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3 years ago

A source from which organisms generally take elements is called a/an

Physics

2 answers:

Advocard [28]3 years ago

8 0

A source from which organisms generally take elements is called C. Exchange Pool.

lubasha [3.4K]3 years ago

3 0

Answer: C. Exchange pool.

Explanation:

The exchange pool can be define as the pool which includes the elements which are required to be exchanged between the abiotic and biotic components of the environment. This is the part of every biochemical cycle.

On the basis of the above explanation, an exchange pool is the source from which the organisms generally take elements so as to lead their life cycle. For example, the atmosphere is the exchange pool of gaseous elements such as water vapors, oxygen, carbon dioxide and nitrogen. The organisms take oxygen from the atmosphere for respiration and release out carbon dioxide which becomes the part of the atmosphere.

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A glass plate (n = 1.64) is covered with a thin, uniform layer of oil (n = 1.28). A light beam of variable wavelength from air i

Makovka662 [10]

Answer:

Explanation:

This is case of interference in thin films

for constructive interference in thin film the condition is

2μ t = (2n+1)λ/2 ; μ is refractive index of oil , t is thickness of oil , λ is wave length of light .

2 x 1.28 x t = λ/2 , if n = 0

2 x 1.28 x t = 605 /2

t = 118.16 nm .

the minimum non-zero thickness of the oil film required = 118.16 nm.

8 0

3 years ago

A 19.12 g mixture of Ca(NO3)2 and KCl is dissolved in 149 g of water. The freezing point of the solution was measured as −5.77 ∘

hichkok12 [17]

Answer:

The mass percentage of calcium nitrate is 31.23%.

Explanation:

Let the the mass of calcium nitrate be x and mass of potassium chloride be y.

Total mass of mixture = 19.12 g

x + y = 19.12 g..(1)

Mass of solvent = 149 g = 0.149 kg

Freezing point of the solution, $T_f$ = -5.77 °C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:

i = 3

i' = 2

Freezing point of water = T = 0°C

$\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC$

$\Delta T_f=i\times K_f\times m$

$Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}$

$5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})$

On solving we get:

$\frac{3x}{164 g/mol}+\frac{2x}{74.5 g/mol}=0.4622 mol$ ....(2)

Solving equation (1)(2) for x and y:

x =5.973 g

y = 13.147 g

Mass percent of $Ca(NO_3)_2$ in the mixture:

$\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%$

The mass percentage of calcium nitrate is 31.23%.

5 0

3 years ago

While traveling home at dusk, a motorcyclist gets on the highway and increases the combined mass (400 kg

alexdok [17]

The acceleration of one of those bugs is equal to 305mi/s.

<h3>Acceleration calculation</h3>

To calculate the insect's acceleration, the action and reaction force of the impact must be considered.

As the insect will hit the helmet, the force it hits is the same force it receives, so we can make the following expression:

$m_m \times a_m = m_b \times a_b$

$550 \times 0.0027 = 5 \times 10^{-3} \times a_b$

<em>Speed has been converted to miles per second</em>

$a_b = 305 mi/s$

So, the acceleration of one of those bugs is equal to 305mi/s.

Learn more about acceleration calculation: brainly.com/question/390784

8 0

2 years ago

A golf ball strikes a hard, smooth floor at an angle of 28.2 ° and, as the drawing shows, rebounds at the same angle. The mass o

WITCHER [35]

Answer:

I = 1.06886 N s

Explanation:

The expression for momentum is

I = F t = Δp

therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor

Let's find the components of the initial velocity

sin 28.2 = v_y / v

cos 28.2= vₓ / v

v_y = v sin 282

vₓ = v cos 28.2

v_y = 42.8 sin 28.2 = 20.225 m / s

vₓ = 42.8 cos 28.2 = 37.72 m / s

since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making

θ = -28.2

v_y = -20.55 m / s

v_x = 37.72 m / s

X axis

Iₓ = Δpₓ = $p_{fx} - p_{ox}$

since the ball moves in the x-axis without changing the velocity, the change in moment must be zero

Δpₓ = m $v_{fx}$ - m v₀ₓ = 0

v_{fx} = v₀ₓ

therefore

Iₓ = 0

Y axis

I_y = Δp_y = p_{fy} -p_{oy}

when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards

v_{fy} = - v_{oy}

Δp_y = 2 m v_{oy}

Δp_y = 2 0.0260 (20.55)

$\Delta p_{y}$ = 1.0686 N s

the total impulse is

I = Iₓ i ^ + I_y j ^

I = 1.06886 j^ N s

7 0

3 years ago

An 1120 kg car traveling at 17.2 m/s is brought to a stop while skidding 40m. Calculate the work done on the car by the friction

Nesterboy [21]

Answer:

Work = 165670.4 J = 165.67 KJ

Explanation:

First, we will find the deceleration of the car, using the 3rd equation of motion:

$2as = v_{f}^2 - v_{i}^2\\$

where,

a = deceleration = ?

s = skid distance = 40 m

vf = final speed = 0 m/s

vi = initial speed = 17.2 m/s

Therefore,

$2a(40\ m) = (0\ m/s)^2 - (17.2\ m/s)^2\\a = - 3.698\ m/s^2$

the negative sign indicates deceleration here.

Now, we will calculate the braking force applied by the brakes on the car:

$F = ma\\F = (1120\ kg)(-3.698\ m/s^2)\\F = - 4141.76\ N$

the negative sign indicates braking force.

Now, we will calculate the work done using the magnitude of this force:

$Work = |F|s\\Work = (4141.76\ N)(40\ m)\\$

8 0

3 years ago