / (375 N · m × 2π)
N = 0.167 revolutions

(c) The force to stop the merry-go-round in seven revolutions can be calculated using the concept that the work done by the father's force (W) to stop the merry-go-round equals the negative change in the rotational kinetic energy. Thus, W = -KErot, and the rotational kinetic energy is initially computed in part (a). This gives:
W = -394.5834 J
Force (F) = W / (2π×7)
F = -394.5834 J / (2π × 7)
F = -8.975 N (negative sign indicates that the force is applied in the opposite direction of motion)

Answer:

[tex]t_2=\sqrt{2}(t_1)[/tex]

Explanation:

The kinetic energy of a body is that energy it possesses due to its motion. In classical mechanics, this energy depends only on its mass and speed, as follows:

[tex]K=\frac{mv^2}{2}[/tex]

The speed in an uniformly accelerated motion is given by:

[tex]v=at[/tex]

Replacing this expression in the formula for the kinetic energy, we have:

[tex]K=\frac{ma^2t^2}{2}\\[/tex]

So, if we have [tex]K_2=2K_1[/tex]:

[tex]K_1=\frac{ma^2t_1^2}{2}(1)\\K_2=\frac{ma^2t_2^2}{2}\\2K_1=\frac{ma^2t_2^2}{2}\\K_1=\frac{ma^2t_2^2}{4}(2)\\[/tex]

Equaling (1) and (2) and solving for [tex]t_2[/tex]:

[tex]\frac{ma^2t_1^2}{2}=\frac{ma^2t_2^2}{4}\\t_2=\frac{4t_1^2}{2}\\t_2=\sqrt{2t_1^2}\\t_2=\sqrt{2}(t_1)[/tex]

The intensity level in dB at a distance of 100 meters from the train is 98.46 dB.

To calculate the intensity level in dB at a distance of 100 meters from the train, we need to first calculate the intensity of sound at that distance.

Since the sound energy spreads out spherically, the intensity decreases with the square of the distance from the source.

Using the formula for intensity, I = P/4πr², where P is the acoustic power output and r is the distance from the source, we can calculate the intensity at 100 meters:

I = 100W / (4π * (100m)²) = 0.0796 W/m².

Now, we can use the formula for the intensity level in dB, β = 10log(I/I₀), where I₀ is the reference intensity (10⁻¹² W/m²).

Substituting the values, β = 10log(0.0796 / 10⁻¹²) = 98.46 dB.

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Answer:4.58 m/s

Explanation:

Given

mass of Particle [tex]m=4 kg[/tex]

[tex]F=-cx^3[/tex]

[tex]a=\frac{F}{m}[/tex]

[tex]a=-\frac{cx^3}{m}[/tex]

[tex]a=-\frac{8x^3}{4}[/tex]

[tex]a=-2x^3[/tex]

[tex]v\frac{\mathrm{d} v}{\mathrm{d} x}=-2x^3[/tex]

[tex]vdv=-2x^3dx[/tex]

integrating

[tex]\int_{6}^{v_b}vdv=\int_{1}^{-2}-2x^3dx[/tex]

[tex]\frac{v_b^2-6^2}{2}=-\frac{1}{2}\left [ \left ( -2\right )^4-\left ( 1\right )^4\right ][/tex]

[tex]\frac{v_b^2-36}{2}=-0.5\times 15[/tex]

[tex]v_b^2=36-15[/tex]

[tex]v_b=\sqrt{21}[/tex]

[tex]v_b=4.58 m/s[/tex]

A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx³ The speed of the particle at point B is 5.3 m/s.

