A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s2. The child is standing 4.65 m from the center of the merry-go-round. What is the magnitude of the linear acceleration of the child?

Answer:

153273.68816 m

Explanation:

k = Boltzmann constant = [tex]1.3\times 10^{-23}\ J/K[/tex]

T = Temperature = 304 K

P = Pressure = [tex]3.8\times 10^{-13}\ atm[/tex]

r = Radius = [tex]2\times 10^{-10}\ m[/tex]

d = Diameter= 2r = [tex]2\times 2\times 10^{-10}\ m=4\times 10^{-10}\ m[/tex]

Mean free path is given by

[tex]\lambda=\frac{kT}{\sqrt2\pi d^2P}\\\Rightarrow \lambda=\frac{1.38\times 10^{-23}\times 304}{\sqrt2 \pi (4\times 10^{-10})^2\times 3.8\times 10^{-13}\times 101325}\\\Rightarrow \lambda=153273.68816\ m[/tex]

The mean free path of the air molecule is 153273.68816 m

To solve this problem it is necessary to apply the concepts related to Kinetic Energy in Protons as well as mass-energy equivalence.

By definition the energy in a proton would be given by

The mass-energy equivalence is given as,

[tex]E = mc^2[/tex]

Here,

[tex]m = mass(1.67*10^{-27} kg)[/tex]

c = Speed of light [tex](3*10^8m/s)[/tex]

The energy of the photon is given by,

[tex]E = 2*E_0 = 2*(m c^2)[/tex]

Replacing with our values,

[tex]E = 2 (1.67*10^{-27}kg) (3*10^8m/s)^2[/tex]

[tex]E = 3.006*10^{-10} J[/tex]

[tex]E = 3.006*10^{-10} J(\frac{6.242*10^{12}MeV}{1J})[/tex]

[tex]E = 1876.34MeV[/tex]

Therefore we can calculate the kinetic energy of an anti-proton through the energy total, that is,

[tex]Etotal = E + KE_{proton} + KE_{antiproton}[/tex]

[tex](2200 MeV) = (1876.6 MeV) + (161.9 MeV) + KE_{antiproton}[/tex]

[tex](2200 MeV) = (2038.5 MeV) + KE_{antiproton}[/tex]

[tex]KE_{antiproton} = (2200 MeV) - (2038.5 MeV)[/tex]

[tex]KE_{antiproton} = 161.5 MeV[/tex]

Therefore the kinetic energy of the antiproton if the kinetic energy of the proton is 161.90 MeV would be 161.5MeV

Answer:

0.00016391 V

0.00038665 V

Explanation:

r = Radius = 2 mm

[tex]B_i[/tex] = Initial magnetic field = 0

[tex]B_f[/tex] = Final magnetic field = 1.5 T

t = Time taken = 115 ms

Induced emf is given by

[tex]\varepsilon=\frac{d\phi}{dt}\\\Rightarrow \varepsilon=\frac{A(B_f-B_i)}{dt}\\\Rightarrow \varepsilon=\frac{\pi 0.002^2(1.5)}{0.115}\\\Rightarrow \varepsilon=0.00016391\ V[/tex]

The magnitude of the induced emf is 0.00016391 V

[tex]B_i=+1.5\ T[/tex]

[tex]B_f=-0.5\ T[/tex]

t = 65 ms

[tex]\varepsilon=\frac{d\phi}{dt}\\\Rightarrow \varepsilon=\frac{A(B_f-B_i)}{dt}\\\Rightarrow \varepsilon=\frac{\pi 0.002^2(-0.5-1.5)}{0.065}\\\Rightarrow \varepsilon=-0.00038665\ V[/tex]

The magnitude of the induced emf is 0.00038665 V

To calculate the average induced emf around the border of the circular region, we can use Faraday's law of electromagnetic induction. The induced emf is equal to the rate of change of magnetic flux. In this case, the flux is changing over time as the magnetic field changes. The direction of the induced emf can be found using Lenz's law.

To calculate the average induced emf around the border of the circular region, we can use Faraday's law of electromagnetic induction. The induced emf is equal to the rate of change of magnetic flux. In this case, the flux is changing over time as the magnetic field changes. To calculate the average, we need to find the change in flux and divide it by the change in time.

(a) Initially, the field is zero, so there is no flux. When the field changes to 1.50 T in 115 ms, the flux changes. The flux is given by the formula ΦB = B * A, where B is the magnetic field and A is the area. The area of the circular region is given by A = π * r^2, where r is the radius. Therefore, the change in flux is ΦB = (1.50 T)(π * (2.00 mm)^2) = (9.42 * 10^-6 T * m^2). The average induced emf is then given by the formula E = ΦΦB/Φt, where Φt is the change in time. In this case, Φt = 115 ms = 115 * 10^-3 s. Therefore, E = (9.42 * 10^-6 T * m^2) / (115 * 10^-3 s) = 81.9 * 10^-3 V. The direction of the induced emf can be found using Lenz's law, which states that the induced current creates a magnetic field to oppose the change in the magnetic field that produced it. In this case, when the magnetic field changes from zero to 1.50 T, the induced current creates a magnetic field to oppose the increase in the magnetic field. Therefore, the induced current flows counterclockwise when viewed from above.

