Are all solid natural non living matter on earth rock?

An ice-making machine inside a refrigerator operates in a Carnot cycle. It takes heat from liquid water at 0.0 ∘C and rejects he

Agata [3.3K]

Answer:

2.36 x 10^6 J

Explanation:

Tc = 0°C = 273 K

TH = 22.5°C = 295.5 K

Qc = heat used to melt the ice

mass of ice, m = 85.7 Kg

Latent heat of fusion, L = 3.34 x 10^5 J/kg

Let Energy supplied is E which is equal to the work done

Qc = m x L = 85.7 x 3.34 x 10^5 = 286.24 x 10^5 J

Use the Carnot's equation

$\frac{Q_{H}}{Q_{c}}=\frac{T_{H}}{T_{c}}$

$Q_{H}=286.24\times 10^{5}\times \frac{295.5}{273}$

QH = 309.8 x 10^5 J

W = QH - Qc

W = (309.8 - 286.24) x 10^5

W = 23.56 x 10^5 J

W = 2.36 x 10^6 J

Thus, the energy supplied is 2.36 x 10^6 J.

8 0

3 years ago

Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.

ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

4 0

2 years ago

In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s

sladkih [1.3K]

Answer:

quality factor (Q) = 69.99

inductor = 1.591 x 10⁻⁴ H

capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R = 10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW) $= \frac{R}{2\pi L}$

$BW = \frac{R}{2\pi L}$

make L (inductor) the subject of the formula

$L = \frac{R}{2\pi *BW} = \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH$

$F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}$

make C (capacitor) the subject of the formula

$C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F$

quality factor (Q) $= \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99$

quality factor (Q) = 69.99

5 0

3 years ago

A 5kg box is being pulled for a force of 20 n and is sliding with an acceleration of 2 m/s. Find the coefficient of friction

RUDIKE [14]

Answer:

Coefficient of friction = 0.5

Explanation:

Given:

Mass of box = 5 kg

Force applied = 20 N

Acceleration = 2 m/s²

Find:

Coefficient of friction

Computation:

Friction force = Mass x Acceleration.

Friction force = 5 x 2

Friction force = 10 N

Coefficient of friction = Friction force / Force applied

Coefficient of friction = 10 / 20

Coefficient of friction = 0.5

3 0

2 years ago

How do the cities in the northern hemisphere compare to those in the southern hemisphere?

a_sh-v [17]

By the rate of the hemisphere

4 0

2 years ago