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Consider the system shown in the figure. Block A has weight 4.91 N and block B has weight 2.94 N. Once block B is set into downward motion, it descends at a constant speed. Assume that the mass and friction of the pulley are negligible.

Calculate the coefficient of kinetic friction mu between block A and the table top.

I'm not sure where to begin.

Answer:

[tex]f=81.96 \ Hz[/tex]

Explanation:

Givens

[tex]L=95cm[/tex]

[tex]m_{sculpture} =13kg[/tex]

[tex]m_{wire}=5g[/tex]

The frequency is defined by

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

Where [tex]v[/tex] is the speed of the wave in the string and [tex]\lambda[/tex] is its wave length.

The wave length is defined as [tex]\lambda = 2L = 2(0.95m)=1.9m[/tex]

Now, to find the speed, we need the tension of the wire and its linear mass density

[tex]v=\sqrt{\frac{T}{\mu} }[/tex]

Where [tex]\mu=\frac{0.005kg}{0.95m}= 5.26 \times 10^{-3}[/tex] and the tension is defined as [tex]T=m_{sculpture} g=13kg(9.81 m/s^{2} )=127.53N[/tex]

Replacing this value, the speed is

[tex]v=\sqrt{\frac{127.53N}{5.26 \times 10^{-3} } }=155.71 m/s[/tex]

Then, we replace the speed and the wave length in the first equation

[tex]f=\frac{v}{\lambda}\\f=\frac{155.71 m/s}{1.9m}\\ f=81.96Hz[/tex]

Therefore, the frequency is [tex]f=81.96 \ Hz[/tex]

Explanation:

Below is an attachment containing the solution.

Answer: The power is 156 watt

Explanation:

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Explanation:

Below is an attachment containing the solution.

a. The maximum angular speed of the roller is approximately **7.17 rad/s**.

b. The maximum tangential speed of a point on the rim of the roller is approximately **3.59 m/s**.

c. The driven force should be removed at **t ≈ 2.78 seconds** to avoid reversal of the roller's direction.

d. The roller has completed **approximately 1.75 rotations** between t=0 and the time found in part c (2.78 seconds).

Analyzing the Rotating Roller:

**a. Maximum Angular Speed:**

To find the maximum angular speed, we need to determine the maximum value of the angular velocity (ω). Angular velocity (ω) is the rate of change of angular position (θ) and is given by:

ω = dθ/dt = 5t - 1.8t^2

The maximum value of ω will occur at the extremum point (critical point) of this function. We can find this by setting the derivative of ω (dω/dt) equal to zero:

dω/dt = 5 - 3.6t = 0

t = 5/3 ≈ 1.67 seconds

Now, plug this t back into the original equation for ω:

ω_max = 5 * (5/3) - 1.8 * (5/3)^2 ≈ 7.17 rad/s

Therefore, the maximum angular speed of the roller is approximately **7.17 rad/s**.

**b. Maximum Tangential Speed:**

The tangential speed (v_t) of a point on the rim of the roller is related to its angular speed (ω) and radius (r) by:

v_t = ω * r

Since the maximum angular speed was found in part (a), we can calculate the maximum tangential speed using the radius (0.5 m):

v_t_max = 7.17 rad/s * 0.5 m ≈ 3.59 m/s

Therefore, the maximum tangential speed of a point on the rim of the roller is approximately **3.59 m/s**.

**c. Time to Remove Driven Force:**

To prevent the roller from reversing its direction, we need to find the time at which its angular velocity drops to zero. So, we set ω = 0 in the original equation and solve for t:

5t - 1.8t^2 = 0

Factor the equation: t(5 - 1.8t) = 0

Hence, t = 0 or t = 5/1.8 ≈ 2.78 seconds

Therefore, the driven force should be removed at **t ≈ 2.78 seconds** to avoid reversal of the roller's direction.

**d. Number of Rotations:**

The number of rotations completed by the roller can be calculated by integrating the angular velocity (ω) over the time interval (t=0 to t=2.78):

θ = ∫ω dt = ∫(5t - 1.8t^2) dt

Solving the integral gives:

θ = 2.5t^2 - 0.6t^3

At t = 0, the roller is at its initial position (θ = 0). At t = 2.78, we can plug the value into the equation to find the total angular displacement:

θ = 2.5 * (2.78)^2 - 0.6 * (2.78)^3 ≈ 11.03 radians

Since one complete rotation equals 2π radians, the number of rotations (N) is:

N = θ / 2π ≈ 11.03 radians / 2π radians/rotation ≈ 1.75 rotations

Therefore, the roller has completed **approximately 1.75 rotations** between t=0 and the time found in part c (2.78 seconds).

a) The maximum angular speed of the roller is[tex]\( 5.00 \, \text{rad/s} \).[/tex]

b) The maximum tangential speed of the point on the rim of the roller is [tex]\( 20.0 \, \text{m/s} \).[/tex]

c) The driven force should be removed from the roller at [tex]\( t = 2.50 \, \text{s} \).[/tex]

d) The roller has turned through [tex]\( 9.38 \)[/tex]rotations between [tex]\( t = 0 \)[/tex] and [tex]\( t = 2.50 \, \text{s} \).[/tex]

Therefore, the correct answer is all of these.