The potential energy at any point x is given by:

F(x) = -cx³

U(x) = -∫F(x) dx

U(x)= ∫cx³dx

U(x) = c/4 × x⁴ + C

The total mechanical energy E remains constant:

E = K + U

E_A = K(A) + U(A)

E_A = (1/2)mv(A)²+ c÷4 × x(A)⁴ + C

At point B, the potential energy U(B) is:

U(B) = c/4 × x(B)⁴ + C

The total mechanical energy at point B is:

E(B) = K(B) + U(B)

Since the total mechanical energy is conserved, we have:

E(A) = E(B)

(1/2)mv(A)² + c÷4 × x(A)⁴ + C = (1/2)mv(B)² + c÷4 × x(B)⁴ + C

(1/2)(4.0)(6.0)² + ((8.0) ÷ 4) × (1.0)⁴ = (1/2)(4.0)v(B)² + ((8.0) ÷ 4) × (-2.0)⁴

72.0 + 2.0= 2.0 × v(B)² + 16.0

56.0 J = 2.0× v(B)²

v(B)² = 28.0

v(B) = ±5.3 m/s

Since speed cannot be negative, the speed of the particle at point B is 5.3 m/s.

To know more about the speed:

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Answer:

[tex]T_H = 1335\ J[/tex]

Explanation:

Given,

Work done by engine = 19900 J

rejected energy of heat =  5300 J

temperature = 285 K

efficiency of the engine

[tex]\eta = \dfrac{W}{Q_H}[/tex]

where

[tex]W = Q_H - Q_C[/tex]

[tex]19900 = Q_H - 5300[/tex]

[tex]Q_H = 25200\ J[/tex]

efficiency in terms of temperature

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

from the above two equation

[tex]\dfrac{Q_H-Q_C}{Q_H}= \dfrac{T_H-T_C}{T_H}[/tex]

[tex]1-\dfrac{Q_C}{Q_H}= 1- \dfrac{T_C}{T_H}[/tex]

[tex]\dfrac{Q_C}{Q_H}= \dfrac{T_C}{T_H}[/tex]

[tex]T_H=T_C\ \dfrac{Q_H}{Q_C}[/tex]

[tex]T_H=285\times \dfrac{25200}{5300}[/tex]

[tex]T_H = 1335\ J[/tex]

The smallest possible temperature of the hot reservoir, given the work done by the engine, heat rejected into the cold reservoir, and temperature of the cold reservoir, is approximately 1357.14 Kelvin.

The problem is based on the Carnot heat engine, a theoretical model used for understanding the thermodynamics of heat conversion into work. The efficiency of Carnot's engine depends on the temperatures of the hot and cold reservoirs (Th and Tc respectively). The efficiency, represented as e, is calculated as e = (Th-Tc) / Th.

Before we do the calculation, it's important to note that temperatures must be expressed in Kelvin for this formula to be accurate. Tc, the temperature of the cold reservoir, is given as 285 K. The work done by the engine, represented as W, is 19900 J, and the heat rejected by the engine into the cold reservoir, represented as Qc, is 5300J. The heat absorbed by the engine from the hot reservoir, represented as Qh, can be calculated using the first law of thermodynamics, which states Qh = W + Qc. Therefore, Qh = 19900 J + 5300 J = 25200 J.

Now, to find Th, we rearrange the efficiency formula to Th = Tc / (1 - e), where e is now W / Qh = 19900 J / 25200 J ≈ 0.79. Substituting Tc = 285 K and e ≈ 0.79 into the formula, Th = 285 K / (1 - 0.79) ≈ 1357.14 K. This is the smallest possible temperature of the hot reservoir for the given conditions.

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Answer:

11m

Explanation:

Given:

Resistivity ρ = 5.6e-8 Ωm

Radius r = 0.045 mm [tex]=\frac{0.045}{1000}[/tex] = 4.5 x 10⁻⁵ m

Voltage V = 120V

Current I = 1.24A

From Ohm's law, [tex]R = \frac{V}{I}[/tex]

                            [tex]R = \frac{120}{1.24}[/tex]

                            R = 96.77 Ω

Resistivity = (Resistance × Area)/ length

      ρ = (RA)/L

Therefore, the length of a wire is given by;

       L = (RA)/ρ

Calculating the area A of the wire;

       A = πr²

       A = π × (4.5 x 10⁻⁵)²

       A = 6.36 x 10⁻⁹ m²

Substituting area of the wire A =  6.36 x 10⁻⁹ m² into the equation of the length of wire

       L = (96.77 × 6.36×10⁻⁹ ) / 5.6×10⁻⁸

       L = 10.9977m

       L = 11m (approximately)

Answer:

Acceleration, [tex]a=1.2\ m/s^2[/tex]

Explanation:

Given that,

The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope, m = 100 kg

Force exerted by the doges on the rope attached to the front sled, F = 240 N

To find,

The acceleration of the sleds.

