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Kazeer [188]
3 years ago
10

. A water heater is operated by a solar power. If the solar collector has an area of 6 m2, and the intensity delivered by sunlig

ht is 550 W/m2, how long does it take to increase the temperature of 1000 kg of water from 200C to 60 0C?
Physics
1 answer:
NeTakaya3 years ago
4 0

Answer:

50 h

Explanation:

The heat required (Q) to heat the water  can be calculated using the following expression.

Q = c × m × ΔT = (4.184 kJ/kg.°C) × 1000 kg × (200°C - 60°C) = 5.9 × 10⁵ kJ

where,

c: specific heat capacity

m: mass

ΔT: change in the temperature

The solar collector has an area of 6 m² and the intensity delvered by sunlight is 550 W/m². The power (P) submitted by the collector is:

6 m² × 550 W/m² = 3 × 10³ W

The energy collected is transformed into heat. We know that:

P = E/t = Q/t

where

t: time

t = Q/P = 5.9 × 10⁸ J / (3 × 10³ J/s) = 2 × 10⁵ s × (1 h / 3600 s) = 50 h

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The most useful unit of measurement for measuring the thickness of a side of a lab beaker would be a(n)
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A car has a mass of 8000kg and a kinetic energy of 24,000j. What is the speed?
Oxana [17]

Answer:

2.5m/s

Explanation:

Kinetic energy = 1/2 mv²

Mass = m= 8000kg

Velocity(speed) = v =

K. E = 24000J

K. E = 1/2 x 8000 x v²

24000 = 4000 x v²

Divide both sides by 4000

V²= 24000/4000

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V = √6

V = 2.5 m/s

The S. I unit of energy is joules with symbol J

I hope this was helpful, Please mark as brainliest

7 0
3 years ago
What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire
Marianna [84]

Answer:

The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

When two point are at same height so ,

P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

Q=A_{1}v_{1}

v_{1}=\dfrac{Q}{A_{1}}

v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}

v_{1}=56.58\ m/s

For output velocity,

v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}

v_{2}=6.28\ m/s

Put the value into the formula

P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)

\Delta P=1.58\times10^{6}\ N/m^2

(b). We need to calculate the maximum height

Using formula of height

\Delta P=\rho g h

Put the value into the formula

1.58\times10^{6}=1000\times9.8\times h

h=\dfrac{1.58\times10^{6}}{1000\times9.8}

h=161.22\ m

Hence, The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

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4 years ago
Can anyone help fill out the table?
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Answer:


What table I don’t see any table
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