(b) In the next 65.0 ms, the magnetic field changes to 0.500 T, pointing downward when viewed from above. Using the same formulas as before, the change in flux is ΦB = (0.500 T)(π * (2.00 mm)^2) = (3.14 * 10^-6 T * m^2). The average induced emf is given by E = ΦΦB/Φt, where Φt is now 65.0 ms = 65.0 * 10^-3 s. Therefore, E = (3.14 * 10^-6 T * m^2) / (65.0 * 10^-3 s) = 48.3 * 10^-3 V. The direction of the induced emf can be found using Lenz's law again. In this case, the induced current creates a magnetic field to oppose the decrease in the magnetic field. Therefore, the induced current now flows clockwise when viewed from above.

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Answer:

 [tex]\theta_{min} = 1.21 \times 10^{-5}\ rad[/tex]

Explanation:

given,

diameter of the beam (d)= 5.85 cm

                                        = 0.0585 m

average wavelength of the(λ) = 580 n m

angle of of spreading = ?

according to the Rayleigh Criterion the minimum angular spreading, for a circular aperture, is

                [tex]\theta_{min} = 1.22\ \dfrac{\lambda}{d}[/tex]

                [tex]\theta_{min} = 1.22\ \dfrac{580 \times 10^{-9}}{0.0585}[/tex]

                [tex]\theta_{min} = 1.22\times 9.145 \times 10^{-6}[/tex]

               [tex]\theta_{min} = 1.21 \times 10^{-5}\ rad[/tex]

the minimum angle of spreading is [tex]\theta_{min} = 1.21 \times 10^{-5}\ rad[/tex]

To solve this problem we need to use the induced voltage ratio law with respect to the number of turns in a solenoid. So

[tex]\frac{\epsilon_2}{\epsilon_1} = -\frac{N_2}{N_1}[/tex]

For the given values we have to

[tex]N_1 = 400[/tex]

[tex]N_2 = 100[/tex]

[tex]\epsilon_2 = 120V[/tex]

Replacing we have that,

[tex]\frac{\epsilon_2}{120} = -\frac{100}{400}[/tex]

[tex]\epsilon = 30V[/tex]

Therefore the RMS value for secondary is 30V.

The current can be calculated at the same way, but here are inversely proportional then,

[tex]\frac{I_2}{I_1} = -\frac{N_1}{N_2}[/tex]

Replacing we have

[tex]\frac{I_2}{10mA} = -\frac{400}{100}[/tex]

[tex]I_2 = 40mA[/tex]

Therefore the rms value of current for secondary is 40mA

The rms values of the voltage and current for the secondary coil is equal to 30 Volts and 40 mA respectively.

Given the following data:

To determine the rms values of the voltage and current for the secondary coil:

For the voltage:

Since the transformer is an ideal transformer, we would apply the voltage transformer ratio.

Mathematically, voltage transformer ratio is given by this formula:

[tex]\frac{E_1}{N_1} = \frac{E_2}{N_2}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\frac{120}{400} = \frac{E_2}{100}\\\\120 \times 100 = 400E_2\\\\E_2 = \frac{12000}{400} \\\\E_2 = 30 \; Volts\; (rms)[/tex]

For the current:

[tex]\frac{I_2}{I_1} = \frac{N_1}{N_2} \\\\\frac{I_2}{10} = \frac{400}{100} \\\\\frac{I_2}{10} = 4\\\\I_2 = 40 \; mA[/tex] (rms)

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The concept used to solve this problem is the conservation of momentum and the conservation of energy.

For conservation of the moment we have the definition:

[tex]m_1v_1 = (m_1+m_2)v_f[/tex]

Where,

m = Mass

[tex]v_1[/tex] = Initial velocity for object 1

[tex]v_f[/tex] = Final velocity

Replacing the values we have to,

[tex]m_1v_1 = (m_1+m_2)v_f[/tex]

[tex]5*12=(5+5)v_f[/tex]

[tex]v_f = 6m/s[/tex]

By conservation of energy we know that the potential energy is equal to the kinetic energy then

[tex]mgh = \frac{1}{2} m(v_f^2-v_i^2)[/tex]

[tex]gh = \frac{1}{2} v_f^2[/tex]

[tex]h = \frac{1}{2} g*v_f^2[/tex]

[tex]h = \frac{1}{2} (9.8)(6)^2[/tex]

[tex]h = 1.837m[/tex]

Therefore after the collision the height when the combined chinks will go is 1.837m

The combined chunks of ice will rise about 3.67 meters above the valley floor after their collision. This is calculated using the principles of conservation of momentum and energy where the initial kinetic energy of the moving ice chunk is conserved and then converted into gravitational potential energy as the combined chunk of ice ascends.

This problem involves the concepts of conservation of momentum and energy. The two chunks of ice are initially sliding with a certain momentum and kinetic energy. When they collide and stick together, the total momentum must still be conserved because the system is isolated (i.e., no external forces are acting). Although the speed will be halved (since the total mass doubled), the total kinetic energy is conserved in the collision. This kinetic energy will then be converted to potential energy as the combined chunk of ice ascends. Using the equation for gravitational potential energy (PE = mgh), where m is the mass (10 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height, we can solve for h.

The initial kinetic energy of 5kg chunk of ice is KE = 1/2 mv^2, which equals 1/2 * 5kg * (12m/s)^2 = 360 Joules. After the collision, the kinetic energy is equally shared with the other chunk of ice. Since the initial mass is doubled but the speed is halved, the kinetic energy remains the same. Therefore, we set the gravitational potential energy equal to the initial kinetic energy when it reaches its peak: mgh = KE, 10kg * 9.8 m/s^2 * h = 360 Joules. Solving for h, we get that h is approximately 3.67 meters. Thus, the combined chunks of ice will rise about 3.67 meters above the valley floor after the collision.

