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1 month ago

6

A student wants to determine the coefficient of static friction μ between a block of wood and an adjustable inclined plane. Of t

he following, the minimum additional equipment the student needs to determine a value for μ is

1 answer:

serg [3.5K]1 month ago

7 0

Response:

A protractor to gauge the angle between the inclined plane and the horizontal

Explanation:

The student must elevate the free end of the adjustable inclined plane until the object just begins to slide and record the angle at that precise moment. At this juncture, the frictional force is balanced by the weight component aligned with the incline. That is:

$f=\mu\,* N = \mu * m g\, cos(\theta)$

and $w_{//}= m\,g\,sin(\theta)$

Consequently, the coefficient of static friction can be entirely established by calculating the tangent of the angle formed by the incline with the horizontal.

$f = w_{//}\\\mu *\,m \,g\,cos(\theta) = m\,g\,sin(\theta)\\\mu = tan(\theta)$

For this, the sole additional tool needed is a protractor for angle measurement.

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In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t

ValentinkaMS [3465]

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

$T\cos\theta=mg$

$T=\dfrac{mg}{\cos\theta}$

Vertical component

$T\sin\theta=m\omega^2 r$

Substituting the tension value

$mg\tan\theta=m\omega^2(d+L\sin\theta)$

$\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}$

Substituting the value into the equation

$\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}$

$\omega=1.34\ rad/s$

Thus, the angular speed of rotation computes to 1.34 rad/s.

7 0

3 months ago

3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

serg [3582]

Response:

(a) $W=-19.25J$

(b) $W=-52.8J$

Clarification:

Greetings.

(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:

$W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J$

This value is negative as it expands against the given pressure.

(b) Furthermore, if the process is conducted reversibly, the pressure might change, hence, we need to calculate the work using:

$W=nRTln(\frac{V_1}{V_2} )$

The moles are calculated based on the provided mass of argon:

$n=6.56g*\frac{1mol}{39.95g}=0.164mol$

Consequently, the work amounts to:

$W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J$

Best regards.

4 0

1 month ago

When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (

inna [3103]

The answer is (C) 4 beats per second. The number of beats is computed as the difference between the frequencies of the two tuning forks. Plugging in the frequency values yields a result. Thus, the number of observable beats per second will be 4.

3 0

2 months ago

A beaker of negligible heat capacity contains 456 g of ice at -25.0°C. A lab technician begins to supply heat to the container a

Maru [3345]

The response is 176 minutes. The translation of 456 g equals 0.456 kg. The specific heat of ice is 2093 J kg⁻¹, used to calculate heat required for a 25-degree rise, determined by mass multiplied by specific heat and temperature increase. The necessary calculations yield a total heat load of 176164 J. Finally, by dividing heat required by heat supply rate, we ascertain that it will take approximately 176.16 minutes.

4 0

1 month ago

A block of mass m, initially held at rest on a frictionless ramp a vertical distance H above the floor, slides down the ramp and

kicyunya [3294]

Answer:

$\displaystyle W=-m.g.H$

Explanation:

Transformation of Energy

Also known as energy conversion, this refers to the process in which energy shifts from one type to another. In this context, three energy forms are involved. When the object is stationary at the ramp's peak, it possesses gravitational potential energy, calculated as

$U=m.g.H$

As the object descends the frictionless ramp, it converts all its potential energy into kinetic energy, represented as

$\displaystyle K=\frac{m.v^2}{2}$

Thus,

$K=m.g.H$

Ultimately, when the object encounters a rough surface, all energy converts to thermal energy. The work performed by the friction force corresponds to the alteration in kinetic energy, as all velocity is lost in this process:

$\displaystyle W=\Delta E=K_f-K=0-K=-\frac{m.v^2}{2}$

Given the kinetic energy equals the initial potential energy:

$\boxed{\displaystyle W=-m.g.H}$

The negative sign indicates that the work acted against the direction of movement, meaning the force and displacement are 180° apart.

This outcome is independent of the distance D needed to halt the block or the kinetic friction coefficient.

7 0

3 months ago