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strojnjashka [21]
3 years ago
13

what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe

ed cm/s squared? what was the acceleration of the cart with high fan speed cm/s squared?

Physics
2 answers:
ArbitrLikvidat [17]3 years ago
8 0

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2

Hence, this is the required solution.

mixer [17]3 years ago
5 0

Answer:

Explanation:

Low fan speed: 18.0 cm/s2

Medium fan speed: 24.0 cm/s2

High fan speed: 32.0 cm/s2

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t = 3.516 s

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The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:

v(t) = v_0 +at

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If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.

This can be done knowing that

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Therefore

1 km/h = (1000/3600) m/s = 0.2777... m/s

100 km/h = 27.777... m/s

Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:

27.777 m/s = 7.9 m/s^2 t

Solving for t

t = (27.7777 / 7.9) s = 3.516 s

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Answer:

B) Friction

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Friction is a force that acts when an object is sliding along a surface. Microscopically, this force is due to the fact that the two surfaces are not perfectly smooth, but they have "imperfections" that cause a force that opposes the motion of the object.

For an object sliding on a flat surface, the force of friction has magnitude:

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The direction of the force of friction is always opposite to the direction of motion of the object.

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Option (4)

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Answer:

I) c=1385.667\frac{J}{kg K}

II)The difference from the value obtained on part I is: 2000-1385.67 =614.33 \frac{J}{Kg K}

The possible reason of this difference is that the vibrational motion can increase the value, since if we take in count this factor we will have a higher heat capacity, because molecules with vibrational motion require more heat to vibrate and necessary higher specific heat capacity.

Explanation:

From the problem we have the molar mass given M=18\frac{gr}{mol} of water vapor and at constant volume condition. It's important to say that the vapour molecules have 3 transitionsl and 3 rotational degrees of freedom and the rotational motion no contribution.

Part I

Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three translational and three rotational degrees of freedom and that vibrational motion does not contribute. The molar mass of water is 18.0 g/mol=0.018kg/mol.

Let C_v (\frac{J}{Kg K}) the molar heat capacity at constant volume and this amount represent the quantity of heat absorbed by mole.

Let C (\frac{J}{Kg K}) the specific heat capcity this value represent the heat capacity aboserbed by mass.

For the problem we have a total of 6 degrees of freedom and from the thoery we know that for each degree of freedom the molar heat capacity at constant volume is given by C_v =\frac{R}{2} so the total for the 6 degrees of freedom would be:

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If we replace all the values we have:

c=\frac{24.942\frac{J}{mol K}}{0.018\frac{kg}{mol}}=1385.667\frac{J}{kg K}

So on this case the specific heat capacity with constant volume and with three translational and three rotational degrees of freedom is c=1385.667\frac{J}{kg K}

Part II

The actual specific heat of water vapor at low pressures is about 2000 J/(kg * K). Compare this with your calculation.

The difference from the value obtained on part I is: 2000-1385.67 =614.33 \frac{J}{Kg K}

The possible reason of this difference is that the vibrational motion can increase the value, since if we take in count this factor we will have a higher heat capacity, because molecules with vibrational motion require more heat to vibrate and necessary higher specific heat capacity.

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