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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?

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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An object's inertia is its tendency to maintain contact:

Gala2k [10]

hello! it is velocity.

i say this because, Inertia is the tendency of an object to resist changes in its state of motion. ... The state of motion of an object is defined by its velocity - the speed with a direction.

3 0

3 years ago

An object has a mass of 10grams and a volume of 20m3. what is the density of the object

Pavlova-9 [17]

Explanation:

density = mass/ volume

in the question the mass is given in grams so will convert it into kg,

10 g = 0.01 kg

density= 0.01/ 20 = 1 / 2000 = 0.005 kg/m^3

7 0

3 years ago

A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of i

kirza4 [7]

Answer:

2.5 x 10⁷ J

Explanation:

F = thrust of the engine = 2.3 x 10⁵ N

d = distance traveled = 87 m

Work done by the engine is given as

W = F d = (2.3 x 10⁵) (87) = 200.1 x 10⁵ J

W' = Net work done

W'' = work done by catapult

KE₀ = initial kinetic energy = 0 J

KE = final kinetic energy = 4.5 x 10⁷ J

Net work done is given as

W' = KE - KE₀

W' = 4.5 x 10⁷ J

We know that

W' = W + W''

4.5 x 10⁷ = 2.001 x 10⁷ + W''

W'' = 2.5 x 10⁷ J

8 0

3 years ago

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis o

mihalych1998 [28]

Answer:

θ₁ = 0.5 revolution

Explanation:

We will use the conservation of angular momentum as follows:

$L_1=L_2\\I_1\omega_1=I_2\omega_2$

where,

I₁ = initial moment of inertia = 18 kg.m²

I₂ = Final moment of inertia = 3.6 kg.m²

ω₁ = initial angular velocity = ?

ω₂ = Final Angular velocity = $\frac{\theta_2}{t_2} = \frac{2\ rev}{1.2\ s}$ = 1.67 rev/s

Therefore,

$(18\ kg.m^2)\omega_1 = (3.6\ kg.m^2)(1.67\ rev/s)\\\\\omega_1 = \frac{(3.6\ kg.m^2)(1.67\ rev/s)}{(18\ kg.m^2)}\\\\\omega_1 = \frac{\theta_1}{t_1} = 0.333\ rev/s\\\\\theta_1 = (0.333\ rev/s)t_1$

where,

θ₁ = revolutions if she had not tucked at all = ?

t₁ = time = 1.5 s

Therefore,

$\theta_1 = (0.333\ rev/s)(1.5\ s)\\$

<u>θ₁ = 0.5 revolution</u>

4 0

3 years ago

A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. Th

Nostrana [21]

To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.

Energy in a capacitor can be defined as

$E = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}$

Here,

V = Potential difference across the capacitor plates

Q = Charge stored on the capacitor plates

At the same time capacitance can be defined as,

$C = \epsilon_0 (\frac{A}{d})$

Here,

$\epsilon_0 =$ Vacuum permittivity constant

A = Area

d = Distance

Replacing with our values we have that,

$C = (8.85*10^{-12})(\frac{0.0451}{2.51*10^{-3}})$

$C = 1.5901*10^{-10}F$

PART A) Energy stored in the capacitor is

$E = \frac{1}{2} CV^2$

$E = \frac{1}{2} (1.5901*10^{-10})(575)^2$

$E = 2.628*10^{-5}J$

PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore

$Q = CV$

$Q = (1.59*10^{-10})(575)$

$Q = 9.1425*10^{-8}C$

Now if $d = 10.04*10^{-3}m$ we have that the capacitance is

$C = \epsilon_0 (\frac{A}{d})$

$C = (8.85*10^{-12})(\frac{0.0451}{10.04*10^{-3}})$

$C = 3.9754*10^{-11}F$

Then the energy stored is

$E = \frac{1}{2} \frac{Q^2}{C}$

$E = \frac{1}{2} (\frac{(9.1425*10^{-8})^2}{3.9754*10^{-11}})$

$E = 1.051*10^{-4} J$

PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore

$W = 1.051*10^{-4} J -2.628*10^{-5}J$

$W = 7.882*10^{-5} J$

6 0

3 years ago