Which of these describe life sustaining functions that all organisms and cells perform\

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

Paul [167]

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.

6 0

3 years ago

In an incompressible three-dimensional flow field, the velocity components are given by u = ax + byz; υ = cy + dxz. Determine th

Ksivusya [100]

An incompressible flow field F in a 3D cartesian grid with components u,v,w:

F = u + v + w

where u,v,w are functions of x,y,z

Must satisfy:

∇·F = du/dx + dv/dy + dw/dz = 0

We have a field F defined:

F = u+v+w, u = ax+byz, v = cy+dxz

du/dx = a, dv/dy = c

Recall ∇·F = 0:

∇·F = du/dx + dv/dy + dw/dz = 0

a + c + dw/dz = 0

dw/dz = -a-c

Solve for w by separation of variables:

w = ∫(-a-c)dz

w = -az - cz + f(x,y)

f(x,y) is some undetermined function of x and y

The question states that w is not a function of x and y, therefore f(x,y) = 0...

w = -az - cz

8 0

3 years ago

The gravitational force between two objects has a magnitude of F. If both masses were doubled and the distance between them doub

Fofino [41]

Answer:

F' = F

Explanation:

The gravitational force of attraction between two objects can be given by Newton's Gravitational Law as follows:

$F = \frac{Gm_1m_2}{r^2}$

where,

F = Force of attraction

G = Universal gravitational costant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects

Now, if the masses and the distance between them is doubled:

$F' = \frac{G(2m_1)(2m_2)}{(2r)^2}\\\\F' = \frac{Gm_1m_2}{r^2}$

7 0

3 years ago

The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t

jek_recluse [69]

Answer:

$e=3367.2J$

% $e=1.43$ %

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

$v_{r}=10km/h=2.77m/s$

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

$v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s$

Knowing that we can calculate Kinetic energy for the real and experimental speed

$E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J$

$E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J$

Now, the potential error in her calculated kinetic energy is:

$e=E_{r}-E_{e}=(234023-230656)J=3367.2J$

% $e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43$ %

4 0

4 years ago

A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c

satela [25.4K]

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

Given the data in the question;

First course : $a_{x}$ = 0.75 m/s², $d_{x}$ = 20 m, $u_{x}$ = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

$v_{x}$ ² = (10)² + (2 × 0.75 × 20)

$v_{x}$ ² = 100 + 30

$v_{x}$ ² = 130

$v_{x}$ = √130

$v_{x}$ = 11.4 m/s

for the Second Course:

$u_{y}$ = 11.4 m/s, $a_{y}$ = -1.15 m/s², $v_{y}$ = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) × $d_{y}$ )

0 = 129.96 - 2.3 $d_{y}$

2.3 $d_{y}$ = 129.96

$d_{y}$ = 129.96 / 2.3

$d_{y}$ = 56.5 m

so;

|d| = √( $d_{x}$ ² + $d_{y}$ ² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

7 0

3 years ago