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Elena L [17]
3 years ago
14

Cold water flows to the solar panels at 15°C. During the day, the panels supply 3.8 kg of hot water

Physics
1 answer:
Yuliya22 [10]3 years ago
5 0

Answer:

The average energy that the solar panel delivers to the water in one hour is 798,000 Joules

Explanation:

The given parameters are;

The temperature of the water that flows to the solar panels, T₁ = 15°

The mass of the hot water the panel supplies during the day, m = 3.5 kg

The temperature of the hot water supplied by the panel, T₂ = 65 °C

The heat capacity of the water, c = 4,200 J/(kg·°C)

The heat delivered to the solar panels in one hour, ΔQ, is given by the following formula

ΔQ = m·c·ΔT

m = 3.8 kg, c = 4,200 J/(kg·°C), and ΔT = T₂ - T₁ = 65 °C - 15 °C = 50 °C

∴ ΔQ = 3.8 kg × 4,200 J/(kg·°C) × 50°C = 798,000 joules

∴ΔQ = 798,000 joules

The average energy that the solar panel delivers to the water in one hour, ΔQ = 798,000 joules.

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Multiply the following numbers, using scientific notation and the correct amount of significant digits. 1.003 m⋅3.09 = _____ 3.0
lozanna [386]

Answer

correct answer is 3.10

Explanation:

in this question we have to multiply  two numbers 1.003 and 3.09.

1.003 has 4 significant digits and 3.09 has 3 significant digits so answer must have 3 significant digits.

1.003\times 3.09=3.09927

1.003\times 3.09=3.10              


Hope it will help you

3 0
3 years ago
Read 2 more answers
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

6 0
3 years ago
1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.
marta [7]

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB

<h3>Speed of sound at the given temperature</h3>

v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )

where;

  • -v₀ is velocity of the observer moving away from the source
  • -vs is the velocity of the source moving towards the observer
  • fs is the source frequency
  • fo is the observed frequency
  • v is speed of sound

f_0 = f_s(\frac{v-v_0}{v- v_s} )

f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz

Learn more about intensity of sound here: brainly.com/question/17062836

3 0
1 year ago
Which statement best describes the forces in this picture
garik1379 [7]

Answer:

D.

Explanation:

Force is strength and energy as an attribute of physical action or movement.

6 0
3 years ago
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A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
Delvig [45]

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

<h3>Work done by the applied force</h3>

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

<h3>Work done by frictional force</h3>

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

<h3>Net work done by all the forces on the tractor</h3>

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

4 0
1 year ago
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