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koban [17]
3 years ago
13

The density of ice can help preserve the habitats of aquatic organisms, but it can also cause the death of an organism. Which st

atement describes how this could happen?
A. The contraction of water as it freezes increases the amount of nutrients that can dissolve in liquid water, but it could cause cells of an organism to shrink.

B. The expansion of water as it freezes causes water to become solid from the top down, but it could cause cells of an organism to burst.

C. The contraction of water as it freezes causes water to become solid from the top down, but it could cause fluid in cells to dissolve fewer helpful substances.

D. The expansion of water as it freezes increases the amount of nutrients that can dissolve in liquid water, but it could cause fluid in cells to dissolve more harmful substances.
Physics
1 answer:
Paul [167]3 years ago
6 0

Answer:

I think it's D. The expansion of water as it freezes increases the amount of nutrients that can dissolve in liquid water, but it could cause fluid in cells to dissolve more harmful substances.

Explanation:

I know when water freezes, it expands and between the two answers that discuss the expansion of water, D sounds the most logical to me lol.

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3 years ago
What are the laws of physics?
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4 years ago
Ten identical steel wires have equal lengths L and equal "spring constants" k. The Young's modulus of each wire is Y. The wires
svlad2 [7]

Answer:

option (B)

Explanation:

Young's modulus is defined as the ratio of longitudinal stress to the longitudinal strain.

Its unit is N/m².

The formula for the Young's modulus is given by

Y=\frac{F \times L}{A\times \delta L}

where, F is the force applied on a rod, L is the initial length of the rod, ΔL is the change in length of the rod as the force is applied, A is the area of crossection of the rod.

It is the property of material of solid. So, when the 10 wires are co joined together to form a new wire of length 10 L, the material remains same so the young' modulus remains same.

8 0
3 years ago
g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the
Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

T^2 = \frac{4\pi^2 d^3}{2GM}

The given data are given with respect to known constants, for example the mass of the sun is

m_s = 1.989*10^{30}

The radius between the earth and the sun is given by

r = 149.6*10^9m

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

m = 8.2*1.989*10^{30}

d = 6.2*149.6*10^6

Substituting in Kepler's third law:

T^2 = \frac{4\pi^2 d^3}{2}

T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}

T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}

T = 120290789.7s

T = 120290789.7s(\frac{1year}{31536000s})

T \approx 3.8143 years

Therefore the period of this star is 3.8years

7 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
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