Solution,

Let a is the acceleration of the sleds. The product of mass and acceleration is called force. Its expression is given by :

F = ma

[tex]a=\dfrac{F}{m}[/tex]

[tex]a=\dfrac{240\ N}{2\times 100\ kg}[/tex] (m = 2m)

[tex]a=1.2\ m/s^2[/tex]

So, the acceleration of the sleds is [tex]1.2\ m/s^2[/tex].

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = [tex]854\times 10^{3}\ Hz[/tex]

Power, P = 50 kW = [tex]50\times 10^{3}\ W[/tex]

[tex]I_{p} = 1\ photon/s/m^{2}[/tex]

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

[tex]n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s[/tex]

Area of the sphere, A = [tex]4\pi r^{2}[/tex]

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is [tex]1\ photon/s/m^{2}[/tex]

[tex]I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}[/tex]

[tex]1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}[/tex]

[tex]r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m[/tex]

Now,

We know that:

1 ly = [tex]9.4607\times 10^{15}\ m[/tex]

Thus

[tex]r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly[/tex]

Answer:

[tex]0.836cm[/tex] in the positive direction.

Explanation:

The equation that describes the motion of this mass-spring system is given by;

[tex]y=Asin\omega t...................(1)[/tex]

Where A is the amplitude, which defined as the maximum displacement from the equilibrium position for a body in simple harmonic motion.

[tex]\omega[/tex] is the angular velocity measured in [tex]rads^{-1[/tex], this is the angle turned through per unit time.

[tex]y[/tex] is the displacement along the axis of the amplitude, and [tex]t[/tex] is any instant of time in the motion.

Given; A = 8.8cm = 0.088m

The angular velocity is given by the following relationship also;

[tex]\omega=2\pi/T[/tex]

Where T is the period, which is defined as the time taken for a body in simple harmonic motion to make one complete oscillation.

Given; T=0.66s

Therefore;

[tex]\omega=2\pi/0.66\\\omega=3.03\pi rads^{-1[/tex]

Substituting all values into equation (1) we obtain the following;

[tex]y=0.088sin3.03\pi t................(2)[/tex]

Equation (2) is the equation describing the motion of the mass on the spring.

At an instant of time t = 2.3s, the displacement is therefore given as follows;

[tex]y=0.088sin[3.03\pi(2.3)]\\y=0.088sin6.97\pi\\[/tex]

By conversion, [tex]6.97\pi rad=6.97*180^o=1254.55^o[/tex]

Therefore

[tex]y=0.088sin1254.54=0.00836m\\[/tex]

[tex]y=0.00836m=0.836cm[/tex]

Answer:

20 metric tons

Explanation:

Using the Principle of Conservation of Linear Momentum which states that the total momentum of a system is constant in any direction in which no external force acts.

momentum before collision= momentum after collision

mathematically expressed as

m1u1 + m2u2 = m1v1+m2v2 =0

where:

m1= mass of rolling freight car = 10 metric tons = 10Mg

1 ton = 1000kg= 1Mg

u1 =initial velocity of freight car = 108km/h,

m2 = mass of stationary freight car = ?, this the unknown we are finding

u2 = initial velocity of

m2 = 0, because it is at rest,

v1= final velocity of m1,

v2= final velocity of m2

since they move together in the same direction they share a common final velocity as such

v1=v2= 36km/h

substituting valves in expression give

m1u1 + m2u2 = m1v1 + m2v2= 0

10Mgx 108km/hr + m2 x 0 = 10Mg x 36km/h + m2x36km/h

1080( Mg x km/hr) + 0 = 360 (Mg x Km/h) + 36m2 (km/h)

