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Fofino [41]
3 years ago
13

A refrigerator operates on average for 10.0 hours an day. If the power rating is the refrigerator is 709 w how much electrical e

nergy does the fridge use in one day
Engineering
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer:

61257.6 kW per day

Explanation:

The rating of the refrigerator is given as : 709 w

This means the power consumption is 709 joules per second

The operating time is given as 10 hours.

Change hours to second as

1 hour = 3600 seconds

10 hours = 36000 seconds

Apply the rating as;

709 w = 1 s

?         = 36000 seconds

perform cross product to get;

709 * 36000= 25524000 w

25524000 = 25524kW

For a day, 24 hours will be;

25524 kW = 10 hours

?                 =24 hours

={24*25524}/10 = 61257.6 kW per day

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Explain how voltage level are interpreted by a digital circuit ​
lisabon 2012 [21]

Answer:

  the state of the circuit is a function of the voltage level. The interpretation is up to the user.

Explanation:

A binary digital circuit adopts one of two states, depending on whether the voltage level is above or below some threshold that depends on the design of the circuit. Within each state, the voltage may have some typical range. When the voltage is near the threshold, the state of the circuit may actually be "indeterminate".

The internal/output voltage is a function of the state of the circuit. The interpretation of that voltage as a true/false or 1/0 or other meaning is up to the user of the circuit.

The circuit interprets a given input voltage as intending to convey a particular input signal state according to the circuit specifications. Input voltages near the threshold between states may cause unexpected or even destructive results.

__

In order to conserve space, some digital circuits use more than 2 different voltage levels to signify more than 2 different states.

5 0
3 years ago
A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream
Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}

Substituting figures in the equation we have:

= \frac{1}{1+0.1(2.5-1)+0(2-1)}

= 0.75

Let's now calculate equivalent flow rate of the car using:

Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}

= \frac{7500}{0.9*4*0.75*1}

= 2777.7 pc/h/en

Calculating for traffic density, we have:

D = \frac{Vp}{FFS}

D = \frac{2777.7}{70}

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0
3 years ago
A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol
insens350 [35]

<u>Explanation:</u>

Determine the weight of footing

W_{f}=\gamma(L)(B)(D)

Where W_{f} is the weight of footing, γ is the unit weight of concrete,  L is the length of footing is the width of footing, and D is the depth of footing

Substitute 2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3} for γ in the equation

\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

\sigma_{z p}^{\prime}=\gamma(D+B)-u

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute 18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0 for u in the equation

\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

\sigma_{z D}^{\prime}=\gamma D

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute \(18 kN / m ^{3}\) for \(\gamma, 0.5 m\) for \(D\) and 0 for \(u\) in the equation.

\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}

Here I_{e p} is the influence factor at the midpoint of soil layer  \sigma_{z^{p}}^{\prime} is initial vertical stress, \sigma_{z^{p}}^{\prime} is vertical effective stress, and Q is bearing pressure

Substitute 36 kPa for \(\sigma_{z p}^{\prime}, 228.47\) kPa for \(q,\) and 9 kPa for \(\sigma_{z D}^{\prime}\) in the equation\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}

Therefore the influence factor at the midpoint of the soil layer is 0.693

6 0
3 years ago
Application 1: Consider the following beam under combined loading of a distributed pressure and applied bending moment. Complete
Ivan

Answer:

Explanation:

kindly check he attached file below

7 0
3 years ago
You plan to install an active, liquid-based solar heating system for hot water. There are four candidate collector systems. Your
olchik [2.2K]

Solution:

The given formula,

x=F_{R} U_{L} \times \frac{P l}{F R_{1}} \times\left(T_{r e f}-\bar{T}_{a}\right) \Delta t \times \frac{A_{c}}{L}

y=F_{R}(\tau \alpha)_{n} x \frac{F_{R}^{\prime}}{F_{R}} \times \frac{(\bar{\tau} d)}{(T d)_{n}} \times \bar{H}_{T} N \times \frac{A C}{L}

\frac{x}{y}=\frac{ u_{L} \times\left(T_{x t}-\bar{T}_{a}\right) \times \Delta t}{\left(\tau_{x}\right)_{h} \times\left(\frac{\bar{\tau}_{d}}{\left.| \tau_{d}\right)_{n}}\right) \times \bar{H}+N}

From the table,

1) \(\quad x=2 \cdot 87, \quad y=0.96\)\\\(\frac{x}{y}=\frac{2187}{0.96}\)22895\\\\2) \(x=3 \cdot 466 \cdot y=6 \cdot 998\)\\\(\frac{x}{y}=\frac{3 \cdot 466}{0.898}\)\(=3 \cdot 4729\)

3\(x=3 \cdot 229, y=1 \cdot 08\)\\\(\frac{x}{x}=\frac{3 \cdot 229}{1 \cdot 08}\)\\=2.9898\)\\\\4) \(x=6.525, y=1.094\)\\\(\frac{x}{y}=\frac{5.625}{1.094}\)\\=5.0502

8 0
3 years ago
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