Is there any possibility to make 100% efficient system

PLEASE HELP ME 20 POINTS AND I WILL MARK BRAINLIEST!!!

Andreyy89

<span>B. a homogeneous mixture in which one substance is dissolved into another</span>

7 0

3 years ago

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What was David Berkowitz best known for? How did arson play a role in his past?

tankabanditka [31]

David Berkowitz, who became known as Son of Sam, went on a murderous rampage in New York City during the 1970s, taunting and insulting police, until they captured him. On August 10, 1977, 11 days after his last murder, David Berkowitz, known as Son of Sam, was arrested and later sentenced to six consecutive 25-years-to-life terms. Arson has evolved, since the 18th century, from a wrongful individual act into an effective means of collective violence, too. From 1750, the privatisation of common land in England limited peasants' access to resources such as firewood and game. Hope this helped!

5 0

2 years ago

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The motor in a refrigerator has a power output of 294 W. If the freezing compartment is at 271 K and the outside air is at 310 K

dusya [7]

Answer:

Maximum amount of heat = 10002151.38J

Explanation:

Workdone by motor in 86.1 minutes I given by:

W = power × time

W= 294 × 86.1×60

W= 1439424 Joules

W= 1.4 ×10^6Joules

The amount of heat extracted is given by:

/QL /= K/W/ = TL/W/ /(TH - TL)

Where TL= freezing compartment temperature

TH = Outside air temperature

/QL /= 271 × 1439424 / (310 - 271)

/QL/ = 390083904/39

/QL/ = 10002151.38 Joules

5 0

2 years ago

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Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain

Luden [163]

Answer:

$P=1362\ W$

$t'=251.659\ s$ is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, $T_i=18^{\circ}C$

time taken to vapourize half a liter of water, $t=18\ min=1080\ s$

desity of water, $\rho=1\ kg.L^{-1}$

So, the givne mass of water, $m=1\ kg$

enthalpy of vaporization of water, $h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}$

specific heat of water, $c=4180\ J.kg^{-1}.K^{-1}$

Amount of heat required to raise the temperature of given water mass to 100°C:

$Q_s=m.c.\Delta T$

$Q_s=1\times 4180\times (100-18)$

$Q_s=342760\ J$

Now the amount of heat required to vaporize 0.5 kg of water:

$Q_v=m'\times h_{fg}$

where:

$m'=0.5\ kg=$ mass of water vaporized due to boiling

$Q_v=0.5\times 2256.4$

$Q_v=1.1282\times 10^{6}\ J$

Now the power rating of the boiler:

$P=\frac{Q_s+Q_v}{t}$

$P=\frac{342760+1128200}{1080}$

$P=1362\ W$

Now the time required to heat to boiling point form initial temperature:

$t'=\frac{Q_s}{P}$

$t'=\frac{342760}{1362}$

$t'=251.659\ s$

6 0

3 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago