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3 years ago

Explain how competition between organisms affects the availability of resources in an area.

Physics

1 answer:

Fynjy0 [20]3 years ago

7 0

Answer:

Completion between two organism is when they fight over resources inadvertently causing the use and destruction of recourses making it more scares. if they don't adapt they die out.

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For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency

Angelina_Jolie [31]

Answer:

2.7 J

Explanation:

The energy of one photon is given by

$E=hf$

where

h is the Planck constant

f is the frequency

For the photons in this problem,

$f=6.8\cdot 10^9 Hz$

So the energy of one photon is

$E_1=(6.63\cdot 10^{-34})(6.8\cdot 10^9 )=4.5\cdot 10^{-24} J$

The number of photons contained in 1.0 mol is

$N_A = 6.022\cdot 10^{23} mol^{-1}$ (Avogadro number)

So the total energy of $N_A$ photons contained in 1.0 mol is

$E=N_A E_1 =(6.022\cdot 10^{23})(4.5\cdot 10^{-24})=2.7 J$

3 0

3 years ago

What is the volume of a 1.2kg and displaced 1.0g/cm3

Zinaida [17]

Mass = 1.2 kg = 1200 grams.

Volume = mass/density = 1200 cm3.

Hope this helps!

4 0

3 years ago

Plz help me to find my science matching... <br>

s2008m [1.1K]

Answer:

fibrous =potato

taproot =radish

stilt =maize and sugar cane

5 0

3 years ago

A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9

11111nata11111 [884]

Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

$m_{1} v_{1}=m_{2} v_{2}$

$\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }$

$\mathrm{v}_{1} \text { velocity before the collision }$

$\mathrm{v}_{2} \text { velocity after the collision }$

$\mathrm{m}_{1}=600 \mathrm{kg}$

$\mathrm{m}_{2}=900 \mathrm{kg}$

$\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}$

$600 \times 5=900 \times v_{2}$

$3000=900 \times v_{2}$

$\mathrm{v}_{2}=\frac{3000}{900}$

$\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

5 0

4 years ago

Carlos pushes a 3 kg box with a force of 9 newtons. The force of friction on the box is 3 newtons in the opposite direction. Wha

Ivenika [448]

The 'net' force acting on the box is (9 - 3) = 6 newtons
in the direction that Carlos is pushing.

Force = (mass) x (acceleration)

6 = (3) x (acceleration)

Divide each side by 3 :

<em>2 m/s² = acceleration</em>

4 0

3 years ago

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