1080 (Mgx km/hr) - 360(Mg x km/h) = 36m2 (km/h)

720(Mg x km/h) = 36m2 (km/h)

divide both side by  36 (km/h) gives

m2 = 20Mg =20 metric tons

Answer: B). 20 metric tons

Explanation: If a freight car with a mass of 10 metric tons is going at 108 km/h collides with another freight car, that is parked. If the speed of the cars, after they crash together, is 36 km/h, the mass of the second car is 20 metric tons.

Answer:

111.6 g

Explanation:

Given

m + M = 2.58 / 9.8

= 0.2632 kg

When the can of fruit juice is balance from scale , we get following relation

M x ( 50 - 21.2 ) = m x 21.2 ( balancing the torque due to weight of scale and can about the balancing point )

M x 28.8 = 21.2 m

= 21.2 ( 0.2632 - M )

= 5.58 - 21.2 M

M ( 28.8 + 21.2 ) = 5.58

M = .1116 kg

= 111.6 g

Answer:

[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]

Explanation:

Given:

We have:

Assuming that heat loss is neither from any of the components before mixing nor from the mixture just after mixing. Also the container does not absorbs any heat.

Therefore,

heat gained by the water = heat lost by the quartz

[tex]Q_w=Q_q[/tex]

[tex]m_w.c_w. (T_f-T_{wi})=m_q.c_q.(T_{qi}-T_f)[/tex]

    where: [tex]c_q=[/tex] specific heat capacity of quartz

[tex]250\times 4.186\times (27.9-25)=58.1\times c_q\times (98.4-27.9)[/tex]

[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]

To calculate the specific heat capacity of quartz, use the equation q = mcΔT and set it equal to the equation for the heat transferred to the water. Solve for c to find the specific heat capacity of quartz.

The specific heat capacity of quartz can be calculated using the equation:

q = mcΔT

where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The heat transferred to the water can be calculated using the equation:

q = mcΔT

where q is the heat transferred, m is the mass, c is the specific heat capacity of water (4.184 J/g °C), and ΔT is the change in temperature.

By setting the two equations equal to each other and solving for c, the specific heat capacity of quartz can be determined.

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Answer:0.906 N/m

Explanation:

Given

time period [tex]T=0.66 s[/tex]

mass [tex]m=0.01 kg[/tex]

System can be considered as spring mass system

Time Period of spring mass system is given by

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

squaring

[tex]k=\frac{4\pi ^2m}{T^2}[/tex]

[tex]k=\frac{4\times \pi ^2\times 0.01}{0.66^2}[/tex]

[tex]k=0.906 N/m[/tex]

Answer:

E)  E₂= 20000 N/C

Explanation:

The electric field between two parallel conductive plates is thus calculated

E= V/d

Where:

E: Electric field (N/C)

V:  voltage (V)

d : distance between the plates (m)

Problem development

E₁= V₁/d₁  = 2000 N/C

E₂= V₂/d₂

[tex]E_{2} = \frac{2V_{1} }{\frac{d_{1} }{5} }[/tex]

[tex]E_{2} = 10(\frac{V_{1} }{d_{1} })[/tex]

E₂= 10* (2000)

E₂= 20000 N/C

Final answer:

When the voltage between two parallel conducting plates is doubled and the distance between them is reduced to 1/5 of the original distance, the magnitude of the new electric field is 20,000 N/C.

Explanation:

The question relates to the electric field between two parallel conducting plates connected to a voltage source. The original electric field's magnitude is given as 2,000 N/C, and we're asked to find the new electric field magnitude when the voltage is doubled and the distance between the plates is reduced to 1/5 of the original distance. The electric field (E) between two parallel plates is given by the equation E = V/d, where V is the potential difference (voltage) between the plates, and d is the distance between them. When the voltage is doubled, V becomes 2V, and when the distance is reduced to 1/5, d becomes d/5. Thus, the new electric field E' = (2V)/(d/5) = 10(V/d) = 10E. So, the new electric field strength is 10 * 2,000 N/C = 20,000 N/C, which corresponds to option E).

a typical asteroid in the asteroid belt, a typical oort cloud object, a typical kuiper belt object

Although Oort cloud comets are now located far beyond the Kuiper belt, they are thought to have formed in the region of the jovian planets. They were then "kicked out" to their current orbits by gravitational encounters with the jovian planets.

