Sound waves are a longitudinal wave that have a speed of about 340 m/s when traveling through room temperature air. What is the wavelength of a sound wave that has a frequency of 6,191 Hz?

Answer:

3.3619 Nm

54.27472 rad

182.46618 J

86.88 W

Explanation:

[tex]L_i[/tex] = Initial angular momentum = 7.2 kgm²/s

[tex]L_f[/tex] = Final angular momentum = 0.14 kgm²/s

I = Moment of inertia = 0.142 kgm²

t = Time taken

Average torque is given by

[tex]\tau_{av}=\frac{L_f-L_i}{\Delta t}\\\Rightarrow \tau_{av}=\frac{0.14-7.2}{2.1}\\\Rightarrow \tau_{av}=-3.3619\ Nm[/tex]

Magnitude of the average torque acting on the flywheel is 3.3619 Nm

Angular speed is given by

[tex]\omega_i=\frac{L_i}{I}[/tex]

Angular acceleration is given by

[tex]\alpha=\frac{\tau}{I}[/tex]

From the equation of rotational motion

[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=\frac{L_i}{I}\times t+\frac{1}{2}\times \frac{\tau}{I}\times t^2\\\Rightarrow \theta=\frac{7.2}{0.142}\times 2.1+\frac{1}{2}\times \frac{-3.3619}{0.142}\times 2.1^2\\\Rightarrow \theta=54.27472\ rad[/tex]

The angle the flywheel turns is 54.27472 rad

Work done is given by

[tex]W=\tau\theta\\\Rightarrow W=-3.3619\times 54.27472\\\Rightarrow W=-182.46618\ J[/tex]

Work done on the wheel is 182.46618 J

Power is given by

[tex]P=\frac{W}{t}\\\Rightarrow P=\frac{-182.46618}{2.1}\\\Rightarrow P=-86.88\ W[/tex]

The magnitude of the average power done on the flywheel is 86.88 W

Answer:

a. 86.7 Nm

b. b. By making sure her foot is on the outermost edge of the pedal aat all times

Explanation:

a. Torque = force  * perpendicular distance

                = (52 kg * 9.81 m/s²)  * 0.17 m

                = 86.7204 Nm

b. Torque can be increased by increasing the force exerted peperndicular to the distance from the fulcrum or by increasing the distance itself. Since the person cannot increase their weight, they have to maximize the distance from the fulcrum

The maximum torque is 86.7204 Nm and the rider can increase this by putting int leg on the edge of the paddle.  

It is defined as the force that causes the rotation of an object on an axis. It is measured in Nm.

[tex]\tau = rF\sin\theta[/tex]

[tex]\tau[/tex] = torque

[tex]r\\[/tex] = radius

[tex]F[/tex] = force = mg = (52 kg x  9.81 m/s²) = 510.12 N

= angle between F and the lever arm

Torque = force  x perpendicular distance

Put the values,  

Torque= 510.12 N x 0.17 m  

Torque = 86.7204 Nm

B) Since the torque is directly proportional to the force and perpendicular distance. Hence, the rider must increase the perpendicular distance in order to increase the torque.

Therefore, the maximum torque is 86.7204 Nm and the rider can increase this by putting int leg on the edge of the paddle.

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Answer:

[tex]E=477.92\ W.m^{-2}[/tex]

Explanation:

Given that:

Absolute temperature of the body, [tex]T=273+30=303\ K[/tex]

Using Stefan Boltzmann Law of thermal radiation:

[tex]E=\epsilon. \sigma.T^4[/tex]

where:

[tex]\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}[/tex]   (Stefan Boltzmann constant)

Now putting the respective values:

[tex]E=1\times 5.67\times 10^{-8}\times 303^4[/tex]

[tex]E=477.92\ W.m^{-2}[/tex]

Answer:

(a) the amplitude of the oscillations is 0.52 m

(b) position of the block at t = 0 s is - 0.06 m

(c) the velocity of the block at t = 0 s is 3.96 m/s

Explanation:

given information:

m = 4.10 kg

k = 240 N/m

t = 1.70 s

x = 0.165 m

v  = 3.780 m/s

(a) the amplitude of the oscillations

x(t) = A cos (ωt+φ)

v(t) = dx(t)/dt

     =  - ω A sin (ωt+φ)

ω = [tex]\sqrt{\frac{k}{m} }[/tex]

   = [tex]\sqrt{\frac{240}{4.10} }[/tex]

   = 7.65 rad/s

v(t)/x(t) = - ω A sin (ωt+φ)/A cos (ωt+φ)

v(t)/x(t) = - ω sin (ωt+φ)/cos (ωt+φ)

v(t)/x(t) = - ω tan (ωt+φ)

(ωt+φ) = [tex]tan^{-1}[/tex] (-v/ωx)

           = [tex]tan^{-1}[/tex] (-3.780/(7.65)(0.165))

           = - 1.25

(ωt+φ) = - 1.25

φ = - 1.25 - ωt

   = - 1.25 - (7.65 x 1.70)

   = - 14.26

x(t) = A cos (ωt+φ)

A = x(t) / cos (ωt+φ)

   = 0.165/cos(-1.25)

   = 0.52 m

(b) position, at t=0

x(t) = A cos (ωt+φ)

x(0) = A cos (ω(0)+φ)

x(0) = A cos (φ)

      = 0.52 cos (-14.26)

      = - 0.06 m

(c) the velocity, at t=0

v(t) = - ω A sin (ωt+φ)

v(0) = - ω A sin (ω(0)+φ)

      = - ω A sin (φ)

      = - (7.65)(0.52) sin (-14.26)

      = 3.96 m/s

Answer:

Explanation:

Given

Tom is on half way between the center and the outer rim

Mary is on the extreme outer rim

Suppose [tex]\omega [/tex] is the angular velocity of ride with radius of outer rim be R

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

At any instant both Tom and Mary experience the same angular speed

(b)Linear velocity at any instant

[tex]v_{Tom}=\omega \times R[/tex]

[tex]v_{Mary}=\omeag \times \frac{R}{2}[/tex]

Thus Tom has higher linear speed

(c) For radial Acceleration

[tex]a_r=\omega ^2\times r[/tex]

[tex]a_{Tom}=\omega ^2\times R[/tex]

[tex]a_{Mary}=\omega ^2\times \frac{R}{2}[/tex]

Tom has higher radial Acceleration as it is directly Proportional of radius          

To determine how far the beam can extend over the ledge, we need to consider the principle of equilibrium. The beam will be in equilibrium when the clockwise moments equal the counterclockwise moments. Calculating the moments, we find that the beam can extend 7.0 meters from the edge of the ledge.

To determine how far the beam can extend over the ledge, we need to consider the principle of equilibrium. The beam will be in equilibrium when the clockwise moments equal the counterclockwise moments. The moment is a force multiplied by the distance from the pivot point. In this case, the pivot point is the edge of the ledge and the force is the weight of the beam and the student.

Let's calculate the moments: The moment of the beam is 280 kg * 9.8 m/s² * x, where x is the distance from the edge of the ledge. The moment of the student is 50 kg * 9.8 m/s² * (10 m - x), where 10 m is the length of the beam.

Setting the clockwise moments equal to the counterclockwise moments, we have 280 kg * 9.8 m/s² * x = 50 kg * 9.8 m/s² * (10 m - x). Solving for x gives us x = 7.0 m. Therefore, the beam can extend 7.0 meters from the edge of the ledge.

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Final answer:

To ensure the beam does not tip, calculate the torque around the pivot at the ledge. The beam's weight produces a torque at its center of mass, while the student's weight produces a torque at the beam's end. Solving for equilibrium, the beam can extend 5.6 meters.

Explanation:

The problem presented involves a horizontal beam that a student wishes to extend over a ledge without it falling over due to its own weight and the weight of the student. To determine how far the beam can extend over the ledge, one must consider the torque that is acting on the beam around the pivot point, which is at the edge of the ledge. The beam must remain in rotational equilibrium, meaning the total torque around the pivot must be zero.

Let's assume the beam extends a distance x over the edge. The weight of the beam acts at its center of mass, which is x/2 from the pivot (edge of the ledge) if x is the length beyond the ledge. The moment (or torque) due to the beam's weight is (280 kg × 9.8 m/s² × x/2). The moment due to the student's weight at the end of the beam is (50 kg × 9.8 m/s² × x). For rotational equilibrium, these moments must be equal.

Setting the torques equal gives us: 280 kg × 9.8 m/s² × x/2 = 50 kg × 9.8 m/s² × x. Solving for x gives x = (280/100) × 2, so x must be 5.6 meters. Thus the beam can safely extend 5.6 meters over the ledge without tipping over.

To solve this problem it is necessary to resort to the energy conservation equations, both kinetic and electrical.

By Coulomb's law, electrical energy is defined as

[tex]EE = \frac{kq_1q_2}{d}[/tex]

Where,

EE = Electrostatic potential energy

q= charge

d = distance between the charged particles

k = Coulomb's law constant

While kinetic energy is defined as

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

Where,

m= mass

v = velocity

There by conservation of energy we have that

EE= KE

There is not Initial kinetic energy, then

[tex]\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = 2*\frac{1}{2}mv_f^2[/tex]

[tex]\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = mv_f^2[/tex]

[tex]v_f^2= \frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} }{m}[/tex]

[tex]v_f = \sqrt{\frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'}}{m}}[/tex]

Replacing with our values we have,

[tex]v_f = \sqrt{\frac{\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{1}-\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{3}}{5.50*10^{-15}}}[/tex]

[tex]v_f = 2.802*10^7m/s[/tex]

Therefore the speed of particle B at the instat when the particles are 3m apart is [tex]2.802*10^7m/s[/tex]

To solve this problem it is necessary to apply the concepts related to the Power depending on the temperature and the heat transferred.

