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Bezzdna [24]
4 years ago
13

A winch is a mechanical device that is used to adjust the tension in a rope or line. A weekend sailor works the manual winch to

trim the sail during an outing on the water. The radius of the winch drum is 4.0 cm and the man turns the winch at the rate of 2.9 complete revolutions every 1 second.
Physics
1 answer:
Andrei [34K]4 years ago
5 0

Answer:

. A weekend sailor works the manual winch to trim the sail during an outing on the water. The radius of the winch drum is 4.0 cm and the man turns the winch at the rate of 2.9 complete revolutions every 1 second.

(a) What is the tangential speed of the line as he brings it in? m/s

Update 2:

(b) How would the value of the tangential speed change if the radius of the wheel were tripled?

The tangential speed would increase by a factor of 3.

The tangential speed would decrease by a factor of 3.

The tangential speed would increase by a factor of 9.

The tangential speed would decrease by a factor of 9.

The answer to the questions are as follows

a. The tangential speed = 0.729 m/s

b. The tangential speed would increase by a factor of 3.

Explanation:

a. The appropriate relations to help us find the tangential speed are

w = 2•pi•N and

w = v/r,

Where r = radius = 4.0 cm

v = velocity

N = rotational speed

From the question, N = 2.9 rotations/s or 2.9 Hz

Therefore w = 2•pi•2.9 = 18.22 r/s

Therefore,

v = w × r = (18.22 r/s) × (4.0 cm) = (18.22 r/s) × (0.040 m) = 0.729 m/s

The tangential speed = 0.729 m/s

b. Tripling the radius of the wheel we have, new radius = 4 × 3 = 12 cm

v = w × r = 18.22×0.12 = 2.1865 m/s

Finding the ratio of the two velocities, we have

(2.1865 m/s)/(0.729 m/s) = 3

Hence the tangential speed would increase by a factor of 3

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Tpy6a [65]

Answer:

The value is  F  =  34.3 \  N

Explanation:

From the question we are told that

    The mass of the shed is  m  =  25 \  kg

    The length of the rope is  l = 5.0 \  m

    The angular speed of the shed is w = 5 rev/minute = \frac{2* \pi * 5}{60 }= 0.524 \  rad/sec

Generally the force exerted on the shed is mathematically represented as

     F  =  m  *  w^2 *  l

=>  F  =  25  *  0.524^2 *  5

=>  F  =  34.3 \  N    

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4 years ago
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You could get sick by breathing throw your mouth and you have a less chance of getting sick by breathing throw your nose.
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A common laboratory reaction is the neutralization of an acid with a base. When 50.0 mL of 0.500 M HCl at 25.0°C is added to 50.
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Answer:

B

Explanation:

In this calorimetry problem, the heat released by the reaction is equal to the heat absorbed by the solution (assumed to have the same specific heat capacity as water, 4.19 Jg⁻¹°C⁻¹).

The formula Q = mcΔt will be used to calculate the heat energy, where m is the mass, c is the specific heat capacity, and Δt is the change in temperature from final to initial.

The volume of solution is (50.0 + 50.0)mL = 100.0mL = 100.0g, since water has a density of 1.00g/mL.

The heat absorbed by the solution is then calculated.

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The closest answer is B) 1300 J. This answer is obtained by including only two significant figures in the answer.

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3 years ago
8. A 20 newton object is pulled along a surface at a constant speed of 3.0 meters per second using a force of 4.5 newtons. The k
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To calculate the kinetic coefficient of friction is closest to: 1. 0.23.

<u>Given the following data:</u>

  • Horizontal force = 4.5 Newton
  • Normal force = 20 Newton
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To calculate the kinetic coefficient of friction:

Mathematically, the kinetic force is given by this formula;

F_k = uF_n

<u>Where;</u>

  • F_k represents the kinetic force.
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  • μ represents the kinetic coefficient of friction.

Making μ the subject of formula, we have:

u=\frac{F_k}{F_n}

Substituting the given parameters into the formula, we have;

u=\frac{4.5}{20}

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Read more on kinetic coefficient of friction here: brainly.com/question/13940648

6 0
3 years ago
A driver slammed on her brakes and came to a stop with constant acceleration. Measurements on her tires and skid marks on the pa
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Answer:

u = 29.22 m/s

Explanation:

distance (s) = 58.52 m

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final velocity (v) = 0 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

How fast was she driving (u)

we can get how fast she was driving by using the formula below

s = ut - \frac{1}{2}.at^{2}  ...equation 1

where

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  • u = her initial velocity
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  • k = coefficient of kinetic friction
  • g = acceleration due to gravity
  • t = time

        from v = u - at  (recall that v = 0)

        0 = u - at, therefore t = u/a = u/kg

now substituting the required values above into equation 1 we have

s = \frac{u^{2}}{kg} - \frac{u^{2}}{2kg}

s = \frac{u^{2}}{2kg}

u = \sqrt{2kgs}

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u = 29.22 m/s

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