Answer:
They are both unsegmented worms
Also:
They are both multicellular, invertebrates, heterotrophs, bilateral symmetry
Hope this Helps
Answer:
182 to 3 s.f
Explanation:
Workdone for an adiabatic process is given as
W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)
where γ = ratio of specific heats. For carbon dioxide, γ = 1.28
For an adiabatic process
P₁V₁ʸ = P₂V₂ʸ = K
K = P₁V₁ʸ
We need to calculate the P₁ using ideal gas equation
P₁V₁ = mRT₁
P₁ = (mRT₁/V₁)
m = 2.80 g = 0.0028 kg
R = 188.92 J/kg.K
T₁ = 27°C = 300 K
V₁ = 500 cm³ = 0.0005 m³
P₁ = (0.0028)(188.92)(300)/0.0005
P₁ = 317385.6 Pa
K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89
W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)
V₁ = 0.0005 m³
V₂ = 2.10 dm³ = 0.002 m³
1 - γ = 1 - 1.28 = - 0.28
W =
18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)
W = -67.47 (5.698 - 8.4)
W = 182.3 = 182 to 3 s.f
Answer:
Smoke detectors
Explanation:
This is because the ionization chamber in smoke detectors uses radiation, a form of heat transfer to detect smoke.
.The NRC allows beneficial use of smoke detectors which is a radioactive material because a smoke detector has the ability to save lives . it's ability to save lives outweighs any health risk from the radiation. The smoke detector use a very minimal amounts of radioactive materials.
Answer:
21.30 m2
Explanation:
First of all, we calculate the energy necessary to increase the temperature of the water from 20°C to 49°C. We use the specific heat in order to achieve this:
Where m_w is the mass of water, sh is the specific heat.
The mass of the water can be found using the volume of water and density:
Then:
You want to heat the water in 2.6h, this means, you will need to supply a power equal to the total amount of energy required divided by the time in seconds. In 2.6h there are 2.6*3600 = 9360s:
But remember, you have an efficiency of 28%, so the actual amount of power that you need to recieve from the sun is equal to:
Now, for the power of a collector is:
Where Psun is the power of the sun, S is the intensity of incident sunlight and A is the area. Solving for Area:
RA 4h 21m 39s | Dec +22° 59′ 54″