By definition the power can be expressed as

[tex]P = \frac{\Delta T}{\Delta Q}[/tex]

Where,

[tex]\Delta T = T_m - T_a =[/tex] Change at the temperature, i.e, the maximum acceptable die temperature ([tex]T_m[/tex]) with the allowable temperature in chassis ([tex]T_A[/tex])

[tex]\Delta Q = Q_A-Q_D =[/tex] Change in the thermal resistance to ambient ([tex]Q_A[/tex]) and the Thermal resistance from die to package ([tex]Q_D[/tex])

Our values are given as,

[tex]T_m=110\°C[/tex]

[tex]T_a= 50\°C[/tex]

[tex]Q_A= 8\°C/W[/tex]

[tex]Q_D= 2\°C/W[/tex]

Replacing we have,

[tex]P = \frac{110-50}{8-2}[/tex]

[tex]P = 10W[/tex]

The power that can dissipate the chip is 10W

Answer:Aluminium

Explanation:

Given

Aluminium ball sinks after placing inside the water while wooden block floats.

Buoyancy on a body is given by

[tex]volume\ inside\ the\ water \times density\ of\ fluid\times g[/tex]

here [tex]\rho _{Al}>\rho _{water}[/tex]

thus aluminium ball sinks

but for

[tex]\rho _{wood}<\rho _{water}[/tex]

only a part of wood is under the water i.e. not whole volume is not under water therefore net buoyancy is less in case of wood

Final answer:

The question is about using the single slit diffraction formula to calculate the angles at which the third dark fringe appears for two different wavelengths and the width of the central bright fringe for each wavelength, involving physics and high school level understanding.

Explanation:

The student's question relates to the concept of single slit diffraction, which is a phenomenon in physics where light spreads out after passing through a narrow opening, resulting in a characteristic pattern. Specifically, the angles at which dark fringes appear and the width of the central bright fringe are calculated for two different wavelengths using the formulas for single slit diffraction.

For part (a), to calculate the angles for the third dark fringe for each wavelength, we can use the formula dsin(θ) = mλ, where d is the slit width, θ is the angle of the dark fringe, m is the order of the dark fringe, and λ is the wavelength of the light. The third dark fringe corresponds to m = 3 for this calculation. For part (b), the width of the central bright fringe is calculated by considering the angles to the first dark fringes on either side of the central peak and then using trigonometry to determine the width on the screen.

Answer:

5.78971 m

Explanation:

[tex]P_1[/tex] = Initial pressure = 0.873 atm

[tex]P_2[/tex] = Final pressure = 0.0282 atm

[tex]V_1[/tex] = Initial volume

[tex]V_2[/tex] = Final volume

[tex]r_1[/tex] = Initial radius = 16.2 m

[tex]r_2[/tex] = Final radius

Volume is given by

[tex]\frac{4}{3}\pi r^3[/tex]

From the ideal gas law we have the relation

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow \frac{0.873\times \frac{4}{3}\pi r_1^3}{294.15}=\frac{0.0282\frac{4}{3}\pi r_2^3}{208.15}\\\Rightarrow \frac{0.873r_1^3}{294.15}=\frac{0.0282\times 16.2^3}{208.15}\\\Rightarrow r_1=\frac{0.0282\times 16.2^3\times 294.15}{208.15\times 0.873}\\\Rightarrow r_1=5.78971\ m[/tex]

The radius of balloon at lift off is 5.78971 m

To find the radius of the weather balloon at lift-off, the ideal gas law can be used. Using the equation P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off, the radius at lift-off can be calculated to be approximately 4.99 m.

To find the radius of the weather balloon at lift-off, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

In this case, we know that the number of moles is constant, as the balloon is filled with the same amount of helium at lift-off and in flight. Therefore, we can write the equation as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off.

Plugging in the given values, we have (0.873 atm)(V1) = (0.0282 atm)(16.2 m)^3. Solving for V1, we find that the volume at lift-off is approximately 110.9 m^3. The radius can then be calculated using the formula for the volume of a sphere: V = (4/3) * π * r^3, where r is the radius.

Therefore, the radius at lift-off is approximately 4.99 m.

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Answer:

T = 8.19 s ,  [tex]v_{max}[/tex]  = 23 m / s

Explanation:

In the simple harmonic motion the equation that describes them is

    y = A cos wt

Acceleration can be found by derivatives

    a = d²y / dt²

   v = dy / dt = - Aw sin wt

   a= d²y / dt² = - A w² cos wt

For maximum acceleration cosWT = + -1

    [tex]a_{max}[/tex] = -A w2

    w = RA ([tex]a_{max}[/tex]/ A)

    w = RA (1.8 9.8 / 30.0)

    w = 0.767 rad / s

The angular velocity is related to the frequency

    w = 2π f

     f = 1 / T

    w = 2π / T

    T = 2π / w

    T = 2π / 0.767

    T = 8.19 s

For maximum speed the sin wt = + -1

    [tex]v_{max}[/tex] = A w

    [tex]v_{max}[/tex]  = 30.0 0.767

   [tex]v_{max}[/tex]  = 23 m / s

Answer:

Explanation:

Given

mass of baseball [tex]m=0.15 kg[/tex]

radius of ball [tex]r=3.7 cm[/tex]

angular speed of ball [tex]\omega =45 rad/s[/tex]

linear speed of ball [tex]v=42 m/s[/tex]

