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MissTica
3 years ago
14

An observation is _______.

Physics
1 answer:
Mice21 [21]3 years ago
8 0
An observation can basically be described as analyzing something closely and/or taking notice of something.
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In a coal-burning power plant, a condenser is used to cool down the steam and convert it back to liquid water true or false
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Two round rods, one steel andthe other copper, are joined end to end. Each rod is 0.750 mlong and 1.50 cmin diameter. The combin
nikitadnepr [17]

Answer: a) Strain on Steel rod = 0.0001078

b) elongation on the steel rod = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain on Copper Rod = 0.000189

Explanation: a) To obtain the strain of the steel rod, we invoke Hooke's law which states that, provided the elastic limit of A material isn't exceeded, the stress it undergoes is directly proportional to its strain.

(Stress, σ) ∝ (Strain, ε)

The constant of proportionality is called Young's modulus, E.

σ = Eε

For steel, Younger Modulus as obtained from literature = 210GPa.

Strain = Stress/Young's Modulus

Stress = (Force or Load applied)/ Cross sectional Area.

Force applied For the steel = 4000N

Cross sectional Area = (π(D^2))/4

D = 1.50cm = 0.015m

A = 0.0001767 m2

σ = 4000/0.0001767 = 22637238.257 N/m2

ε = σ/E = 22637238.257/(210 × (10^9)) = 0.0001078.

b) To solve for elongation.

Strain, ε = (elongation, dl)/(original length, lo)

Elongation, dl = strain × original length

dl = 0.0001078 × 0.75 = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain in Copper

ε = σ/E; σ = 22637238.257 N/m2

Young's modulus of Copper, from literature, = 120GPa

ε = 22637238.257/(120 × (10^9)) = 0.000189

4 0
3 years ago
A swimming pool is 25.0 ft. long, 18.5 ft. wide, and 9.0 ft. deep. When filled, the water level is 7.0 inches from the top. Disi
GarryVolchara [31]

Answer:

1.9841256 kg

Explanation:

Given;

Length of the swimming pool = 25.0 ft = 7.62 m   ( 1 ft = 0.3048 m )

Width of the swimming pool = 18.5 ft =  5.64 m

Depth of the pool = 9.0 ft =

Total depth of the water in the pool when filled = 9 ft - 7 inches = 2.56 m

now,

Volume of the water in the pool = Length × Width × Depth

or

Volume of the water in the pool = 7.62 × 5.64 × 2.56 = 110.2292 m³

also,

1 m³ = 1000 L

thus,

110.2292 m³ = 110229.2 L

also it is given that 18 mg of Cl is added to 1 liter of water

therefore,

In 110229.2 L of water Cl added will be = 110229.2 × 18 = 1984125.6 mg

or

= 1.9841256 kg

8 0
4 years ago
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