The objects in the solar system can be ranked from nearest to farthest based on their formation distance from the Sun: Mercury, Venus, Earth, Mars, and Jupiter.

The objects in the solar system can be ranked based on their distance from the Sun as follows:

Therefore, the ranking from nearest to farthest would be Mercury, Venus, Earth, Mars, and Jupiter.

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Answer:

option (D)

Explanation:

Here initial rotation speed is given, final rotation speed is given and asking for time.

If we use

A) θ=θ0+ω0t+(1/2)αt2

For this equation, we don't have any information about the value of angular displacement and angular acceleration, so it is not useful.

B) ω=ω0+αt

For this equation, we don't have any information about angular acceleration, so it is not useful.

C) ω2=ω02+2α(θ−θ0)

In this equation, time is not included, so it is not useful.

D) So, more information is needed.

Thus, option (D) is true.

The equation to directly solve for time t in this context is ω = ω₀ + αt, as it includes all necessary variables except for t, making it easy to solve algebraically.  Hence, option(b) is correct.

The correct equation to use in this problem is ω = ω₀ + αt.

By rearranging this equation, you can directly find the value of t using given angular velocities and acceleration. This is applicable in analyzing rotational motion in physics.

To solve for the time t, we can rearrange the equation algebraically:

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

This equation directly relates

Given the initial and final angular velocities and the angular acceleration, you can plug these values in to find t.

Answer:

Explanation:

Time period of oscillation of spring mass system

T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]

m = [tex]\frac{T^2\times k}{4\pi^2}[/tex]

= [tex]\frac{(2.6)^2\times 13.2}{4\pi^2}[/tex]

= 2.26 kg

Amplitude is the maximum displacement of the mass from the _equilibrium point .

maximum displacement from equilibrium of +14.8 cm at time, t = 0.00 s

x(t) = 14.8cosωt

ω = 2πn = 2π x 9.73 = 61.1 rad /s

x(t) = 14.8cos61.1 t

when t = 1.25

x(t) = 14.8cos61.1 x 1.25

= 14.8cos61.1 x 1.25

= 14.8cos76.38

= 8.22 cm

A ball of mass 1.55 kg is hung from a spring which stretches a distance of 0.3800 m

spring constant

k = mg / x

= 1.55 x 9.8 / .38

= 40 N / m

T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]

so  if mass is hung from a spring with a spring constant of 3k

T decreases by a factor of the square root of 3

For the first question, the unknown mass is approximately 0.71 kg. For the second question, the term 'equilibrium point' completes the statement. For the third question, the displacement at time 1.25 seconds is -14.2 cm. For the fourth question, the spring constant k is approximately 40.0 N/m. For the fifth question, the period of the motion will be decreased by a factor of the square root of 3.

The unknown mass hung from a spring can be calculated using the formula for the period of a mass-spring system, T = 2π√(m / k), where T is the period, m is the mass, and k is the spring constant. Rearranging for m gives us m = (T²k) / (4π²). Given that T = 2.60 seconds and k = 13.2 N/m, we find m = (2.60^2 * 13.2) / (4*π^2), which gives a mass of approximately 0.71 kg.

The amplitude is the maximum displacement of the mass from the equilibrium point.

The displacement of a harmonically oscillating mass can be found using the equation:    x(t) = A*cos(2πft + φ), where A is the amplitude, f is the frequency, and φ is the phase angle. Given the information we have a displacement at 1.25 seconds of -14.2 cm.