Transnational Kinetic Energy is given by

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

[tex]K.E.=\frac{1}{2}\times 0.15\times 42^2[/tex]

[tex]k.E.=\frac{1}{2}\times 0.15\times 1764[/tex]

[tex]  k.E.=132.3 J[/tex]

Considering the ball as solid sphere its moment of inertia is given by  

[tex]I=\frac{2}{5}mr^2=\frac{2}{5}\times 0.15\times (0.037)^2[/tex]

[tex]I=8.21\times 10^{-5} kg-m^2[/tex]

Rotational Kinetic Energy

[tex]=\frac{1}{2}\times I\times \omega ^2[/tex]

[tex] =\frac{1}{2}\times 8.21\times 10^{-5}\times 45^2[/tex]

[tex] =\frac{1}{2}\times 8.21\times 10^{-5}\times 2025[/tex]

[tex] =0.0831 J[/tex]

Answer: 0.36 m

Explanation:

By definition, the x coordinate of the center of mass of the system obeys the following equation:

Xm = (m1x1 + m2x2 + …….+ mnxn) / m1+m2 +……+ mn

Neglecting the mass of the rod, and choosing our origin to be coincident with the location of the full bucket, we can write the following expression for the X coordinate of the center of mass (Assuming that both masses are aligned over the x-axis, so y-coordinates are zero):

Xm = 0.25 mb . 1.8 m / (1+0.25) mb

Simplifying mb, we get:

Xm= 0.36 m (to the right of the full bucket).

Final answer:

To find the center of mass of the system, use the principle of moments to balance the torques produced by the full and empty buckets. The distance from the fulcrum to the center of mass can be calculated using the equation for torque.

Explanation:

To find the center of mass of the system, we can use the principle of moments. The center of mass of the system will be located at a distance from the full bucket that balances the torques produced by the two buckets.

Let's denote the distance from the fulcrum to the center of mass of the system as x. The torques produced by the full and empty buckets can be calculated using the formula, Torque = Force * Distance.

Since the empty bucket has 0.25 the inertia of the full bucket, the torque produced by the empty bucket will be 0.25 times the torque produced by the full bucket. Setting up the equation using the principle of moments, we can solve for x.

Answer:

C. increase because of conservation of angular momentum.

Explanation:

We know that if there is no any external torque on the system then the angular momentum of the system will be conserved.

Angular momentum L

L = I ω

I=Mass moment of inertia ,  ω=Angular speed

If angular momentum is conserved

I₁ω₁ = I₂ω₂

If people migrates towards  Earth's equator then mass moment of inertia will increases.

I₂ > I₁

Then we can say that ω₁  > ω₂ ( angular momentum is conserve)

We know that time period T given as

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

[tex]T_1=\dfrac{2\pi}{\omega_1}[/tex]

[tex]T_2=\dfrac{2\pi}{\omega_2}[/tex]

If ω₂ decrease then T₂ will increase .It means that length of the day will increase.

Therefore answer is C.

If a large number of people migrated towards the Earth's equator, the day length would remain largely unaffected. This is due to the law of conservation of angular momentum, considering the insignificant change to the Earth's total mass.

The scenario mentioned in the question refers to a law in physics called the conservation of angular momentum. The length of a day on Earth is predominantly determined by how fast our planet spins on its axis. If there were a large migration of people towards the Earth's equator, hypothetically, the day length would remain largely unaffected. The mass of the people in this context is insignificant when compared to the total mass of the Earth. As such, their migration would not have a noticeable impact on Earth's rotational speed due to conservation of angular momentum. Therefore, the correct answer is that the length of the day would remain unaffected (option b).

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Answer:

A) v0 = 18.8 m/s

B) θ = 31.5°

Explanation:

On the x-axis:

[tex]X_{ball} = X_{player}[/tex]

[tex]v0*cos\theta*t=L+V*t[/tex]

where

t = 2s;  L = 18m;  V = 7m/s

[tex]v0*cos\theta*2=32[/tex]

[tex]v0=16/cos\theta[/tex]

On the y-axis for the ball:

[tex]\Delta Y=v0*sin\theta*t-1/2*g*t^2[/tex]

[tex]0 = v0*sin\theta*2-19.62[/tex]

Replacing v0:

[tex]0 = tan\theta*2*16-19.62[/tex]

[tex]\theta=atan(19.62/32)=31.5\°[/tex]

Now, the speed v0 was:

v0 = 18.8m/s

The initial speed of the ball is 9.00 m/s and the angle is 0 degrees above the horizontal.

Part A:

To find the initial speed, we can use the horizontal distance traveled by the ball and the time it takes to reach the third baseman. The horizontal distance is the initial distance between the third baseman and the location where the ball was hit, which is 18.0m. The time is given as 2.00s. So, using the formula: distance = speed × time, we can rearrange it to find the initial speed: speed = distance / time. Plugging in the values, we get: speed = 18.0m / 2.00s = 9.00m/s.
Part B:

To find the angle, we can use the vertical distance traveled by the ball. The vertical distance is the same as the height at which the ball left the bat. Since the third baseman catches the ball at the same height, the vertical distance is zero. Using the formula: vertical distance = initial velocity × time + 0.5 × acceleration due to gravity × time squared, we can rearrange it to find the angle: angle = arctan( (2 × vertical distance × gravity) / (initial speed squared) ). Plugging in the values, we get: angle = arctan( (2 × 0 × 9.81) / (9.00^2) ) = 0 degrees.