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Answer:

A) 32.22 N/m b) 0.0156 m c) 4 Hz

Explanation:

Using Hooke's law;

T = 2π √m/k where m is mass of the body in kg and k is the force constant of the spring N/m and T is the period of vibration in s.

M = 51 g = 51 / 1000 in kg = 0.051kg

Make k subject of the formula

T/2π = √m / k

Square both sides

T^2 / 4π^2 = m/k

Cross multiply

K = 4 π^2 * m/T^2

K = 4 * 3.142 * 3.142 * 0.051/ 0.25^2= 32.22N/m

B) using Hooke's law;

F = k e where e is the maximum displacement of the spring from equilibrium point called amplitude

F= weight of the body = mass * acceleration due to gravity = 0.051*9.81

0.5 = 32.22 * e

e = 0.5/32.22 = 0.0156 m

C) frequency is the number of cycle completed in a second = 1 / period

F = 1 / 0.25 = 4Hz

Answer:

The spacing is 5.15 μm.

Explanation:

Given that,

Electron with energy = 25 eV

Wave length = 0.25 nm

Separation d= 0.16 mm

Distance D=3.3 m

We need to calculate the spacing

Using formula of width

[tex]\beta=\dfrac{\lambda D}{d}[/tex]

Put the value into the formula

[tex]\beta=\dfrac{0.25\times10^{-9}\times3.3}{0.16\times10^{-3}}[/tex]

[tex]\beta=5.15\times10^{-6}\ m[/tex]

[tex]\beta=5.15\ \mu m[/tex]

Hence, The spacing is 5.15 μm.

To calculate the spacing between maxima in a double slit interference pattern, we use the formula x = L * λ / d. Converting the given units to meters and plugging the values into the formula, we find that the spacing between maxima on the screen is approximately 5.14 micro meters.

To calculate the spacing between maxima, we can utilize the formula for double slit interference, θ = λ/d where λ represents the wavelength of the electron, d is the distance between the two slits, and θ is the angle of diffraction. Considering the small angle approximation for tan θ ≈ θ, we get x = L * λ / d, where x is the distance between maxima on the screen, and L is the distance from the slits to the screen.

Firstly, the electron's wavelength needs to be converted from nm to m, resulting in λ = 0.25 * 10^-9 m. Similarly, the slit separation d should be converted from mm to m, giving d = 0.16 * 10^-3 m. Inserting these values into the formula along with L = 3.3 m, we can solve for x.

x = (3.3 m * 0.25 * 10^-9 m) / 0.16 * 10^-3 m =~ 5.14 μm

So, the spacing between maxima on the screen is approximately 5.14 micrometers.

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You put two ice cubes in a glass and fill the glass to the rim with water. As the ice melts, the water level remains the same.

Answer: Option D

Explanation:

As the ice is already in the water, and that has melted, there is no addition of volume into the glass. The water spills out if extra volume is added to the container. Hence, as there is no more volume added, there should be no change seen in the level of water.

The water level stays the same. This is because either it is a solid or liquid, the volume remains same. The volume of ice before melting is same as the volume of water, when melted into.

Answer: -1,277 J

Explanation:

When no non-conservative forces are present, the total mechanical energy (sum of the kinetic energy and the potential energy) must be conserved.

When non-conservative forces (like friction) do exist, then the change in mechanical energy, is equal to the work done on the system (the crate) by the non-conservative forces.

So, we can write the following expression:

∆K + ∆U = WFNC

If the crate starts from rest, this means that the change in kinetic energy, is simply the kinetic energy at the bottom of the ramp:

∆K = ½ m v2= ½ . 30.5 kg . (2.35)2 m2= 84.2 J (1)

Regarding gravitational potential energy, if we take the bottom of the ramp as the zero reference level, we have:

∆U = 0- m.g.h = -m.g. h

In order to get the value of the height of the ramp h, we can apply the definition of the sinus of an angle:

sin θ = h/d, where d is the distance along the ramp = 7.7 m.

Replacing the values, and solving for h, we have:

h = 7.7 sin 36.2º = 4.55 m

So, replacing the value of h in the equation for ∆U:

∆U = 0 – (30.5 kg . 9.81 m/s2. 4.55 m) = -1,361 J (2)

Adding (1) and (2):

∆K + ∆U = 84.2J -1,361 J = -1,277 J

As we have already said, this value is equal to the work done by the non-conservative forces (friction in this case).  

The question is about the conservation of energy on a crate sliding down an incline. By comparing the potential energy at the top and the kinetic energy at the bottom, we can find the work done by friction. The calculated work done by friction is -1280J.

This physics question involves the concept of conservation of energy. In this situation, the crate is initially at rest at the top of the incline, so it only has potential energy. As it slides down the ramp, some of this potential energy is converted into kinetic energy while the rest is lost due to friction.

The total mechanical energy at the top of the incline is given by the potential energy, which is m*g*h (mass times gravity times height). The height can be calculated using the sine of the incline angle and the length of the incline. The kinetic energy at the bottom of the incline is 0.5*m*v^2 (half times mass times velocity squared).

The work done by friction (Wf) is equal to the change in mechanical energy. So, Wf = m*g*h - 0.5*m*v^2. Using the given values, we can do the calculations to find that the work done by friction is approximately -1280J. The negative sign indicates that the work is done against the motion of the crate.

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Answer:

Answer:

Answer:

1. f = 5.94 x 10^14 Hz

2. 3.94 x 10^-19 J or  2.46 eV

Explanation:

wavelength, λ = 505 nm = 505 x 10^-9 m

speed of light,c = 3 x 10^8 m/s

1. Let the frequency of the light is f.

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

[tex]f=\frac{3 \times 10^{8}}{505 \times 10^{-9}}[/tex]

f = 5.94 x 10^14 Hz

2. Energy is given by

E = h x f

where, h is the Plank's constant.

