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Ber [7]
3 years ago
13

En el mar, la luz visible alcanza una profundidad de aproximadamente 200 metros (zona fotica). ¿De dónde obtienen energías las p

lantas y los animales por debajo de esa profundidad?
Physics
1 answer:
Marizza181 [45]3 years ago
3 0

Answer:

Explanation:

En la zona apótica (profundidad inferior a 200 m); todo lo que queda de la luz solar es una luz tenue, opaca, azul-verde, demasiado impotente para siquiera considerar permitir que ocurra la fotosíntesis. Sin embargo, hay comida para tener; basura, trozos de plantas podridas y derroche de criaturas caen desde arriba para cuidar a los seres vivos en la zona apótica.

Las formas de vida a una profundidad inferior a 200 m dependen de los productos químicos que salen de los respiraderos; el procedimiento que utilizan para hacer los alimentos se llama quimiosíntesis en lugar de fotosíntesis.

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The intensity of light from a star varies inversely as the square of the distance. If you lived on a planet ten times farther aw
frosja888 [35]

Answer:

the intensity of the sun on the other planet is a hundredth of that of the intensity of the sun on earth.

That is,

Intensity of sun on the other planet, Iₒ = (intensity of the sun on earth, Iₑ)/100

Explanation:

Let the intensity of light be represented by I

Let the distance of the star be d

I ∝ (1/d²)

I = k/d²

For the earth,

Iₑ = k/dₑ²

k = Iₑdₑ²

For the other planet, let intensity be Iₒ and distance be dₒ

Iₒ = k/dₒ²

But dₒ = 10dₑ

Iₒ = k/(10dₑ)²

Iₒ = k/100dₑ²

But k = Iₑdₑ²

Iₒ = Iₑdₑ²/100dₑ² = Iₑ/100

Iₒ = Iₑ/100

Meaning the intensity of the sun on the other planet is a hundredth of that of the intensity on earth.

3 0
4 years ago
A machine does 500 j of work in 20 sec. What is the power of this machine?
NNADVOKAT [17]

Explanation:

P=W/t

P=500/20

P=25 W

6 0
3 years ago
QUESTION 10
Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

\theta =37.01^{\circ}

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

T_{1}sin(\theta)-T_{2}sin(\theta)=0    (1)

Where:

  • T(1) is the tension due to the rope 1
  • T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

T_{1}cos(\theta)+T_{2}cos(\theta)-W=0   (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)=m_{steel}g

T(1) must be equal to 5479 N, so we have:

cos(\theta)=\frac{m_{steel}g}{2T_{1}}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=0.80

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

8 0
3 years ago
A golf club with 65J of kinetic energy strikes a stationary golf ball with a mass of 46g. The energy transfer is only 20% effici
umka21 [38]
Kinetic energy of golf club = 65J, 
kinetic energy supplied to golf ball = 20% of 65 = 0.2 * 65 = 13J,
kinetic energy of ball = [mass * Velocity²]/2,
mass = 46gm = 0.046Kg,
[0.046 * V²]/2 = 13, or 0.046 *V² = 26, 
V² = 26/0.046 = 565.22, 
V = 23.77 m/sec = initial velocity of golf ball after hitting.
4 0
3 years ago
Neglecting air resistance, what maximum height will be reached by an arrow launched straight upward with an initial speed of 35
tankabanditka [31]
The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.

The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)

Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
3 0
3 years ago
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