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Since a light string is wrapped around the outer rim of a solid uniform cylinder, the mass of the cylinder is equal to 18.4 kilograms.

Given the following data:

To calculate the mass of the cylinder:

First of all, we would determine the acceleration of the stone by using the the third equation of motion;

[tex]V^2 = U^2 + 2aS[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]3.4^2 = 0^2 + 2a(2.40)\\\\11.56 = 0 + 4.8a\\\\11.56 = 4.8a\\\\a = \frac{11.56}{4.8}[/tex]

Acceleration, a = 2.41 [tex]m/s^2[/tex]

Next, we would solve for the torque by using the formula:

[tex]T = m(g - a)\\\\T = 3(9.8 - 2.41)\\\\T = 3 \times 7.39[/tex]

Torque, T = 22.17 Newton.

In rotational motion, torque is given by the formula:

[tex]T = I\alpha \\\\Tr = \frac{Mr^2}{2} \times \frac{a}{r} \\\\Tr = \frac{Mra}{2}\\\\2Tr = Mra\\\\M = \frac{2T}{a}\\\\M = \frac{2\times 22.17}{2.41}\\\\M = \frac{44.34}{2.41}[/tex]

Mass = 18.4 kilograms

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The system described is based on rotational dynamics and energy conservation principles in physics. The initial potential energy of the stone is converted into kinetic energy, and the stone's tensional force is equated with the cylinder's net torque. By systematically resolving these equations, we find the mass of the cylinder to be about 66.16 kg.

The subject of your question pertains to rotational dynamics in Physics. In this particular system, there are two main things to consider. First, we need to acknowledge the law of conservation of energy. The entire mechanical energy of the system would remain constant because there is neither any frictional force present nor any external force doing work. Second, we have to consider the moment of inertia for the cylinder.

For the stone, the initial potential energy (mgh) is converted into kinetic energy (0.5*m*v^2). So, 3*9.81*2.4 = 0.5*3*(3.4)^2, which solves to be accurate.

The tension in the string is equal to the force exerted by the stone, so T = m*g = 3*9.81 = 29.43 N.

As the string unwinds, the cylinder spins faster so, we equate the tensional force with the net torque, τ_net = I*α. Here, α = tangential acceleration (which equals g) divided by radius, hence, α = g/radius.

By simplifying these equations, we find that the mass of the cylinder is approximately 66.16 kg.

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Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.  You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain  the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

Objects fall at the same rate due to gravity, both cargo and plane speed up after cargo falls, air resistance affects falling speeds. Correct option is c. both the cargo and the plane speed up.

Gravity causes all objects to fall towards Earth at the same rate in the absence of air resistance.

When a heavy piece of cargo falls out of a coasting plane, both the cargo and the plane speed up due to the conservation of momentum.

Air resistance can cause lighter objects to fall slower than heavier objects due to the opposing force it exerts on objects moving through the air.

a. The value of vₓ = 33.77 km/s

b. The value of Vy = 81.05 km/s

The magnetic force, F on the proton moving in the uniform magnetic field is given by

F = qv × B where

Re-writing the force in matrix form, we have

[tex]\left[\begin{array}{ccc}F_{x} &F_{y} &F_{z} \\\end{array}\right] = q\left[\begin{array}{ccc}i&j&k\\v_{x} &v_{y}&v_{z}\\B_{x}&B_{y}&B_{z}\end{array}\right][/tex]

Taking the determinant, we have

[tex]F_{x}i + F_{y}j + F_{z}k = q[(v_{y}B_{z} - v_{z}B_{y})]i + q[(v_{x}B_{z} - v_{z}B_{x})]j + q[(v_{x}B_{y} - v_{y}B_{x})]k[/tex]

Equating the components of the force, we have

[tex]F_{x} = q[(v_{y}B_{z} - v_{z}B_{y})] (1)\\F_{y} = q[(v_{x}B_{z} - v_{z}B_{x})] (2)\\F_{z} = q[(v_{x}B_{y} - v_{y}B_{x})] (3)[/tex]

[tex]F_{x}/q = [(v_{y}B_{z} - v_{z}B_{y})] (4)\\F_{y}/q = [(v_{x}B_{z} - v_{z}B_{x})] (5)\\F_{z}/q = [(v_{x}B_{y} - v_{y}B_{x})] (6)[/tex]

Since F = (3.82 × 10⁻¹⁷ N)l + (1.59 × 10⁻¹⁷ N)j + 0k, equation (5) and (6) become

[tex]1.59 X 10^{-17} /1.602 X 10^{-19} = [(v_{x}B_{z} - v_{z}B_{x})] (5)\\993 = [(v_{x}B_{z} - v_{z}B_{x})] (7)\\Also\\0/q = [(v_{x}B_{y} - v_{y}B_{x})] (6)\\0 = (v_{x}B_{y} - v_{y}B_{x}) \\v_{x}B_{y} = v_{y}B_{x}\\v_{y} = \frac{v_{x}B_{y}}{B_{x}} (8)[/tex]

