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3 years ago

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You and your housemate have an argument over the cost of the electric bill. You want to turn off the outside porch light before

going to bed each night. Your housemate wishes to keep the light on from 6:00 PM each evening to 6:00 AM the next morning and says that this will not add any significant amount to the electric bill. If the porch light bulb is a 65 W bulb, how much will your housemate's plan contribute to the electric bill each month? Take the number of days in a month as 30 and the average cost of electricity is 17.0¢/kW · h.

1 answer:

deff fn [24]3 years ago

5 0

Answer:

Cost = $ 3,995

Explanation:

Let's start by looking for the energy consumed, which is the power over time

E = P t

The power is 65 W, the time is 12 h each day, in a period of 30 days the time is

t = 12 30 = 360 h

E = 65 360

E = 23,500 W h

We reduce to KW h

1 kW = 1000 W

E = 23.5 kW h

Let's use a direct ratio rule (rule of three) to find the cost

Cost = 23.5 17.0

Cost = 399.5 c

Cost = $ 3,995

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regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac

solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=MV_{i}+mU_{i}$

$p_{f}=(M+m) V$

Where:

$M=480 g \frac{1 kg}{1000 g}=0.48 kg$ the mas of the peregrine falcon

$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

$V=\frac{MV_{i}}{M+m}$ (3)

$V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}$ (4)

$V=30 m/s$ (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

$F=\frac{mV}{\Delta t}$ (9)

$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

Finally:

$F=480 N$

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3 years ago

The liquid pressure exerted in one direction only

Wewaii [24]

The pressure exerted by a liquid on an object increases as we go more deep into the liquid and this pressure is called as hydro static pressure . if we consider a part of the static fluid then all the horizontal forces will cancel out while the vertical forces will add vectorilly and due to which a pressure difference is created . so as we go more deep the pressure increase .

Now pressure is a scalar so it does not depend on direction but when two objects are on the same level with respect to a reference level then the pressure exerted on them by fluid is always the same . hope this helps

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2 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

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3 years ago

Speed is determined based on two factors. What are they

a_sh-v [17]

Answer:

distance and time

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2 years ago

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murzikaleks [220]

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