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sammy [17]
3 years ago
8

A hot water heater is operated by using solar power. if the solar collector has an area of 5.3 m2 , and the power delivered by s

unlight is 995 w/m2 , how long will it take to increase the temperature of 1 m3 of water from 20 ◦c to 65 ◦c? the specific heat of water is 4186 j/kg · ◦ c and the density of water is 1000 kg/m3 . answer in units of h.
Physics
1 answer:
OLEGan [10]3 years ago
4 0
Heat absorbed by the solar collector = Area*Irradiance = 5.3*995 = 5273.5 W

Heat Q in joules absorbed in t hours = Heat used to heat water. That is,

5273.5*t = mCΔT; where mass = volume*density = 1*1000 = 1000 kg

Therefore;
5273.5t = 1000*4186*(65-20) = 188370000
t = 188370000/5273.5 = 35720.11 seconds = 35720.11/(60*60) hours ≈ 9.92 hours.

It will take approximately 9.92 hours.
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The element selenium (Se) bonds with chlorine (Cl) to make the formula SeCl2 Chlorine is more electronegative than selenium. Wha
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Answer:

Selenium dichloride

Explanation:

Selenium (Se) and Chlorine (Cl) are both elements capable of combining together to form a compound with the chemical formula; SeCl2. Since the chlorine atom is more electronegative than selenium atom, the chlorine pulls more electrons towards itself to form an IONIC bond.

The SeCl2 compound formed is called Selenium dichloride as two atoms of Chlorine are needed to combine with one atom of Selenium to form the compound.

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3 years ago
What type of diagram is this?
FromTheMoon [43]

Answer: That’s a sankey diagram

Explanation:

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3 years ago
The machine which turns in a power station​
Anna [14]

Answer:

generators

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the machine which turns in a power station

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3 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
3 years ago
Numerical Problems:
dangina [55]
  • Displacement=1200m
  • Time=4min=4(60)=240s

\boxed{\sf Velocity=\dfrac{Displacement}{Time}}

\\ \sf\longmapsto Velocity=\dfrac{1200}{240}

\\ \sf\longmapsto Velocity=5m/s

6 0
3 years ago
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