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denis23 [38]
3 years ago
14

What volume of water at O°C can a freezer make into ice cubes in 1.0 h, if the coefficient of performance of the cooling unit is

6.0 and the power input is 1.7 kilowatt? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
masha68 [24]3 years ago
5 0

Answer:

volume of water is 0.1092 m3

Explanation:

We know that heat energy required to convert water into ice is given as

Q =mL

where L is Latent heat of fusion of ice is 336000 Jkg-1

Mass = \rho *V

           =1000*V

Q = 1000V*336*10^3

   Q = 336*10^6 V

ENERGY used by the freezer is given as

= 6.0*1.7*10^3 W = 10,200 w

energy provided per hour is Q = 10,200 W *1*60*60 s = 3.67*10^7 J

THEREFORE we have V

336*10^6 V = 3.67*10^7 j

v = .109 m3

volume of water is 0.1092 m3

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S Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to
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The diameter of the wire is 2.8 * 10^-3 m.

<h3>What is the length?</h3>

Mass of the wire = 1.0 g or 1 * 10^-3 Kg

Resistance = 0.5 ohm

Resistivity of copper = 1.7 * 10^-8 ohm meter

Density of copper = 8.92 * 10^3 Kg/m^3

V = m/d

But v = Al

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l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3

l = 1.82 m

A = πr^2

Also;

A = m/ld

A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3

A = 6.2 * 10^-5 m^2

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Learn more about resistivity:brainly.com/question/14547003

#SPJ4

Missing parts;

Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?

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6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
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Answer:

a. 10.5 s b. 6.6 s

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a. The driver's perception/reaction time before drinking.

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v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

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So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

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Option A looks like the best definition
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