The maximum angular speed of the roller is obtained by taking the first derivative of the angular position equation with respect to time [tex](\( \theta = 2.50t^2 - 0.600t^3 \))[/tex], resulting in [tex]\( \omega = 5.00 \, \text{rad/s} \)[/tex]. To find the maximum tangential speed, we use the formula [tex]\( v = r \cdot \omega \),[/tex]where [tex]\( r \)[/tex] is the radius of the roller. Given the diameter of the roller is [tex]\( 1.00 \, \text{m} \),[/tex] the radius is [tex]\( 0.50 \, \text{m} \),[/tex] leading to a maximum tangential speed of[tex]\( 20.0 \, \text{m/s} \).[/tex]

The removal of the driven force occurs when the angular speed is at its maximum, which corresponds to [tex]\( t = 2.50 \, \text{s} \).[/tex] At this point, the roller is at its peak rotation speed, and removing the force ensures it does not change direction. To calculate the number of rotations, we integrate the angular speed equation over the given time interval, resulting in[tex]\( 9.38 \)[/tex]rotations.

In conclusion, understanding the dynamics of the roller's angular position and speed allows us to determine critical points in the manufacturing process.

Therefore, the correct answer is all of these.

Answer:

a. Shorter

b. Same mass

Explanation:

a.- From the question, lo has an orbit size of 0.002819 AU while Europa has an orbit size of  0.004473 AU.

-It is evident from this information that Europa's orbit size is greater than that of lo.

-Hence, Lo's orbital period is shorter than Europa's.

b. The mass of Jupiter's moons in their orbit around Jupiter  depends upon the mass of Jupiter and the inverse square of the moon's distance from Jupiter.

-As such, any object around Jupiter can be used to determine Jupiter's mass and results in the same mass.

Final answer:

Using Kepler's third law of planetary motion, Io's orbital period is shorter than Europa's because it is closer to Jupiter. Calculating Jupiter's mass using Io's or Europa's orbit should yield the same value for Jupiter's mass.

Explanation:

The question pertains to the orbital dynamics of Jupiter's moons, Io and Europa, and the calculation of Jupiter's mass based on their orbits. According to Kepler's third law of planetary motion, a moon's orbital period is related to its distance from the planet. Since Io has a smaller semi-major axis compared to Europa, Io's orbital period is shorter than Europa's. Kepler's third law states that the square of the orbital period (P) of a moon is directly proportional to the cube of its semi-major axis (a) when referring to orbits around the same central body. Therefore, using Kepler's third law and comparing the semi-major axes of Io and Europa, one can derive the masses of Jupiter. However, since both moons obey the same law, calculating Jupiter's mass using either moon's orbit should theoretically give the same result.

Explanation:

we know that,

linear speed = circumference × revolution per minute

linear speed of belt = 2πr × revolution per minute

now we will compute the linear speed of a belt for 2 inch pulley that is,

linear speed for 2 inch pulley = (2π×2)×( 3 revolutions per minute)  ∵ r =2

                                              = 4π × 3 revolution per minute    (1)

again we will compute the linear speed of a belt for 8 inch pulley,

linear speed of 8 inch pulley = (2π×8) × (x revolution per minute) ∵ r =8

                                            = 16π×x revolutions per minute      (2)

As the linear speed is same for both pulleys. by comparing equations (1) and (2).

                           4π×3 = 16π×x

                              x = 3/4

Thus, the revolutions per minute for the 8 inch pulley is 3/4.

Answer:

(v, 0) or (0, v)

Explanation:

The law of conservation of linear momentum states total initial momentum equal total final momentum.

Total initial momentum = [tex]mv + m\times 0 = mv[/tex]

Total final momentum = [tex]mv_1 + mv_2 = m(v_1+v_2)[/tex]

Equating both,

[tex]mv = m(v_1+v_2)[/tex]

[tex]v = v_1+v_2[/tex]

For an elastic collision, kinetic energy is conserved i.e. total initial kinetic energy = total final kinetic energy

[tex]\frac{1}{2}mv^2 + \frac{1}{2}m\times0^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2[/tex]

[tex]v^2 = v_1^2+v_2^2[/tex]

From the first equation, [tex]v_1 = v-v_2[/tex].

Substituting this in the second equation,

[tex]v^2 = (v-v_2)^2+v_2^2[/tex]

[tex]v^2 = v^2-2vv_2 +v_2^2+v_2^2[/tex]

[tex]0 = 2v_2^2-2vv_2[/tex]

[tex]v_2(v_2-v) = 0[/tex]

[tex]v_2 = 0[/tex] or [tex]v_2 = v[/tex]

From [tex]v_1 = v-v_2[/tex],

[tex]v_1 = v-v = 0[/tex]

OR

[tex]v_1 = v-0 = v[/tex]

Answer:

Sound waves are classified as Longitudinal waves. This means they move longitude, or sideways. A good way to think of it would be like a slinky moving sideways.

Sound waves are longitudinal waves because particles of the medium through which the sound waves is transported vibrate parallel to the direction of propagation.

Longitudinal waves can be described as the waves in which the vibration of the medium is parallel to the direction the wave travels and displacement of the medium is in the same direction of propagation.

Mechanical longitudinal waves are also known as compression waves as they form compression and rarefaction when traveling through a medium, and pressure waves. Sound waves and seismic P-waves are the examples of longitudinal waves.

For sound waves, the amplitude of the wave can be described as the difference between the pressure of the undisturbed air and the maximum pressure lead by the wave.

The propagation of sound speed depends on the type, temperature, and composition of the medium through which wave propagates.

Learn more about longitudinal waves, here:

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Answer:

a) F = 2.2275 * 10^-4 N , upwards

b) F = 2.2275 * 10^-4 N , east to west

c) F = 0

Explanation:

Given:

- The straight wire of length  L = 2.7 m

- The current I in the wire I = 1.50 A

- The magnetic Field B = 0.55 * 10^-4 T

Find:

Find the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if it is oriented so that the current in it is running

a) from west to east

b) vertically upward

c) from north to south

d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

Solution:

- If current runs from west to east the angle between the magnetic field B is θ = 90°, so the magnitude of force due to magnetic field is given by Lorentz force.

                           F = B*I*L*sin(θ)

                           F = 0.55*1.5*2.7*sin(90)*10^-4

                           F = 2.2275 * 10^-4 N

- From Figure B points north and current I points east. From right hand rule, the direction of force is out of page, so its upward.

- If current runs upward the angle between the magnetic field B is θ = 90°, so the magnitude of force due to magnetic field is given by Lorentz force.

                           F = B*I*L*sin(θ)

                           F = 0.55*1.5*2.7*sin(90)*10^-4

                           F = 2.2275 * 10^-4 N

- From Figure B points north and current I points out of page . From right hand rule, the direction of force is out of page, so its east to west.

- If current runs north to south the angle between the magnetic field B is θ = 0°, so the magnitude of force due to magnetic field is given by Lorentz force.

                           F = B*I*L*sin(θ)

                           F = 0.55*1.5*2.7*sin(0)*10^-4

                           F = 0 N

The magnitude of the force that our planet's magnetic field exerts is mathematically given as

F=1.82* 10^{-4}N

Question Parameter(s):

A straight 2.70 m wire carries a typical household current of 1.50

the earth's magnetic field is 0.550 gauss

Generally, the equation for the  force exerted by the magnetic field   is mathematically given as

F=ILB

Where

B=0.550 G

B= 0.55 *10^{-4}T

In conclusion,force exerted by the magnetic field is

F=(1.50 A)(2.20 m)(0.55* 10^{-4} T)

F=1.82* 10^{-4}N

Read more about Force

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Complete Question

A straight 2.70 m wire carries a typical household current of 1.50 an (in one direction) at a location where the earth's magnetic field is 0.550 gauss from south to north.

Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east

Answer:

b. unity

Explanation:

In the images I added you can observe Atsuko Tanak's works, in both of them you can observe how the sum of colorful and repetitive elements creates the sensation of unity.

I hope you find this information useful and interesting! Good luck!

Answer:

Explanation:

Given that

x m/s is equivalent to y km/hr

We know that 1000m=1km

Also, 1hour=60mins

And, 1mins=60sec

Therefore,

xm/s=y (km/hr) × (1000m/1km) × (1hr/60mins) × (1mins/60s)

Then, the km cancels out, also the hours cancels out and the minutes cancel out

Then, we see left with

xm/s=y × 1000m/3600s

xm/s=(1000/3600)y m/s

xm/s=5y/18 m/s

Then, make y the subject of the formula

Cross multiply

18xm/s=5y

Divide both side by 5

y=18x/5

The correct answer is C

Answer:

0 K or -273.15°C

Explanation:

Temperature is a measure of the kinetic energy of the molecules of a body. At the absolute zero or 0 K temperature, the kinetic energy is 0 J. Kinetic energy is a measure of the velocity of the molecules, hence, theoretically, the molecules stop moving.

Answer:

0K or -273°C

Explanation:

The temperature at which the motion of  particles ceases is 0K or -273°C. At  this temperature the motion of particles is theatrically at rest. This temperature was explained in Charles law

Charles' Law: Charles' Law states that the volume of a given mass of gas is directly proportional to it absolute temperature.

Charles law can be expressed mathematically as

V/T = V'/T'

Where V and V' = Are initial and Final volume respectively, T and T' = Initial and Final Temperature respectively