The spring constant can be calculated using the equation  k = mg / x, where m is the mass, g is gravity (9.81 m/s²), and x is the distance stretched. Given that the mass is 1.55 kg and the distance stretched is 0.3800 m, we find that the spring constant k is approximately 40.0 N/m.

If the same mass is hung from a spring with a spring constant of 3k, the period of the motion will be decreased by a factor of the square root of 3.

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Answer:

7.8 mph

Explanation:

Rate of cycling = 1.1 rev/s

Rear wheel diameter = 26 inches

Diameter of sprocket on pedal = 6 inches

Diameter of sprocket on rear wheel = 4 inches

Circumference of rear wheel =  [tex]\pi d=26\pi[/tex]

Speed would be

[tex]\text{Rate of cycling}\times \frac{\text{Diameter of sprocket on pedal}}{\text{Diameter of sprocket on rear wheel}}\times{\text{Circumference of rear wheel}}\\ =1.1\times \frac{6}{4}\times 26\pi\\ =134.77432\ inches/s[/tex]

Converting to mph

[tex]1\ inch/s=\frac{1}{63360}\times 3600\ mph[/tex]

[tex]134.77432\ inches/s=134.77432\times \frac{1}{63360}\times 3600\ mph=7.65763\ mph[/tex]

The Speed of the bicycle is 7.8 mph

To determine the bicycle speed, calculate the rear wheel's circumference, use the gear ratio to find the rear wheel's rotation rate, and then convert the distance traveled per second into miles per hour. The resulting speed is 17.3 mph when pedaling at 1.1 revolutions per second.

The student's question involves calculating the speed of a bicycle based on the diameters of the sprockets on the pedal and rear wheel, the diameter of the rear wheel, and the pedaling rate in revolutions per second. To find the speed in miles per hour, we first calculate the circumference of the rear wheel, which gives us the distance traveled per revolution. We then use the gear ratio to find the effective revolutions of the rear wheel based on the pedaling speed. The total distance traveled per second is then converted to miles per hour.

The diameter of the rear wheel is 26 inches, so its circumference is π(26 inches) ≈ 81.68 inches. The gear ratio is the diameter of the pedal sprocket (6 inches) over the diameter of the rear wheel sprocket (4 inches), giving us a 1.5:1 gear ratio. For every pedal revolution, the rear wheel rotates 1.5 times. Pedaling at 1.1 revolutions per second, the rear wheel rotates at 1.5 * 1.1 = 1.65 revolutions per second. The distance traveled per second is then 81.68 inches * 1.65 ≈ 134.77 inches per second.

To convert inches per second to miles per hour, we use the conversion factors 12 inches per foot and 5280 feet per mile, and 3600 seconds per hour. Thus, the bike speed is (134.77 inches/second) * (1 foot/12 inches) * (1 mile/5280 feet) * (3600 seconds/hour) ≈ 17.26 miles per hour. Rounded to the nearest tenth, the bicycle is traveling at 17.3 mph.

Answer:

Both A and B are correct

Explanation:

Thus both the technicians are correct.

Answer:

Technician A is correct

Magnetic lines of force can be seen by placing iron filings on a piece of paper and then holding them over a magnet.

Explanation:

As we know that the magnetic field lines are always originated out from the North pole of magnet and then terminate at south pole of the magnet

Here we know that when magnetic material aligns itself in the direction of external magnetic field

So here when we put the iron filings on the cardboard and then place a magnet near it then all the iron filing will align itself in the direction of the magnetic field

Now if we place a compass near the magnet then it will aligns itself in the tangential direction of the magnetic field at the same position

So here we can say that Technician A is correct

Answer

A molecule with seesaw molecular geometry has a bond angle of 90° and 120°.

Seesaw is a type of molecular geometry where there are four bonds attached to a central atom.

Bond is observed in the shape of playground seesaw that is why it is known as a seesaw bond.

Four bonds attached to the central bond formed results as tetrahedral or square planar geometry.

In see-saw, shape maximizes the bond angle of the lone pair and the other atoms in the molecule.

lone pair in the equatorial position offer 120 and 90-degree bond angles whereas bond angle will be 90-degree bond angles if placed at axial position.