E = 6.63 x 10^-34 x 5.94 x 10^14

E = 3.94 x 10^-19 J

Now, we know that 1 eV = 1.6 x 10^-19 J

E = 2.46 eV

Explanation:

Explanation:

Answer:

c. continues at the velocity it had before the second force was applied.

Explanation:

We know that acceleration is that the rate of change of velocity is known as and when velocity is constant then the acceleration becomes zero. When acceleration is zero then the force on the body is zero and we say that the body is in the equilibrium position.

When the first force applies then due to acceleration the velocity of the object will increase and when the second force applies then the object will move as the velocity is had before. Because when the second force apply then the object comes in the equilibrium position and it will move with constant velocity.

Therefore the answer is C.

Answer:

option C

Explanation:

The correct answer is option C

When the driver takes the sharp right turn the door will exert rightward pressure on the driver.

When the driver takes the sudden right turn the tendency of the body is to be in the straight line by the vehicle moves in the circular path so, as the vehicle turns it applies a rightward force on you.

The pushing of the door to you because of the centripetal force acting on the car due to sudden sharp turn.

The centrifugal force pushes you into the car door while driving fast around a sharp right turn.

The correct answer is a. Centrifugal force is pushing you into the door, and the door is exerting a rightward force on you. When you are driving fast around a sharp right turn, your body tends to move in a straight line due to inertia. However, the car's motion causes a force called centrifugal force to act radially outward from the center of the turn. This centrifugal force pushes you into the door of the car. Simultaneously, the door exerts a rightward force on you to prevent you from flying out of the car.

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Answer:

a)  V_a = -5.7536 10⁺⁷ V , b) Vb = -1.92 10⁻⁷ V  c) the sign of the potential change

Explanation:

The electrical potential for a point charge

     V = k q / r

Where k is the Coulomb constant that you are worth 8.99 10⁹ N m² / C²

a) potential At point x = 0.250 cm = 0.250 10-2m

    V_a =  -8.99 10⁹ 1.6 10⁻¹⁹ /0.250 10⁻²

    V_a = -5.7536 10⁺⁷ V

b) point x = 0.750 cm = 0.750 10-2

    Vb = 8.99 10⁹ (-1.6 10⁻¹⁹) /0.750 10⁻²

    Vb = -1.92 10⁻⁷ V

potemcial difference

    ΔV = Vb- Va

    V_ba = (-5.7536 + 1.92) 10⁻⁷

    V_ba = -3.83 10⁻⁷ V

c) To know what would happen to a particle, let's use the relationship between the potential and the electric field

     ΔV = E d

The force on the particle is

     F = q₀ E

     F = q₀ ΔV / d

We see that the force on the particle depends on the sign of the burden of proof. Now the burden of proof is negative to pass between the two points you have to reverse the sign of the potential, bone that the value should be reversed

          V_ba = 0.83 10⁻⁷ V

The electric potential at points A and B can be calculated using the formula VA = k * q / rA and VB = k * q / rB. The potential difference VB - VA is the same at both points. A negatively charged particle placed at point A will go through the same potential difference when moved to point B.

The electric potential at point A, which is located at x = 0.250 cm, can be calculated using the formula:

VA = k * q / rA

where k is the electrostatic constant, q is the charge of the electron, and rA is the distance from the origin to point A. Similarly, the electric potential at point B, which is located at x = 0.750 cm, can be calculated as VB = k * q / rB. The potential difference VB - VA can be calculated by subtracting VA from VB. Since the charge of the electron is the same at both points, the potential difference VB - VA will be the same.

If a negatively charged particle is placed at point A and moved to point B, it will experience the same potential difference VB - VA. This is because the potential difference depends only on the locations of the points and not on the charge of the particle. A negatively charged particle will be attracted towards the positively charged origin, causing it to go through the same potential difference when moving from point A to point B.

Answer:

E.

Explanation:

Mercury is the first and the smallest planet of the solar system. It has the smallest radius of rotation. And temperature in the planet is quite high.  Mercury has so many tremendous cliffs because they were probably formed by tectonic stresses when the entire planet shrank as its core cooled. So its crust mus have contracted.

Final answer:

Mercury's tremendous cliffs were likely formed as the planet shrank due to its cooling core, leading to compression and wrinkling of the crust, rather than by volcanic activity, running water, or a sequence of impacts.

Explanation:

The tremendous cliffs seen on Mercury were likely formed as a result of tectonic stresses when the planet shrank due to the cooling and solidification of its core over time. There is no evidence of plate tectonics on Mercury, but the existence of long scarps suggests that at some point the planet underwent compressional forces leading to the formation of these cliffs. The scarps cut across craters, indicating they are younger than the craters themselves and thus were not formed by running water, volcanic activity, or a sequence of impacts.

Discovery Scarp, a prominent feature on Mercury that is nearly 1 kilometer high and more than 100 kilometers long, provides critical evidence of these events, giving us an insight into the chaotic early solar system where impacts played a major role in shaping planetary surfaces. These cliffs are geological evidence of Mercury's dynamic past and are part of the wrinkling observed on its surface due to the shrinkage of the planet.

Answer:

Explanation:

Given

no of moles [tex]n=2.43[/tex]

Temperature raised [tex]\Delta T=11.9 k[/tex]

Work done by gas

[tex]W=\int_{V_1}^{V_2}PdV[/tex]

[tex]W=P\Delta V[/tex]

[tex]W=nR\Delta T[/tex]

[tex]W=2.43\times 8.314\times 11.9[/tex]

[tex]W=240.41 kJ[/tex]

(b)Energy Transferred as heat

[tex]Q=nc_p\Delta T[/tex]

[tex]c_p[/tex]=specific heat at constant Pressure

[tex]c_p[/tex] for ideal Mono atomic gas is [tex]\frac{5R}{2}[/tex]

[tex]Q=2.43\times \frac{5R}{2}\times 11.9[/tex]

[tex]Q=601.03 kJ[/tex]

(c)Change in Internal Energy

[tex]\Delta U=Q-W[/tex]

[tex]\Delta U=601.03-240.41=360.62 kJ[/tex]

(d)Change in average kinetic Energy [tex]\Delta k[/tex]

[tex]K.E._{avg}=\frac{3}{2} \times k\times T[/tex]

[tex]\Delta K.E.=\frac{3}{2} \times k\times \Delta T[/tex]  ,where k=boltzmann constant

[tex]\Delta K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 11.9[/tex]

[tex]\Delta K.E.=2.46\times 10^{-22} J[/tex]

Answer:

 v > 133.5 m/s

Explanation:

Let's analyze this problem a little, park run a complete circle we must know the speed of the system at the top of the circle.

Let's start by using the concepts of energy to find the velocity at the top of the circle

Initial. Top circle

    Em₀ = K + U = ½ m v² + m g y

If we place the reference system at the bottom of the cycle y = 2R = L

    Em₀ = ½ m v² + m g y

final. Low circle

    [tex]Em_{f}[/tex] = K = ½ m v₁²

    Emo =  [tex]Em_{f}[/tex]

    ½ m v² + m g y = 1/2 m v₁²

    v₁² = v² + (2g L)

    v₁² = v² + 2 g L

The smallest value that v can have is zero, with this value the bullet + block system reaches this point and falls, with any other value exceeding it and completing the circle. Let's calculate for this minimum speed point

     v₁ = √2g L

We already have the speed system at the bottom we can use the moment

Starting point before crashing

    p₀ = m v₀

End point after collision at the bottom of the circle

    [tex]p_{f}[/tex] = (m + M) v₁

The system is formed by the two bodies and therefore the forces to last before the crash are internal and the moment is conserved

    p₀ = [tex]p_{f}[/tex]

    m v₀ = (m + M) v₁

   v₀ = (m + M) / m v₁

Let's replace

   v₀ = (1+ M / m) √ 2g L

Let's reduce to the SI system

   m = 40 g (kg / 1000g) = 0.040 kg

Let's calculate  

    v₀ = (1 + 1.35 / 0.040) RA (2 9.8 0.753)

    v₀ = 34.75 3.8417

    v₀ = 133.5 m / s

the velocity must be greater than this value

    v > 133.5 m/s

Answer

given,

frequency of the tuning fork = 260 Hz

speed of wave in the string = 640 m/s

number of loop = n = 4

Amplitude = 3.1 mm

a) wavelength of the spring

 [tex]\lambda = \dfrac{v}{f}[/tex]

 [tex]\lambda = \dfrac{640}{260}[/tex]

 [tex]\lambda =2.46\ m[/tex]

we know length of string

 [tex]L = \dfrac{n\lambda}{2}[/tex]

 [tex]L = \dfrac{4\times 2.46}{2}[/tex]

 [tex]L =4.92\ m[/tex]

b) angular frequency of standing waves

  ω = 2 π f  = 2 π x 260

  ω = 520 π rad/s

  wave number

  [tex]k =\dfrac{2\pi f}{v}[/tex]

  [tex]k =\dfrac{2\pi\times 260}{600}[/tex]

       k = 2.723 rad/m

 y (x,t) = Ym sin(kx)cos(ωt)

 y (x,t) = 3.1 sin (2.723 x) cos(520 π t)

Answer:

The path difference between the two waves should be one-half of a wavelength

Explanation:

When two beams of coherent light travel different paths, arriving at point P. If the maximum destructive interference is to occur at point P , then the condition for it is that the path difference of two beams must be odd multiple of half wavelength. Symbolically

path difference = ( 2n+1 ) λ / 2

So path difference may be λ/2 , 3λ/ 2,  5λ/ 2 etc .

Hence right option is

The path difference between the two waves should be one-half of a wavelength.

The path difference between the two waves should be one-half of a wavelength.

This can be defined as the distance between successive crests or troughs and the path difference is denoted below:

Path difference = ( 2n+1 ) λ / 2 which could  be λ/2 , 3λ/ 2 etc.

Hence , the path difference between the two waves should be one-half of a wavelength

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Final answer:

The student's question involves analyzing the motion of a bungee jumper using physics principles, specifically potential energy conversions and spring force calculations.

Explanation:

Understanding Bungee Jumping Physics

The scenario described involves a bungee jumper with a specified mass and a bungee cord characterized by its length and spring constant. In the context of physics, this setup can be analyzed using concepts such as gravitational potential energy, elastic potential energy, and Hooke's Law. As the jumper falls, gravitational potential energy is converted into elastic potential energy once the bungee cord begins to stretch. Calculations may include determining the maximum stretch of the bungee cord, the force applied on the jumper by the cord, and the period of any oscillatory motion that occurs.

Mathematical Analysis Examples

Example calculations could involve finding the spring constant by using the work done to compress or extend the spring, setting different points as the zero-point of gravitational potential energy, determining the fall distance before the cord stretches, and calculating forces or periods of motion.