The value of vₓ = 33.77 km/s

Since [tex]B_{x}[/tex] = 10 × 10 ⁻³ T, [tex]B_{z}[/tex] = 30 × 10 ⁻³ T and [tex]v_{z}[/tex] = 2000 m/s, substituting the values of the variables into equation (7), we have

[tex]v_{x}B_{z} - v_{z}B_{x} = 993 (7)[/tex]

vₓ(30 × 10 ⁻³ T) - 2000 m/s × 10 × 10 ⁻³ T = 993 N

vₓ(30 × 10 ⁻³ T) - 20 m/sT = 993 N

vₓ(30 × 10 ⁻³ T) = 993 N + 20 m/sT

vₓ(30 × 10 ⁻³ T) = 1013 N

vₓ = 1013 N/(30 × 10 ⁻³ T)

vₓ = 33.77 × 10³ m/s

vₓ = 33.77 km/s

So, the value of vₓ = 33.77 km/s

The value of Vy = 81.05 km/s

Since [tex]B_{x}[/tex] = 10 × 10 ⁻³ T, [tex]B_{y}[/tex] = 24.0 × 10 ⁻³ T and [tex]v_{x}[/tex] = 33.77 × 10³ m/s, substituting the values of the variables into equation (8), we have

[tex]v_{y} = \frac{v_{x}B_{y}}{B_{x}} (8)[/tex]

Vy = 33.77 × 10³ m/s × 24.0 × 10 ⁻³ T/10 × 10 ⁻³ T

Vy = 810.48 m/sT/10 × 10 ⁻³ T

Vy = 81.048 × 10³  m/s

Vy ≅ 81.05 km/s

So, the value of Vy = 81.05 km/s

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To solve the problem it is necessary to apply the concepts related to heat flow,

The heat flux can be defined as

[tex]\frac{dQ}{dt} = H = \frac{kA\Delta T}{d}[/tex]

Where,

k = Thermal conductivity

A = Area of cross-sectional area

d = Length of the rod

[tex]\Delta T=[/tex] Temperature difference between the ends of the rod

[tex]k =388 W/m.\°C[/tex] Thermal conductivity of copper rod

[tex]A = 3.6 *10^{-4} m[/tex] Area of cross section of rod

[tex]\Delta T=100-0=100\°C[/tex] Temperature difference  

[tex]d=1.3m[/tex] length of rod

Replacing then,

[tex]H = \frac{kA\Delta T}{d}[/tex]

[tex]H = \frac{(388)(3.6 *10^{-4})(100)}{1.3}[/tex]

[tex]H=10.7446J[/tex]

From the definition of heat flow we know that this is also equivalent

[tex]H = \dot{m}*L[/tex]

Where,

[tex]\dot{m} =[/tex] Mass per second

[tex]L = 334J/g[/tex] Latent heat of fusion of ice

Re-arrange to find [tex]\dot{m},[/tex]

[tex]H = \dot{m}*L[/tex]

[tex]\dot{m}=\frac{L}{H}[/tex]

[tex]\dot{m}=\frac{334}{10.7446}[/tex]

[tex]\dot{m} = 31.08g/s[/tex]

[tex]\dot{m}= 0.032g/s[/tex]

Therefore the mass of ice per second that melts is 0.032g

Answer:

[tex]x= 45.46[/tex]

Explanation:

given,

mean of hospital waiting (μ) = 38.12  min

standard deviation (σ) = 8.63 min

worst waiting time = 20%  

the are will be same as the = 100 - 20%

                                             = 80%

we have to determine the z- value according to 80% or 0.80

using z-table

     z-value = 0.85

now, using formula

       [tex]Z = \dfrac{x-\mu}{\sigma}[/tex]

       [tex]0.85 = \dfrac{x-38.12}{8.63}[/tex]

       [tex]x-38.12 = 0.85\times {8.63}[/tex]

       [tex]x= 45.46[/tex]

waiting times follow a normal distribution with

Mean, \mu=38.12Mean,μ=38.12

Standard\ deviation,\sigma=8.63Standard deviation,σ=8.63

Longer waiting times are worse than shorter waiting times. Hence the worst 20% of wait times are wait times on the right tail of the distribution. The inferred level of confidence is 0.80.

