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ElenaW [278]
3 years ago
15

A dockworker loading crates on a ship finds that a 25-kg crate, initially at rest on a horizontal surface, requires a 72-N horiz

ontal force to set it in motion. However, after the crate is in motion, a horizontal force of 55 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.
Physics
1 answer:
Gnom [1K]3 years ago
7 0

Answer:

Explanation:

Force of static friction = μs x mg

Given force of static friction = 72 N

μs x mg = 72

μs = 72 / mg

= 72 / 25 x 9.8

= .294

Force of kinetic  friction = μk x mg

Given force of kinetic  friction = 55

μk x mg = 55

μk = 55 / mg

= 55 / 25 x 9.8

= .2245

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elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer (\dot Q), measured in watts, that is, joules per second, by the following formula:

\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t} (1)

Where:

m - Mass of the sphere, measured in kilograms.

c - Specific heat of the material, measured in joules per kilogram-degree Celsius.

T_{o}, T_{f} - Initial and final temperatures of the sphere, measured in degrees Celsius.

\Delta t - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}} (2)

Where:

m_{I}, m_{X} - Masses of the iron and unknown spheres, measured in kilograms.

\Delta t_{I}, \Delta t_{X} - Times of the iron and unknown spheres, measured in seconds.

c_{I}, c_{X} - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}

If we know that \Delta t_{I} = 6.35\,s, \Delta t_{X} = 4.59\,s, m_{I} = 0.515\,kg, m_{X} = 1.263\,kg and c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}, then the specific heat of the unknown material is:

c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)

c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

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