In steering in control on a PWC, the most needed requirement is to have power in which will allows the one using it to control the engine for it to stop or for it to move. Without power, the person using it could be out of balance or fail to start the engine. This is the basic requirement because even though it is shut off or the engine is off, the PWC is still on and it could still move without proper knowledge of the engine.
Answer:
Glass, , is actually neither a liquid supercooled or otherwise nor a solid. When glass is made, the material often containing silica is quickly cooled from its liquid state but does not solidify when its temperature drops below its melting point
Explanation:
Answer:

Explanation:
The heat lost by the water in the cooling process is transferred to the muscles. Therefore, we must calculate this water lost heat, which is defined as:

Where m is the water's mass, c is the specific heat capacity of the water and
is the change in temperature. Replacing the given values:

Answer:
1.25L
Explanation:
Applying Gay-Lussac's Law which states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant.

Hence the volume at 4.0atm is 1.25L
Answer:
a) flow must be doubled
, b) Ф₂ =∛2² Ф₀, c) Ф does not change
, d) flow is constant
, e) flow is constant
Explanation:
For this exercise let's use Gauss's law, which states that the flow is equal to the waxed load divided by the permissiveness of the vacuum
Ф = E. dA =
/ ε₀
Let's apply this equation
a) if the charge inside the Gaussian surface is doubled by the equation the flow must be doubled
Ф₀ =
/ ε₀
Ф₂ = 2
/ ε₀ = 2 Ф₀
b) the volume is V₂ = 2V₀. Area is the volume divided by the length
V = LLL = L³
A = LL = L²
A = V / L
V₂ = 2V₀
L³ = 2 L₀³
L = ∛2 L₀
V₂ = A₂ L
A₂ = V₂ / L
A₂ = 2V₀ / L = 2 L₀³ / (∛2 L₀)
A₂ = 2 /∛2 V₀ / L₀
A₂ = ∛2² A₀
Ф₀ = E A₀
Ф₂ = E A₂
Ф₂ = E ∛2² A₀
Ф₂ =∛2² Ф₀
The flow doubles when the volume is increased by the factor 
c) the flow does not depend on the shape of the surface, so Ф does not change
d) the flow does not depend on the position of the load, while inside the surface the flow is constant
e) when the load leaves the surface, the load inside drops to ero, so the flow changes to zero
Ф = 0