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viva [34]
3 years ago
12

W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively

. Any energy not mentioned in the transformation is assumed to remain constant; if work is not mentioned, it is assumed to be zero.
1. Give a specific example of a system with the energy transformation shown.
W→ΔEth

2. Give a specific example of a system with the energy transformation shown.

a. Rolling a ball up a hill.
b. Moving a block of wood across a horizontal rough surface at constant speed.
c. A block sliding on level ground, to which a cord you are holding on to is attached .
d. Dropping a ball from a height.
Physics
1 answer:
MArishka [77]3 years ago
7 0

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c)  W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

 W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

      W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

    W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

      ΔU = ΔK

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Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to
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Answer:

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Explanation:

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We apply this formal to both configurations

Small face measurements (W W)

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Large face measurements (D L)

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3 years ago
A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (
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\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

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                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

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                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

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4 0
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maks197457 [2]
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6 0
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aleksandrvk [35]

Answer:

(a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

Explanation:

Given that,

Area = 1100 acres

Number of house in 1 acres = 4

\text{Number of house in 1100 acres} = 4\times1100

\text{Number of house in 1100 acres} = 4400

Per house water demand = 800 g/day/house

(a). We need to calculate the average daily demand of this subdivision

Using formula for average daily demand

\text{average daily demand}=house\times\text{Per house water demand}

\text{average daily demand}=4400\times800\ gallon/day

\text{average daily demand}=3520000\ gallon/day

\text{average daily demand}=\dfrac{3520000}{24\times60}\ gallon/min

\text{average daily demand}=2444.44\ gallon/min

The average daily demand of this subdivision is 2444.44 gallon/min.

(b). We need to calculate the design-demand used to design the distribution system

Using formula for the design-demand

\text{design demand}=(Q_{max})daily\times\text{fire flow}

\text{design demand}=1.64\times2444.4\times1000

\text{design demand}=4008816\ gallon/m

Hence, (a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

8 0
2 years ago
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