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Genrish500 [490]
3 years ago
5

The temperature inside a Carnot refrigerator placed in a kitchen at 22 degree Celcius is 2 degree Celcius. The heat extracted fr

om the refrigerator is 89 MJ/h. What power is needed to operate this refrigerator?
Physics
1 answer:
Elanso [62]3 years ago
4 0

Answer: 1.79 kW

Explanation:

Given

The temperature of the refrigerator is T_L=2^{\circ}C\approx 275\ K

The temperature of the kitchen is T_H=22^{\circ}C

COP of the refrigerator is

COP=\dfrac{T_L}{T_H-T_L}=\dfrac{\text{Desired effect}}{\text{Power}}\\\\\Rightarrow COP=\dfrac{275}{22-2}=\dfrac{89}{P}\\\\\Rightarrow P=\dfrac{20}{275}\times 89=6.47\ MJ/h\ \text{or}\ 1.79\ kW

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melomori [17]

The units of G must be C. m³ / ( kg s² )

<h3>Further explanation</h3>

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }

<em>F = Gravitational Force ( Newton )</em>

<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>

<em>m = Object's Mass ( kg )</em>

<em>R = Distance Between Objects ( m )</em>

Let us now tackle the problem !

To find unit of Gravitational Constant can be carried out in the following way:

F = G \frac{m_1 ~ m_2}{R^2}

{[N]}= G\frac{{[kg]}{[kg]}}{{[m^2]}}

{[kg ~ m / s^2]}= G \frac{{[kg^2]}}{{[m^2]}}

G = \frac{{[kg ~ m / s^2]}{[m^2]}} {{[kg^2]} }

G = \frac{{[kg ~ m^3 / s^2]}} {{[kg^2]} }

G = \frac{{[m^3 / s^2]}} {{[kg]} }

\boxed {G = \frac{{[m^3]}} {{[kg ~ s^2]} }}

The unit of G must be \large {\boxed {\frac{m^3} {kg ~ s^2 }}}

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

5 0
3 years ago
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A car travels west at 40 km/h
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Answer:

The car is going 0 km/h more than the bike

Explanation:

3 0
3 years ago
What is Initial temperature and final temperature equations??<br> ...?
Neporo4naja [7]
One that can help you is:
ΔT=<span>T<span>Final</span></span>−<span>T<span>Initia<span>l
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</span></span></span><span><span>Tf</span>=<span>Ti</span>−Δ<span>H<span>rxn</span></span>∗<span>n<span>rxn</span></span>/(<span>C<span>p,water</span></span>∗<span>m<span>water</span></span>)
This one is taking into account that yu can find temperature and that there could be a change with a chemical reaction. Hope this helps</span>
8 0
4 years ago
A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a d
igor_vitrenko [27]

Answer:

Explanation:

potential energy of compressed spring

= 1/2 k d²

= 1/2 x 730 d²

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This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .

Kinetic energy after crossing the rough patch

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= 3.174 J

Loss of energy

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This loss is due to negative work done by frictional force

work done by friction = friction force x width of patch

= μmg d ,   μ = coefficient of friction , m is mass of block , d is width of patch

= .44 x 1.2 x 9.8 x .05

= .2587 J

365 d² - 3.174   = .2587

365 d² = 3.4327

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= .0094

d = .097 m

= 9.7 cm

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Answer:

Given that

speed u=4*10^6 m/s

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distance b/w the plates d=2 cm

basing on the concept of the electrostatices

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acceleration a=qE/m=1.6*10^{-19}*4*10^3/9.1*10^{-31} =0.7*10^{15}=7*10^{14} m/s

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horizontal distance

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=4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}

=3*10^{-2}= 3 cm

5 0
4 years ago
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