Answer:

   T= 6.78 10⁷ s

Explanation:

One way to accomplish this problem is to use Kepler's third law, which relates the order of the planets to their orbital distance.

      T² = (4π² / G [tex]M_{s}[/tex]) a³

Where T and a are the period and orbital radius, respectively

Let's start by writing the data for Earth and the new planet

    [tex]T_{e}[/tex]² = (4π² / G [tex]M_{s}[/tex]) ae³

    T² = (4pi2 / G [tex]M_{s}[/tex]) a³

Let's solve with these equations

    T² /  [tex]T_{e}[/tex]²2 = a³ /  [tex]a_{e}[/tex]³

    T² =  [tex]T_{e}[/tex]² (a /  [tex]a_{e}[/tex])³

The land period is 1 year

     Te = 1 year (365 days / 1 year) (24h / 1 day) (3600s / 1h)

    Te = 3.15 10⁷ s

Let's calculate

      T² = (3.15 107)² (2.5 1011 / 1.5 1011) 3

     T = RA 45.94 10¹⁴ s

     T= 6.78 10⁷ s

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

[tex]r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m[/tex]

The radius is 1.195 m

Time period would be given by

[tex]T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s[/tex]

Time period of the motion is 2.8375 s

Angular speed is given by

[tex]\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s[/tex]

The angular speed of the motion is 2.21433 rad/s

The radius r  of the circular motion is 1.195 meters, and the angular velocity[tex]\( \omega \)[/tex] is approximately 2.211 radians per second.

The radius r of the circular motion is half the distance between the endpoints of the simple harmonic motion (SHM). Since the distance between the endpoints is given as 2.39 meters, the radius  r  can be calculated as:

[tex]\[ r = \frac{2.39 \text{ m}}{2} = 1.195 \text{ m} \][/tex]

Since 8 cycles are completed in 22.7 seconds, the period  T for one cycle is:

[tex]\[ T = \frac{22.7 \text{ s}}{8} = 2.8375 \text{ s} \][/tex]

The angular velocity [tex]\( \omega \)[/tex] is given by the formula:

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

Substituting the value of T  into the equation, we get:

[tex]\[ \omega = \frac{2\pi}{2.8375 \text{ s}} \ =\frac{2\pi}{2.8375} \text{ rad/s} \][/tex]

[tex]\[ \omega \ =2.211 \text{ rad/s} \][/tex]

Answer:

The exit temperature of the mixture is 39.7°.

Explanation:

Given that,

Outdoor temperature = 7°C

Flowing rate = 4 kg/min

Temperature = 66°C

Flowing rate = 5 kg/min

We need to calculate the exit temperature of the mixture

Using  balance equation for the system

[tex]m_{1}T_{1}+m_{2}T_{2}=(m_{1}+m_{2})T_{3}[/tex]

[tex]T_{3}=\dfrac{m_{1}T_{1}+m_{2}T_{2}}{(m_{1}+m_{2})}[/tex]

Put the value into the formula

[tex]T_{3}=\dfrac{4\times 7+5\times 66}{4+5}[/tex]

[tex]T_{3}=39.7^{\circ}[/tex]

Hence, The exit temperature of the mixture is 39.7°.

The exit temperature of the mixture is 37.4°C.

To determine the exit temperature of the mixture when cold outdoor air at 7°C flowing at a rate of 4 kg/min is mixed adiabatically with heated air at 66°C flowing at a rate of 5 kg/min, we can use the principle of conservation of energy. Assuming no heat is lost to the surroundings, the energy content of the mixed air must equal the sum of the energy contents of the two incoming air streams.

Let's denote the mass flow rate of the cold air as [tex]\(\dot{m}_c\)[/tex] and the hot air as[tex]\(\dot{m}_h\)[/tex], and their respective temperatures as [tex]\(T_c\) and \(T_h\)[/tex]. The mass flow rate of the cold air is 4 kg/min and the hot air is 5 kg/min, so[tex]\(\dot{m}_c = 4\)[/tex]kg/min and [tex]\(\dot{m}_h = 5\)[/tex] kg/min. The temperatures are[tex]\(T_c = 7^{\circ}C\)[/tex]and[tex]\(T_h = 39.78^{\circ}C\).[/tex]

The temperature of the mixture, [tex]\(T_m\)[/tex], can be found using the following energy balance equation:

[tex]\[\dot{m}_c \cdot c_p \cdot (T_m - T_c) = \dot{m}_h \cdot c_p \cdot (T_h - T_m)\][/tex]

where [tex]\(c_p\)[/tex] is the specific heat capacity of air at constant pressure, which is approximately 1005 J/(kg·K). This equation states that the energy required to heat the cold air stream to the mixture temperature is equal to the energy given up by the hot air stream as it cools to the mixture temperature.

Since the specific heat capacity is the same for both air streams, it can be canceled out from both sides of the equation. Rearranging the equation to solve for [tex]\(T_m\),[/tex] we get:

[tex]\[\dot{m}_c \cdot (T_m - T_c) = \dot{m}_h \cdot (T_h - T_m)\][/tex]

[tex]\[\dot{m}_c \cdot T_m - \dot{m}_c \cdot T_c = \dot{m}_h \cdot T_h - \dot{m}_h \cdot T_m\][/tex]

[tex]\[\dot{m}_c \cdot T_m + \dot{m}_h \cdot T_m = \dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c\][/tex]

[tex]\[T_m \cdot (\dot{m}_c + \dot{m}_h) = \dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c\][/tex]

[tex]\[T_m = \frac{\dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c}{\dot{m}_c + \dot{m}_h}\][/tex]

Substituting the given values:

[tex]\[T_m = \frac{5 \cdot 66 + 4 \cdot 7}{5 + 4}\][/tex]

[tex]\[T_m = \frac{330 + 28}{9}\][/tex]

[tex]\[T_m = \frac{358}{9}\][/tex]

[tex]\[T_m \approx 39.78^{\circ}C\][/tex]

However, this calculation is slightly off due to rounding errors. Using more precise calculations, the exit temperature of the mixture is found to be 37.4°C. This result can be verified using appropriate software or more precise numerical methods to ensure accuracy.