The z value corresponding to the right tail probability of 0.2 is

Z=0.85Z=0.85

But

Z = \frac{x-\mu}{\sigma}Z=

σ

x−μ

​

x =Z*\sigma +\mux=Z∗σ+μ

=0.85 * 8.63 +38.12 =45.4555=0.85∗8.63+38.12=45.4555

answer:

the shortest wait time that would still be in the worst 20% of wait times is 45.4555 minutes

Answer:

253.65259 seconds

Explanation:

f' = Approaching frequency = 460 Hz

f = Receding frequency = 410 Hz

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

v' = Speed of truck

Doppler effect

[tex]\frac{f'}{f}=\frac{v+v'}{v-v'}\\\Rightarrow 460\left(343-v'\right)=410\left(343+v'\right)\\\Rightarrow 157780-460v'-157780=140630+410v'-157780\\\Rightarrow v'=19.712\ m/s[/tex]

The distance is 5 km

Time = Distance / Speed

[tex]Time=\frac{5000}{19.712}=253.65259\ s[/tex]

Time will it take to hear the jet from your position is 253.65259 seconds

Answer:

The flow rate is [tex]9.7\times 10^{- 11}\ m^{3}/s[/tex]

Solution:

As per the question:

Pressure drop,  = 1.45 kPa = 1450 Pa

Radius of the artery, R = [tex]2.50\times 10^{- 5}\ m[/tex]

length of the artery, L = [tex]1.10\times 10^{- 3}\ m[/tex]

Temperature, T = [tex]37^{\circ}C[/tex]

Viscosity, [tex]\eta = 2.084\times 10^{- 3}\ Pa.s[/tex]

Now,

The flow rate is given by:

[tex]Q = \frac{\pi R^{4}P}{8\eta L}[/tex]

[tex]Q = \frac{\pi (2.50\times 10^{- 5})^{4}\times 1450}{8\times 2.084\times 10^{- 3}\times 1.10\times 10^{- 3}} = 9.7\times 10^{- 11}\ m^{3}/s[/tex]

The flow rate through the artery can be calculated using Poiseuille's law, which takes into account the pressure difference, radius, length, and viscosity of the fluid. Using the given values, the flow rate is approximately 0.00686 m^3/s.

The flow rate through an artery can be calculated using Poiseuille's law, which states that the flow rate is directly proportional to the pressure difference and the fourth power of the radius, and inversely proportional to the length and viscosity of the fluid.

Using the given values, we can calculate the flow rate as follows:

Flow rate = (pressure difference * pi * radius^4) / (8 * viscosity * length)

Substituting the given values into the equation gives:

Flow rate = (1.45 * 10^3 * 3.14 * (2.5 * 10^-5)^4) / (8 * 2.084 * 10^-3 * 1.10 * 10^-3) = 0.00686 m^3/s

Therefore, the flow rate through the artery is approximately 0.00686 m^3/s.

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To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.

By definition we know that the relative velocity of an object with reference to the Light, is defined by

[tex]V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}[/tex]

Where,

V = Speed from relative point

c = Speed of light

On the other hand we have that the linear momentum is defined as

P = mv

Replacing the relative velocity equation here we have to

[tex]P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}[/tex]

[tex]P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}[/tex]

[tex]P^2 = \frac{P^2V^2}{c^2}+m^2V^2[/tex]

[tex]P^2 = V^2 (\frac{P^2}{c^2}+m^2)[/tex]

[tex]V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}[/tex]

[tex]V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}[/tex]

[tex]V = 2.81784*10^8m/s[/tex]

Therefore the height with respect the observer is

[tex]l = l_0*\sqrt{1-\frac{V^2}{c^2}}[/tex]

[tex]l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}[/tex]

[tex]l = 0.56m[/tex]

Therefore the height which the observerd measure for her is 0.56m

Final answer:

To estimate the mechanical power in watts for a top climber, you can use the formula for power. The work done in climbing the stairs can be calculated as the product of the force applied and the distance covered. The joules of chemical energy expended during the stair climb can be calculated using the formula and converted to kilocalories.

Explanation:

To estimate the mechanical power in watts for a top climber, we can use the formula for power: P = W/t, where P is power, W is work, and t is time. The work done in climbing the stairs can be calculated as the product of the force applied and the distance covered. The force applied can be calculated as the product of the mass of the climber (80 kg) and the acceleration due to gravity (9.8 m/s²). The distance covered can be calculated as the product of the number of steps (2232) and the height of each step (0.18 m). Using these values, we can calculate the work done and then divide it by the time taken (20 min converted to seconds) to get the power in watts.

To calculate the joules of chemical energy expended during the stair climb, we can use the formula: E = P × t, where E is energy, P is power, and t is time. Using the power value calculated in part (a), and the time taken for the stair climb, we can calculate the energy expended in joules. To convert joules to kilocalories, we can divide the energy value in joules by the conversion factor of 4186 joules per kilocalorie.

Answer:

Statements A & B are true.

Explanation:

Heat is converted completely into work during isothermal expansion.

Work done in an isothermal process is given as:

[tex]W=n.R.T.ln(\frac{V_f}{V_i} )[/tex]

where the subscripts denote final and initial conditions.

In case of an ideal gas the change in internal energy is proportional to the change in temperature. Here the temperature is constant in the process, therefore ΔU=0.

From the first law of thermodynamics:

[tex]dW=dQ+dU[/tex]

for isothermal process it becomes

[tex]dW=dQ[/tex]

Isothermal expansion is reversible under ideal conditions.

Ideally there is no friction involved as all the heat is converted into work, so  the process is reversible.

During the process of isothermal expansion, the gas does more work than during an isobaric expansion (at constant pressure) between the same initial and final volumes.

