Answer:
<u>Our beaches would be unprotected</u>
In the short-term, these artificial sand hills will be destroyed by the elements. Because sand dunes protect inland areas from swells, tides, and winds, they must be protected and defended like national treasures. ... The ocean and the wind can have an unpredictable, destructive force on coastal regions.
- surfertoday
Natural sand dunes play a vital role in protecting our beaches, coastline and coastal developments from coastal hazards such as erosion, coastal flooding and storm damage. Sand dunes protect our shorelines from coastal erosion and provide shelter from the wind and sea spray.
- Waikato Regional Council
Answer:
B
Explanation:
the electrical charges of nonpolar molecules are evenly distributed across the molecule
Answer:
If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.
Explanation:
Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure
V ∝ 1/P or V₁·P₁ = V₂·P₂
Where:
V₁ = Initial volume
V₂ = Final volume = V₁/2
P₁ = Initial pressure = 832 torr
P₂ = Final pressure = Required
From V₁·P₁ = V₂·P₂ we have,
P₂ = V₁·P₁/V₂ = V₁·P₁/(V₁/2)
P₂ = 2·V₁·P₁/V₁ = 2·P₁ = 2× 832 torr = 1664 torr
Answer:
The answer is 2i on right hand side.
Explanation:
We should star by checking the equation from right.
First we check how many Zn r there in left hand side. Which is 1. Let us check how many Znr there in right hand side, there is 1.So Zn is balanced, and don't worry about Znplus2 on right hand side it is just the ions not how many zinc r there.
Now let us check how many I are there left hand side. Which is 2. Now how many I are there in right hand side? Only 1.
So we put 2 behind I.
Now there r 2 I on both sides.
Its simple actually.
For this problem, we use the formula for sensible heat which is written below:
Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference
Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