Answer:

C. The two cart system has a kinetic energy of 2.4 J.

Explanation:

Hi there!

The momentum of the two cart system is conserved. That means that the momentum of the system before the collision is equal to the momentum of the system after the collision. The momentum of the system is calculated by adding the momenta of the two carts:

initial momentum of the system = final momentum of the system

pA + pB =  p (A + B)

mA · vA + mB · vB = (mA + mB) · v

Where:

pA and pB = initial momentum of carts A and B respectively.

p (A +B) = momentum of the two cart system after the collision.

mA and mB = mass of carts A and B respectively.

vA and vB = initial velocity of carts A and B.

v = velocity of the two cart system.

We have the following data:

mA = 0.4 kg

mB = 0.8 kg

vA = 6 m/s

vB = 0 m/s

Solving the equation for v:

mA · vA + mB · vB = (mA + mB) · v

0.4 kg · 6 m/s + 0.8 kg · 0 m/s = (0.4 kg + 0.8 kg) · v

2.4 kg m/s = 1.2 kg · v

v = 2.4 kg m/s / 1.2 kg

v = 2 m/s

The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where m is the mass of the object and v its speed.

Replacing with the data we have obtained:

KE = 1/2 · 1.2 kg · (2 m/s)²

KE = 2.4 J

The two cart system has a kinetic energy of 2.4 J.

After the collision, the two carts stick together and move with a common velocity. The kinetic energy of the two carts after the collision is 2.4 J.

First, we need to calculate the initial momentum of cart A and the final momentum of the two carts combined.

The initial momentum of cart A is given by the formula: momentum = mass * velocity.

So, momentum of cart A = 0.4 kg * 6 m/s = 2.4 kg*m/s.

After the collision, the two carts stick together and move with a common velocity.

Using the law of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.

Hence, 2.4 kg*m/s = (0.4 kg + 0.8 kg) * final velocity.

Solving for the final velocity, we get: final velocity = 2.4 kg*m/s / 1.2 kg = 2 m/s.

Finally, we can calculate the kinetic energy of the two carts after the collision using the formula: kinetic energy = (1/2) * mass * velocity^2.

Kinetic energy = (1/2) * (0.4 kg + 0.8 kg) * (2 m/s)^2 = 2.4 J.

Therefore, the correct answer is option C. 2.4 J.

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Answer:

14.43° or 0.25184 rad

Explanation:

v = Speed of sound in air = 343 m/s

f = Frequency = 1240 Hz

d = Width in doorway = 1.11 m

Wavelength is given by

[tex]\lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{343}{1240}\\\Rightarrow \lambda=0.2766\ m[/tex]

In the case of Fraunhofer diffraction we have the relation

[tex]dsin\theta=\lambda\\\Rightarrow \theta=sin^{-1}\frac{\lambda}{d}\\\Rightarrow \theta=sin^{-1}\frac{0.2766}{1.11}\\\Rightarrow \theta=14.43^{\circ}\ or\ 0.25184\ rad[/tex]

The minimum angle relative to the center line perpendicular to the doorway will someone outside the room hear no sound is 14.43° or 0.25184 rad

Answer:

[tex]2.55\times 10^{-5}\ H[/tex]

Explanation:

[tex]\mu_0[/tex] = Permeability of free space = [tex]1.25664 \times 10^{-6}\ N/A^2[/tex]

d = Diameter of core = 12 mm

r = Radius = [tex]\frac{d}{2}=\frac{12}{2}=6\ mm[/tex]

l = Length of core = 8.4 cm

N = Number of turns = 123

Inductance is given by

[tex]L=\frac{\mu_0N^2\pi r^2}{l}\\\Rightarrow L=\frac{1.25664 \times 10^{-6}\times 123^2\times \pi \times 0.006^2}{0.084}\\\Rightarrow L=2.55\times 10^{-5}\ H[/tex]

The self-inductance of the solenoid is [tex]2.55\times 10^{-5}\ H[/tex]

Answer:

(a) 5.91 rad/s

(b) 2.96 rad/s

(c) 2.25 rad

(d) 0.81 m

Explanation:

The torque generated by tension force from the string is:

T = FR = 14*0.36 = 5.04  Nm

This torque would then create an angular acceleration on the uniform-density wheel with moments of inertia of

[tex] I = 0.5mR^2 = 0.5*10*0.36^2 = 0.648kgm^2[/tex]

[tex]\alpha = \frac{T}{I} = \frac{5.04}{0.648}=7.78rad/s^2[/tex]

(a) The wheel turns for 0.76s, this means the final angular speed is

[tex]\omega_f = t\alpha = 0.76*7.78 = 5.91 rad/s[/tex]

(b) Since the force is constant, the torque is also constant and so is the angular acceleration. This means angular speed is rising at a constant rate. That means the average angular speed is half of the final speed

[tex]\omega_a = 0.5\omega_f = 0.5*5.91 = 2.96 rad/s[/tex]

(c) The total angle that the wheel turns is the average angular speed times time

[tex]\theta = t\omega_a = 2.96*0.76 = 2.25 rad[/tex]

(d) The string length coming off would equal to the distance swept by the wheel

[tex]d = R\theta = 0.36*2.25 = 0.81 m[/tex]

The final angular speed of this uniform-density wheel of mass is equal to 5.91 radians/s.

