The ship floats in water due to the buoyancy Fb that is given by the equation:
Fb=ρgV, where ρ is the density of the liquid, g=9.81 m/s² is the acceleration of the force of gravity and V is volume of the displaced liquid.
The density of fresh water is ρ₁=1000 kg/m³.
The density of salt water is in average ρ₂=1025 kg/m³.
To compare the volumes of liquids that are displaced by the ship we can take the ratio of buoyancy of salt water Fb₂ and the buoyancy of fresh water Fb₁.
The gravity force of the ship Fg=mg, where m is the mass of the ship and g=9.81 m/s², is equal to the force of buoyancy Fb₁ and Fb₂ because the mass of the ship doesn't change:
Fg=Fb₁ and Fg=Fb₂. This means Fb₁=Fb₂.
Now we can write:
Fb₂/Fb₁=(ρ₂gV₂)/(ρ₁gV₁), since Fb₁=Fb₂, they cancel out:
1/1=1=(ρ₂gV₂)/(ρ₁gV₁), g also cancels out:
(ρ₂V₂)/(ρ₁V₁)=1, now we can input ρ₁=1000 kg/m³ and ρ₂=1025 kg/m³
(1025V₂)/(1000V₁)=1
1.025(V₂/V₁)=1
V₂/V₁=1/1.025=0.9756, we multiply by V₁
V₂=0.9756V₁
Volume of salt water V₂ displaced by the ship is smaller than the volume of sweet water V₁ because the force of buoyancy of salt water is greater than the force of fresh water because salt water is more dense than fresh water.
Answer:
Your answer would be C, Radio waves.
Explanation:
155Ω
Explanation:
R = R ref ( 1 + ∝ ( T - Tref)
where R = conduction resistance at temperature T
R ref = conductor resistance at reference temperature
∝ = temperature coefficient of resistance for conductor
T = conduction temperature in degrees Celsius
T ref = reference temperature that ∝ is specified at for the conductor material
T = 600 k - 273 k = 327 °C
Tref = 300 - 273 K = 27 °C
R = 50 Ω ( 1 + 0.007 ( 327 - 27) )
R = 155Ω
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I hope after I answer that you understand and don't just paste my answer into your assignment!!! (<- read!!!)
Answer \|/
Ice is less dense than water.
Reason why \|/
When water freezes the molecules inside completely stop moving (They still vibrate but don't change their position much). In doing so, they spread out a touch which makes it less dense than liquid water. So ice floats
Answer:
P = 1 (14,045 ± 0.03 ) k gm/s
Explanation:
In this exercise we are asked about the uncertainty of the momentum of the two carriages
Δ (Pₓ / Py) =?
Let's start by finding the momentum of each vehicle
car X
Pₓ = m vₓ
Pₓ = 2.34 2.5
Pₓ = 5.85 kg m
car Y
Py = 2,561 3.2
Py = 8,195 kgm
How do we calculate the absolute uncertainty at the two moments?
ΔPₓ = m Δv + v Δm
ΔPₓ = 2.34 0.01 + 2.561 0.01
ΔPₓ = 0.05 kg m
Δ
= m Δv + v Δm
ΔP_{y} = 2,561 0.01+ 3.2 0.001
ΔP_{y} = 0.03 kg m
now we have the uncertainty of each moment
P = Pₓ /
ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²
ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²
ΔP = 0.006 + 0.0026
ΔP = 0.009 kg m
The result is
P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s