A molecule with seesaw molecular geometry has bond angles that are less than 120 degrees for equatorial positions and less than 90 degrees for axial positions, due to lone pair repulsion. Hence the correct option is 1.

Bond Angles in Seesaw Molecular Geometry

A molecule with a seesaw molecular geometry has four nuclei and one lone pair of electrons. This molecular structure is based on a trigonal bipyramidal geometry but with one equatorial position occupied by a lone pair. As a result, the bond angles are adjusted due to electron pair repulsion.

The bond angles for a seesaw geometry are:

The correct question is:

A molecule with a seesaw molecular geometry has a bond angle of

1. <120 for equatorial bonds and <90 for axial bonds.

2.180

3. <90

4. 120 for equatorial bonds and 90 for axial bonds.

5.120

The principle of conservation of momentum states that in a closed system, an object's momentum before a collision will equal its momentum after the collision.

The principle of conservation of momentum states that in a closed system, the total momentum of the objects before a collision will be equal to the total momentum of the objects after the collision. This means that in a closed system, an object's momentum before a collision will equal its momentum after the collision.

Final answer:

The rod's kinetic energy at its lowest position is 1.26 J, calculated using the formula for rotational kinetic energy. The center of mass of the rod rises to a height of approximately 30.6 cm above the lowest point, using the principle of conservation of mechanical energy to equate potential and kinetic energies.

Explanation:

A thin rod acting as a pendulum is an example of physical principles related to rotational motion and conservation of energy. To find the rod's kinetic energy at its lowest position, we use the formula for rotational kinetic energy, K = (1/2)Iω2, where 'I' is the moment of inertia and 'ω' is the angular speed. For a rod rotating about one end, the moment of inertia, 'I', is given by (1/3)ml2, with 'm' being the mass of the rod and 'l' its length.

We can find the kinetic energy of the rod at its lowest position, where the angular speed is 4.0 rad/s: K = (1/2)Iω2 = (1/2)(1/3)(0.42 kg)(0.75 m)2(4.0 rad/s)2. After calculation, this gives K = 1.26 J (Joules).

To determine how far above the lowest position the center of mass rises, we have to consider the conservation of mechanical energy. At the highest point, all kinetic energy is converted into potential energy (assuming no energy losses), so we have mgh = K, where 'g' is the acceleration due to gravity, 'h' is the height gained, and 'm' is the mass of the rod. Therefore, h = K / (mg). Plugging in the values, we get h = 1.26 J / (0.42kg * 9.81 m/s2) = 0.306 m, or approximately 30.6 cm above the lowest position.

Bio-gas is the naturally produced fossil fuel, a by-product when bacteria decompose organic material under anaerobic conditions.

Organic matter particularly waste material is broken down by bacteria through fermentation in an environmental condition without any presence of oxygen. This process of decomposition leads to formation of bio-gas with "carbon dioxide and methane" in a 2:3 ratio.

The above biological process is termed as bio-digestion or anaerobic digestion. Methane is flammable and thus bio-gas can be used as "energy source", a waste-to-energy transformation. The remaining decomposed matter is ideal as manure for plants due to its rich nutrient level.

Answer:

242 lts de fuel

Explanation:

We know density of fuel (is variable) but we can say that is close to 1 Kg/lts.

If we have 110 pounds over maximun certificated gross weight, we need to get rid of that 110 pounds  or 110 * 2.2  = 242 Kgs

Now    by rule of three

If     1 lt        weight      1   kg.

        x                          242 kgs

x  = 242 lts de fuel

To bring the aircraft within the maximum gross weight limits, approximately 18.33 gallons of fuel need to be drained, calculated by dividing the excess weight (110 pounds) by the weight of fuel per gallon (6 pounds/gallon).

To calculate how much fuel needs to be drained to bring an aircraft weight within the maximum certificated gross weight limits, we need to know two pieces of information: the weight by which the aircraft is over its maximum gross weight and the weight of the fuel that needs to be drained.