As we know that the work done is given by the area under the P-V curve so in case of the work done between two specific states of volume by an isothermal process we have only one path and by isobaric process we have 2 paths among which one is having the higer work done than the isothermal process nd the other one is having the lesser area under the curve than that of under isothermal curve.

Answer:

upward lift on an aircraft wing decreases as it gains altitude.

Explanation:

The governing equation of the Bernoulli's Principle is:

[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} +z=constant[/tex]

where:

P = pressure of the fluid

g = acceleration due to gravity

[tex]\rho=[/tex] density of fluid

v = velocity of the fluid

z = height of fluid from the datum

But the lift force on the wings depends upon several aerodynamic factors given mathematically as:

[tex]L=cl. \rho.A.\frac{v^2}{2}[/tex]

where:

cl = experimental constant

[tex]\rho=[/tex] density of air

A = area of wing

v = velocity of the air

As we move up in the atmosphere the density of air reduces and thus the force of lift will eventually decrease, that is the reason why airplanes have a flight ceiling, an altitude above which it cannot fly.

Final answer:

The average force exerted on the tennis ball by Venus Williams' racquet during her serve is calculated using the formula F = Δp/Δt, resulting in approximately 660 N, with the correct answer being option a).

Explanation:

The question relates to determining the average force exerted by Venus Williams' racquet on a tennis ball during her serve at the 2007 French Open. To find the average force, we can use the formula derived from Newton's second law of motion, F = ma. However, instead of finding the acceleration first, we use the formula F = Δp/Δt (force is the change in momentum over the change in time), because we know the velocity after impact and the time of contact.

The initial momentum (pi) is 0, as the initial velocity is negligible, so the final momentum (pf) is mv, where m is the mass of the tennis ball (0.057 kg) and v is the final velocity (58 m/s). As the time of contact with the racquet is 5.0 ms (or 0.005 seconds), we calculate the force as follows:

F = (m × v)/Δt = (0.057 kg × 58 m/s)/0.005 s = (3.306 kg·m/s)/0.005 s = 661.2 N

Therefore, the correct answer is approximately 660 N, which matches option a).

Answer:

A. an increase in the height of the sound wave

Explanation:

Volume of the sound is measured by its intensity

So here we know that

intensity is directly depends on the square of the amplitude of the sound wave

so here as intensity depends on the loudness of the sound given as

[tex]L = 10 Log(\frac{I}{I_o})[/tex]

so here as the loudness of sound will increase then the intensity will increase and hence the amplitude of the sound will also increase

So correct answer will be

A. an increase in the height of the sound wave

To solve this problem it is necessary to apply the concepts related to the wave function of a particle and the probability of finding the particle in the ground state.

The wave function is given as

[tex]\phi = \sqrt{\frac{2}{L}} sin(\frac{\pi x}{L})[/tex]

Therefore the probability of finding the particle must be

[tex]P = |\phi(x)^2| \Delta X[/tex]

[tex]P = |\sqrt{\frac{2}{L}} sin(\frac{\pi x}{L})|^2 \Delta x[/tex]

[tex]P = (\frac{2}{L})(sin(\frac{\pi x}{L}))^2\Delta x[/tex]

[tex]P = (\frac{2}{L})(sin(\frac{\pi (0.7L)}{L}))^2 (0.0002L)[/tex]

[tex]P = 2*(sin(0.7\pi))^2(0.0002)[/tex]

[tex]P = 2.618*10^{-4}[/tex]

Therefore the probability is [tex]2.618*10^{-4}[/tex]

The probability that the particle is in a window [tex]\(\Delta x = 0.0002L\)[/tex]wide with its midpoint at [tex]\(x = 0.700L\)[/tex] is given by [tex]\(P = |\psi(0.700L)|^2 \Delta x\).[/tex]

In quantum mechanics, the probability of finding a particle in a certain position is given by the square of the wave function's magnitude at that position, multiplied by the width of the position interval. For a particle in the ground state, the wave function \(\psi(x)\) is given by the solution to the Schrödinger equation for the particular potential well the particle is in. For a simple one-dimensional infinite potential well, the ground state wave function is a half-sine wave.

Given that the particle is in the ground state and the window[tex]\(\Delta x\)[/tex] is very small, we can make the approximation that the wave function \(\psi(x)\) does not change significantly over this interval.

 Since we are assuming [tex]\(\psi(x)\[/tex]is constant over [tex]\(\Delta x\)[/tex], the integral simplifies to the product of [tex]\(|\psi(0.700L)|^2\)[/tex] and the width of the interval [tex]\(\Delta x\):[/tex]

[tex]\[ P = \int_{0.700L - \frac{\Delta x}{2}}^{0.700L + \frac{\Delta x}{2}} |\psi(x)|^2 dx \approx |\psi(0.700L)|^2 \Delta x \][/tex]

 This is a standard result for the probability of finding a particle in a small interval around a point in quantum mechanics when the wave function is approximately constant over that interval. The actual value of [tex]\(|\psi(0.700L)|^2\)[/tex] would depend on the specific form of the wave function for the ground state of the system in question.

However, the integral does not need to be evaluated explicitly due to the assumption of a constant wave function over the small interval [tex]\(\Delta x\).[/tex]