two pulleys one with adius 2 inches and the other with radius 8 inches are connected by a belt. If the 2 inch pulley is caused to rotate at 3 revolutions per minuite detirmine te revolutions per minute of the 8 inch pulley
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.
In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 1.00 m and while being driven into rotation around a fixed axis, its angular position is expressed as
θ = 2.50t2 - 0.600t3
where θ is in radians and t is in seconds.
a) Find the maximum angular speed of the roller
b) what is the maximum tangential speed of the point an the rim of the roller?
c) at what time t should the driven force be removed from the roller so that the roller does not reverse its direction of rotation?
d) Through how many rotations has the roller turned between t=0 and the time found in part c?
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.
Explanation: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.
A straight 2.70 m wire carries a typical household current of 1.50 a (in one direction) at a location where the earth's magnetic field is 0.550 gauss from south to north.
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
What is the magnitude of the force that our planet's magnetic field exerts on this cord?
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
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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
What kind of wave is sound and how do the particles move?
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.
What are the longitudinal waves?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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Let two objects of equal mass m collide. Object 1 has initial velocity v, directed to the right, and object 2 is initially stationary.
A. If the collision is perfectly elastic, what are the final velocities v1 and v2 of objects 1 and 2?
Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express each velocity in terms of v.
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]
The temperature at which the motion of particles theoretically ceases is
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
n her abstract works Electric Dress and Untitled, Japanese artist Atsuko Tanaka created ________ through the repetition of lines and circular shapes in bold colors. a. scale b. unity c. balance d. emphasis e. focal point
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!
A 20.0 kg rock is sliding on a rough , horizontal surface at8.00 m/s and eventually stops due to friction .The coefficient ofkinetic fricction between the rock and the surface is 0.20
what average thermal power is produced as the rock stops?
Answer: The power is 156 watt
Explanation:
is in the attachment
Explanation:
Below is an attachment containing the solution.