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pochemuha
2 years ago
7

In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller i

s 1.00 m , and, while being driven into rotation around a fixed axis, its angular position is expressed as
θ =2.50t² - 0.600 t³
where θ is in radians and t is in seconds.(d) Through how many rotations has the roller turned between t=0 and the time found in part (c)?
Physics
1 answer:
kotykmax [81]2 years ago
4 0

The maximum angular speed of the roller is 3.47rad/a,  

the maximum tangential speed of the point an the rim of the roller is 1.47m/s,

the time t should the driven force be removed from the roller so that the roller does not reverse its direction of rotation is t = 2.78s,

5.4 rotation has the roller turned between t=0 and the time found in part c

<h3>What does tangential speed refer to?</h3>

Any item moving along a circular path has a linear component to its speed called tangential velocity. An object's velocity is always pointed tangentially when it travels in a circle at a distance r from the center. The word for this is tangential velocity.

To calculate the tangential speed, divide the circumference by the time required to complete one spin. For instance, if it takes 12 seconds to complete one rotation, divide 18.84 by 12 to get 1.57 feet per second as the tangential velocity.

(a) angular speed w = dθ/dt = 5t - 1.8t^2

dw/dt = 5 - 3.6t = 0 for max w

so max w occurs at t = 5/3.6 s = 1.39s

so w max = 5*1.39 - 1.8*(1.39)^2 = 3.47 rad/s

(b) tangential speed v = r*w

r = D/2 = 0.5m

so v = 0.5*w = 1.74 m/s

(c) w is positive until 5t = 1.8t^2

so t = 5/1.8s = 2.78s (or t = 0 invalid)

After t = 2.87s, w is negative (starts reversing direction of rotation)

Driving force would actually have to be removed some time before t=2.78s because the roller can't stop instantaneously, but insufficient info to calculate this.

(d) Up to t = 2.78s, θ = 2.5*(2.78)^2 - 0.6*(2.78)^3 rad = 33.95 rad = 5.40 rotations

Therefore,

A) 3.47rad/a

B) 1.47m/s

C) t = 2.78s

D) 5.4 rotation

To learn more about tangential speed, refer to:

brainly.com/question/19660334

#SPJ4

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A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of
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Answer:

4.5\times 10^{-5} T

Explanation:

We are given that

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We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

B_{wire}=B_2=\frac{\mu_0I}{2\pi R}

We have R=29 cm=\frac{29}{100}=0.29 m

1 m=100 cm

Substitute the values in the given formula

B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T

The resultant magnetic field is given by

B=\sqrt{B^2_1+B^2_2}

Substitute the values then we get

B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}

B=4.5\times 10^{-5} T

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire=4.5\times 10^{-5} T

Hence, the resultant magnitude of the magnetic field 29 cm above  the wire=4.5\times 10^{-5} T

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The uncertainty in the position of an electron along an x axis is given as 5 x 10-12 m. What is the least uncertainty in any sim
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The least uncertainty in the momentum component px is 1 × 10⁻²³ kg.m.s⁻¹.

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According to Heisenberg's uncertainty principle, the uncertainty in the position of an electron (σx) and the uncertainty in its linear momentum (σpx) are complementary variables and are related through the following expression.

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A steel piano wire, of length 1.150 m and mass 4.80 g is stretched under a tension of 580.0 N.
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A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be  372.77 m/s

<h3>What is a sound wave?</h3>

It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.

For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula

V= √F/μ

where v is the wave velocity of the wave travel on the string

F is the tension in the string of piano

μ is the mass per unit length of the string

As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.

The μ is the mass per unit length of the string would be

μ = 4.80/(1.150×1000)

μ = 0.0041739 kg/m

By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity

V= √F/μ

V=√(580/0.0041739)

V =  372.77 m/s

Thus,  the speed of transverse waves on the wire comes out to be  372.77 m/s

Learn more about sound waves from here

brainly.com/question/11797560

#SPJ1

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