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azamat
2 years ago
5

WHO WANts TO HAVE SOME GLIZZY ACTION

Physics
1 answer:
Lilit [14]2 years ago
6 0
Bruh huh.............
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Convert 26.4 mi to km
expeople1 [14]

Answer:

42.48668

Explanation:

4 0
2 years ago
Give a strong idea to protect , purify and the beautify of the Dal lake​
Sholpan [36]

Answer:

Water in dal lake is test for any heavy metals and pollutant, sewage and drainage system are also monitored for the same.

Explanation:

  • Dal lake is located in Srinagar that is the state capital of Kashmir and is known for recreation and tourism purposes. The area covers about 18 km sq. and forms a part of natural wetlands.
  • The lake is prone to pollution and has recently undergone restoration measures. To address the problems of eutrophication algae and large-scale microplankton have been removed from the water.
  • The government of India has taken various measures to check the pollution by setting up a committee to monitor the proper use of allotted funds.
3 0
3 years ago
What happens to an electromagnetic wave as it passes from space to matter?
alina1380 [7]

Answer:

When an electromagnetic wave passes from space to matter, some part of the energy is absorbed by the matter and it increases its energy. The wave may reflect and some part may pass through the matter depending on the amount of energy they have. The amplitude of the wave decreases if some parts of it are reflected.

4 0
2 years ago
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What are the conditions to increase the stability of a body?​
cricket20 [7]
Training everyday
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5 0
2 years ago
Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at
beks73 [17]

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

8 0
3 years ago
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