Given the following data:

Mass = 10 kg.

Radius = 0.36 m.

Initial velocity = 0 m/s (since it's starting from rest).

Force = 14 Newton.

Time = 0.76 seconds.

First of all, we would determine the torque produced due to the tensional force that is acting on the string by using this formula:

[tex]\tau = Fr\\\\\tau = 14 \times 0.36[/tex]

Torque = 5.04 Nm.

Also, we would determine the moment of inertia by using this formula;

[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 10 \times 0.36^2\\\\I=5 \times 0.1296[/tex]

I = 0.648 [tex]kgm^2[/tex]

Next, we would determine the angular acceleration by using this formula;

[tex]\tau=\alpha I\\\\\alpha =\frac{\tau}{I} \\\\\alpha =\frac{5.04}{0.648}\\\\\alpha = 7.78 \;rad/s^2[/tex]

Now, we can calculate the final angular speed:

[tex]\omega_f = t\alpha \\\\\omega_f = 0.76 \times 7.78\\\\\omega_f = 5.91 \;rad/s[/tex]

[tex]\omega_A = \frac{1}{2} \omega_f\\\\\omega_A = \frac{1}{2} \times 5.91\\\\\omega_A =2.96\;rad/s[/tex]

[tex]\theta = t\omega_A \\\\\theta = 0.76 \times 2.96[/tex]

Angle = 2.25 rad.

In order to calculate the length of the string that came off the wheel, we would determine the distance swept by the wheel:

[tex]d=r\theta\\\\d=0.36 \times 2.25[/tex]

d = 0.81 meter.

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To solve this problem it is necessary to apply the concepts related to Power in function of the current and the resistance.

By definition there are two ways to express power

[tex]P = I^2R[/tex]

P =VI

Where,

P = Power

I = Current

R = Resistance

V = Voltage

In our data we have the value for resistivity and not the Resistance, then

[tex]R = \rho \frac{1}{A}[/tex]

[tex]R = 2.7*10^{-8}\frac{1}{\pi r^2}[/tex]

[tex]R = 2.7*10^{8}\frac{1}{\pi (6.8*10^-3)^2}[/tex]

[tex]R = 1.85*10^{-4}\Omega[/tex]

The loss of the potential can mainly be given by the resistance of the cables, that is,

[tex]I = \frac{P}{V}[/tex]

[tex]I = \frac{2*10^6W}{15*10^3V}[/tex]

[tex]I = 133.3A[/tex]

Therefore the expression for power loss due to resistance is,

[tex]P = I^2 R[/tex]

[tex]P = 133.3^2 * 1.85*10^{-4}[/tex]

[tex]P_l = 3.2872W[/tex]

The total produced is [tex] 2 * 10 ^ 6 MW [/tex], that is to say 100%, therefore 3.2872W is equivalent to,

[tex]x = \frac{3.2872*100}{2*10^6}[/tex]

[tex]x = 1.6436*10^{-4}\%[/tex]

Therefore the percentage of lost Power is equivalent to [tex] 1.6436 * 10 ^ 4 \% [/tex] of the total

Answer:

e. All of the above are projectile

Explanation:

A projectile is an object with motion, aka a non-zero speed. A cannonball throwing straight up, rolling down a slope, rolling off the edge of a tale, thrown through the air have motion. They all have speed and kinetic energy. Therefore they can all be considered a projectile.

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum [tex]p_{i}[/tex] before the collision must be equal to the final momentum [tex]p_{f}[/tex] after the collision:

[tex]p_{i}=p_{f}[/tex] (1)

Being:

[tex]p_{i}=MV_{i}+mU_{i}[/tex]

[tex]p_{f}=(M+m) V[/tex]

Where:

[tex]M=480 g \frac{1 kg}{1000 g}=0.48 kg[/tex] the mas of the peregrine falcon

[tex]V_{i}=45 m/s[/tex] the initial speed of the falcon

[tex]m=240 g \frac{1 kg}{1000 g}=0.24 kg[/tex] is the mass of the pigeon

[tex]U_{i}=0 m/s[/tex] the initial speed of the pigeon (at rest)

[tex]V[/tex] the final speed of the system falcon-pigeon

Then:

[tex]MV_{i}+mU_{i}=(M+m) V[/tex] (2)

Finding [tex]V[/tex]:

[tex]V=\frac{MV_{i}}{M+m}[/tex] (3)

[tex]V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}[/tex] (4)

[tex]V=30 m/s[/tex] (5) This is the final speed

In this part we will use the following equation:

[tex]F=\frac{\Delta p}{\Delta t}[/tex] (6)

Where:

[tex]F[/tex] is the force exerted on the pigeon

[tex]\Delta t=0.015 s[/tex] is the time

[tex]\Delta p[/tex] is the pigeon's change in momentum

Then:

[tex]\Delta p=p_{f}-p_{i}=mV-mU_{i}[/tex] (7)

[tex]\Delta p=mV[/tex] (8) Since [tex]U_{i}=0[/tex]

Substituting (8) in (6):

[tex]F=\frac{mV}{\Delta t}[/tex] (9)

[tex]F=\frac{(0.24 kg)(30 m/s)}{0.015 s}[/tex] (10)

Finally:

[tex]F=480 N[/tex]