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RSB [31]
2 years ago
10

Identrying the speed of the Cart

Physics
2 answers:
FrozenT [24]2 years ago
7 0

The speed of the cart after 3 seconds of Low fan speed is 55 cm/s.

The speed of the cart after 5 seconds of Medium fan speed is 120 cm/s.

The speed of the cart after 2 seconds of High fan speed is 63 cm/s.

<h3>What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its si unit is m/sec.

The graph in the attachment may be used to estimate the cart speed for various fan speeds.

Blue letters indicate the low fan speed. When the line drawn from x hits the blue line, the elapsed time is indicated on the x-axis cart speed after three seconds.

This is equivalent to 55 cm/s.Yellow markings indicate the medium fan speed. After 5 seconds, the line drawn from the point marked 5 on the x-axis crosses the yellow line, marking the time that has passed.

The intersection of the yellow line with the line drawn from the point marked 5 on the x-axis indicates the cart speed five seconds later. The y-axis measurement is roughly 120 cm/s.

Green markings indicate the high fan speed. On the x-axis is the elapsed time. Cart speed at the place where the green line and the line drawn from the x-point axis 2 connect. The y-axis reading is roughly 63 cm/s.

Hence, after the specified seconds, the cart's estimated speeds at low, medium, and high speeds are 55 cm/s, 120 cm/s, and 63 cm/s, respectively.

To learn more about the speed refer to the link;

brainly.com/question/7359669

#SPJ1

aleksklad [387]2 years ago
6 0

The speed of the cart after 3 seconds of Low fan speed is equal to 54 cm/s.

<h3>How to calculate the speed?</h3>

Mathematically, speed can be calculated by using this formula;

Speed = distance/time

At Low fan speed after 3 seconds, the distance covered is 162 cm:

Speed = 162/3

Speed = 54 cm/s.

At Medium fan speed after 5 seconds, the distance covered is 600 cm:

Speed = 600/5

Speed = 120 cm/s.

At High fan speed after 2 seconds, the distance covered is 128 cm:

Speed = 128/2

Speed = 64 cm/s.

Read more on speed here: brainly.com/question/17350470

#SPJ1

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Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three translational
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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:

C_v =6*\frac{R}{2}=3R=3x8.314\frac{J}{mol K}=24.942\frac{J}{mol K}

And by definition we know that the specific heat capacity is defined:

c=\frac{C_V}{M}

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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Please ignore my comment -- mass is not needed, here is how to solve it. pls do the math

at bottom box has only kinetic energy
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s = distance travel = h/sin(a)
h = height at top
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work = Fs = umgh x cot(a)
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(1/2)mv^2 = umgh x cot(a)
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Answer:

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