In this scenario, the aircraft is 110 pounds over the maximum certificated gross weight. In general, aviation gasoline (avgas) weighs about 6 pounds per gallon. To solve this problem, we will divide the excess weight by the weight of the fuel per gallon to find out how many gallons need to be drained.

So, the calculation is as follows:

Excess weight: 110 pounds.

Weight of fuel per gallon: 6 pounds/gallon.

Number of gallons to be drained = Excess weight / Weight of fuel per gallon.

Number of gallons to be drained = 110 pounds / 6 pounds/gallon.

Number of gallons to be drained = 18.33 gallons (approximately).

Therefore, to bring the aircraft back within weight limits, approximately 18.33 gallons of fuel need to be drained.

Answer:

high real rate of interest,  depressing

Explanation:

If output is less than potential and the rate of policy has fall to the ground of zero, the subsequent decline in inflation results in high real rate of interest which further depressing output, causing inflation to even further decline.

If the actual output is less than potential output, demand or supply has fallen, resulting in job and output declines

Answer:

7.65 m

Explanation:

[tex]P_1[/tex] = Initial pressure = 0.03 atm

[tex]P_2[/tex] = Final pressure = 1 atm

[tex]r_1[/tex] = Inital radius = 21 m

[tex]V_1[/tex] = Intial volume of gas = [tex]\frac{4}{3}\pi r_1^3[/tex]

[tex]V_2[/tex] = Final volume of gas = [tex]\frac{4}{3}\pi r_2^3[/tex]

[tex]T_1[/tex] = Initial temperature = 200 K

[tex]T_2[/tex] = Final temperature = 323 K

From ideal gas law we have

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow \frac{P_1\frac{4}{3}\pi r_1^3}{T_1}=\frac{P_2\frac{4}{3}\pi r_2^3}{T_2}\\\Rightarrow \frac{P_1 r_1^3}{T_1}=\frac{P_2r_2^3}{T_2}\\\Rightarrow r_2=\frac{P_1r_1^3T_2}{T_1P_2}\\\Rightarrow r_2=\left(\frac{0.03\times 21^3\times 323}{200\times 1}\right)^{\frac{1}{3}}\\\Rightarrow r_2=7.65\ m[/tex]

The radius at liftoff is 7.65 m

The initial radius of the weather balloon at liftoff can be calculated using the ideal gas law by comparing the conditions at the ground and at the balloon's working altitude.

The subject of this question deals with the physics principle of the behavior of gases, namely the ideal gas law. The ideal gas law links the pressure, temperature and volume of a gas, represented by the equation PV=nRT. Here, 'P' denotes pressure, 'V' is volume, 'n' is the number of moles, 'R' is the universal gas constant, and 'T' is temperature.

In the case of the weather balloon, when it is filled at atmospheric pressure and 323 K on the ground, and then expands to a maximum radius of 21m at a working altitude where the air pressure is 0.030 atm and the temperature is 200K, the principle of the ideal gas law can be applied. Assuming the same quantity of gas (n) we have P1V1/T1=P2V2/T2 where indices 1 and 2 refer to conditions at lift off and at altitude respectively.

We know that the volume of the balloon (V) is related to its radius (r) by the equation for volume of a sphere V=(4/3)*pi*r^3. So we can substitute the volumes in above equation with their expressions in terms of radius and solve for r1 (radius at liftoff).

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Answer:

= 5.1 W

Explanation:

time (t) = 30 ms = 0.03 s

mass (m) = 560 g = 0.56 kg

initial velocity (U) = 0 m/s

final velocity (V) = 0.74 m/s

power = \frac{work done}{t} = \frac{f x d}{t} = f x v = m x a x v

m x a x v = m x \frac{V-U}{t} x \frac{V + U}{2}

m x \frac{V-U}{t} x \frac{V + U}{2} = 0.56 x \frac{0.74 - 0}{0.03} x \frac{0.74+0}{2}

= 5.1 W