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4 years ago

12

Select the correct expression that gives the block's acceleration at a displacement x from the equilibrium position. Note that x

can be either positive or negative; that is, the block can be either to the right or left of its equilibrium position.
a. a=−kx
b. a=kx
c. a=kmx
d. a=−kmx

2 answers:

skad [1K]4 years ago

8 0

Answer:

d. a=−k/mx

Explanation:

To know what is the correct expression for the acceleration you take into account the second Newton law, that is:

$F=ma$ ( 1 )

next, you equal the expression ( 1 ) to the force in a mass-string system, that is F=-kx.

$ma=-kx\\\\a=-\frac{kx}{m}$

hence, the acceleration is:

d. a=−kmx

My name is Ann [436]4 years ago

5 0

Answer:

a = -kmx ( D )

Explanation:

Acceleration = velocity / time

velocity = displacement / time

but since the block is at equilibrium and X which is displacement can be either negative or positive

applying newton law of motion

F = ma

considering the force acting on the object

ma = - kx

hence a = -kx/m

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Read 2 more answers

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

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Answer:

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

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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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