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attashe74
1 month ago
8

A household refrigerator runs one-fourth of the time and removes heat from the food compartment at an average rate of 800 kJ/h.

If the COP of the refrigerator is 2.2, determine the power the refrigerator draws when running.
Physics
1 answer:
Ostrovityanka [3.2K]1 month ago
4 0

Answer:

1454.54 kJ/h or 0.404 kW

Explanation:

Provided:

The heat being removed by the refrigerator is 800 kJ/h.

The refrigerator's coefficient of performance (COP) is 2.2.

Since the refrigerator operates for one-fourth of the time,

it effectively removes heat, Q = 4\times 800 kJ/h = 3200 kJ/h.

Next, the power consumed by the refrigerator during its operation is calculated as:

\frac{Q}{COP}

= \frac{3200 kJ/h}{2.2}.

= 1454.54 kJ/h

Thus, the fridge uses 1454.54 kJ/h

or 0.404 kW (given 1 kW = 3600 kJ/h).

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We fully submerge an irregular lump of material in a certain fluid. The fluid that would have been in the space now occupied by
Ostrovityanka [3204]

Answer:

the lump descends

Explanation:

The full articulation of the query is

Upon fully submerging a 3 kg lump of material in a certain fluid, the fluid that would have occupied the space now filled by the lump weighs 2 kg. (a) When released, does the lump float up, sink, or remain steady

(a)

F_{b} = Buoyant force acting upward on the lump

M = mass of irregular lump = 3 kg

m = mass of fluid displaced = 2 kg

The upward buoyant force on the lump is given by the weight of the displaced fluid, thus

F_{b} = mg = (2) (9.8) = 19.6 N

the weight of the irregular lump of material is represented as

W = mg\\W = (3) (9.8)\\W = 29.4 N

Given that the weight of the lump downward exceeds the upward buoyant force, the lump will indeed descend

4 0
12 days ago
Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2
Ostrovityanka [3204]

The complete question is;

Block 1 sits on the floor with block 2 resting atop it. Block 3, which is stationary on a frictionless table, is attached to block 2 via a string that passes over a pulley depicted in the illustration below. Both the string and pulley have negligible mass.

Once block 1 is taken away without impacting block 2.

Derive an equation for the acceleration of block 3 considering arbitrary values for m3 and m2. Express your answer in terms of m3, m2, and relevant physical constants as needed.

Answer:

a = (m2)g/(m3 + m2)

Explanation:

Examining the attached illustration, by analyzing the free body diagram for block 3 and utilizing Newton's first law of motion, we reach the following formula;

T = (m3)a - - - (eq 1)

where;

T is the tension in the string

a is acceleration

m3 is the mass of block 3

Simultaneously, doing the same for Block 2, the free body diagram yields the equation; (m2)g - T = (m2)a

Rearranging for T results in;

T = (m2)g - (m2)a - - - (eq 2)

where;

g represents acceleration due to gravity

T is the tension in the string

a is acceleration

m2 is the mass of block 2

To deduce the acceleration, we will substitute (m3)a in place of T in eq 2.

Thus;

(m3)a = (m2)g - (m2)a

(m3)a + (m2)a = (m2)g

a(m3 + m2) = (m2)g

a = (m2)g/(m3 + m2)

3 0
8 days ago
Uzupełnij zdania właściwymi sformułowaniami. Wyobraź sobie, że między linę a siodełko karuzeli łańcuchowej wmontowany jest siłom
ValentinkaMS [3465]

Explanation:

Here’s a revised version of the requirements;

Fill in the blanks with the appropriate terms. Picture a force gauge fixed between the rope and the saddle of the chain carousel. If you keep your feet off the ground while the vehicle is not in motion, the dynamometer shows A / B. When the carousel is spinning, you’ll see C / D displayed on the dynamometer.

A. Your weight including the saddle

C. Value of the rope's strength

B. Your weight

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3 0
2 months ago
A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the
ValentinkaMS [3465]

Answer:

The molar mass of the metal in grams per mole is calculated to be 8.87.

Explanation:

Initially, we can consider a sample of the compound weighing 100 g. This results in:

  • 52.92% metal: 52.92 g M
  • 47.80% oxygen: 47.80 g O

 By utilizing the molar mass of oxygen, which is 16 g / mol, we can determine the quantity of moles of oxygen in the sample via the rule of three:

moles of oxygen=\frac{47.8g*1mol}{16g}

moles of oxygen=2.9875

The formula for the metal oxide indicates that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

From the previous equation, it is evident that 3 oxygen ions are necessary to react with 2 metal ions. Hence:

2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M

Given 52.92 g of metal in the sample, the molar mass of the metal is:

molar mass=\frac{52.92 g}{5.975 mol}

molar mass≅ 8.87 g/mol

The molar mass of the metal in grams per mole is 8.87.

The value that most closely corresponds to this is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

3 0
1 month ago
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B = µo*N*I/2r  

Thus, B = 4πx10^-7*150*1.6/2*3.5 = 4.31x10^-5T

7 0
9 days ago
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