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Bingel [31]
3 years ago
10

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

s 6.00 m, and, while being driven into rotation around a fixed axis, its angular position is expressed as

Physics
1 answer:
EastWind [94]3 years ago
3 0

Question:

In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 6.00 m, and, while being driven into rotation around a fixed axis, its angular position is expressed as  θ = 2.70t² − 0.900t3³

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 a point on the rim of the roller?

(c) At what time t should the driving 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)?

Answer:

a. 2.7rad/s

b. 8.1 m/s

c. 2 s

d. 3.6 rad

Explanation:

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3 0
3 years ago
True or Flase The fastest moving traffic on the expressway will be traveling in the right lane
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I think this answer is False because i believe the right lane is the slowest lane you could be in.

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What would be the distance moved if we had a 70 n force and work done is 8j
Lostsunrise [7]

Answer:

0.1143m

Explanation:

W=f×s

8=70s

make s the subject of the formula

s=8/70

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3 0
3 years ago
If 200 ml of tea at 90 °C is poured into a 400 g glass cup initially at 25 °C, what would be the final temperature of the mixtur
sweet-ann [11.9K]

The final temperature of the mixture is 43.62 °C

To solve the question above, we apply the law of calorimetry

The law of calorimetry: which states that if there is no lost of heat the surrounding, heat lost is equal to heat gained, or it can be stated as heat absorbed by a cold body is equal to heat released by a hot body, provided there is no lost of heat to the surrounding.

The law above is expressed mathematically as

Cm(t₁-t₃) = C'm'(t₃-t₂)............. Equation 1

Using equation 1 to solve the question,

Let: C = specific heat capacity of glass cup, m = mass of glass cup, C' = specific heat capacity of tea, m' = mass of tea, t₁ = initial temperature of tea, t₂ = initial temperature of glass cup, t₃ = final temperature of the mixture.

From the question,

Given: m = 400 g = 0.4 kg, C = 840 J/kg°C, m' = 200g (tea is a liquid made of water and the volume of water in ml is thesame a its mass in gram) = 0.2 kg, C' = 4186 J/kg.°C, t₁ = 90°C, t₂ = 25°C

Substitute these values into equation 1 and solve for t₃

0.4(840)(90-t₃) = 0.2(4186)(t₃-25)

336(90-t₃) = 837.2(t₃-25)

30240-336t₃ = 837.2t₃-20930

collect like terms

837.2t₃+336t₃ = 30240+20930

  1173.2t₃ = 51170

t₃ = 51170/1173.2

t₃ = 43.62 °C

Hence, the final temperature of the mixture is 43.62 °C

7 0
3 years ago
Read 2 more answers
if a mass on a spring bobs up nad down completing 2 full cycles every section. WHat are the periouds and frequency of the mass
Alika [10]

Answer:

This question is incomplete

Explanation:

This question is incomplete because of the absence of the time taken to complete one full cycle.

Frequency (<em>f</em>) will be calculated first as

<em>f </em>= <em>N </em>÷<em> t</em>

where <em>N </em>is the number of cycles and <em>t </em>is the time taken to complete one full cycle. The unit for frequency is Hertz (Hz).

To calculate the period, <em>T, </em>the formula below will be used

<em>T </em>= 1 ÷ <em>f</em>

The unit for period is secs

4 0
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