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Marizza181 [45]
3 years ago
7

¿Es posible denominar al creacionismo como modelo científico?

Physics
1 answer:
algol133 years ago
5 0

Answer:

No

Explanation:

El creacinismo se basa en dogmas de fe y creencias que van de la mano con la religion. Mientras que el metodo cientifico se basa en el analisis de fenomenos naturales donde se deben demostrar por metodos matematicos y fisicos el comportamiento de dichos fenomenos.

De esta manera el creacinismo no se relaciona con el metodo cientifico ya que mientras este se basa en la fe como garantia de verdad. El metodo cientifico se basa en una comprobacion fisica-matematica que justifique dicho acontecimiento.

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If you could travel with speed of light (3.0e5 km/s) it would take only 2.5 minutes to reach venus. how far is venus from earth
motikmotik
D=S/T ,S=3.0e5km/s × 2.5min×60
D=4.5×e6km
7 0
3 years ago
A nuclear explosion may release tremendous amounts of energy in the form of noise, heat, visible light, radiation and an atmosph
Nina [5.8K]
Think about how each of noise, heat, visible light, radiation and atmospheric shock waves travel. Which ones require air particles to travel?

A vacuum has no air particles within it, it is completely empty.

Therefore, any of the above that requiring particles to travel will not be able to cross it and the observer will not experience it.
8 0
2 years ago
a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic
algol13

Answer:

Approximately 281.25\; \rm m. (Assuming that the drag on this ball is negligible, and that g = 10\; \rm m \cdot s^{-2}.)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

  • Horizontal: no acceleration, velocity is constant (at v(\text{horizontal}) is constant throughout the descent.)
  • Vertical: constant downward acceleration at g = 10\; \rm m \cdot s^{-2}, starting at 0\; \rm m \cdot s^{-1}.

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: x(\text{horizontal}) = 30\; \rm m. Combine these two quantities to find the duration of this descent:

\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}.

In other words, the ball in this question start at a vertical velocity of u = 0\; \rm m \cdot s^{-1}, accelerated downwards at g = 10\; \rm m \cdot s^{-2}, and reached the ground after t = 7.5\; \rm s.

Apply the SUVAT equation \displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t to find the vertical displacement of this ball.

\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}.

In other words, the ball is 281.25\; \rm m below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be 281.25\; \rm m\!.

5 0
3 years ago
Part 1 :
dybincka [34]

Answer:

1. 218.55 N

2. 30.96^{o}

3. 2.1 m/s^{2}

Explanation:

Part 1;

Net force F=mg sin \theta where m is mass, g is gravitational force and \theta is the angle of inclination

F= 46*9.8*sin 29^{o}= 218.55N

Frictional force, F_{r} is given by

F_{r} = \mu_{s}mg cos \theta where \mu_{s} is the coefficient of static friction

F_{r} = 0.6*46*9.8 cos 29

F_{r}=236.57N

Since F_{r}>F, therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force F_{s} = \mu_{s} mg cos \theta

mg sin \theta =\mu_{s} mg cos \theta

\mu_{s}= \frac {sin \theta}{cos \theta}

\mu_{s}= tan \theta

The maximum angle of inclination \theta = tan^{-1} \mu_{s}

\theta = tan^{-1} (0.6)

\theta= 30.96^{o}  

        

Part 3:

Net force on the object is given by

ma = mg sin 38 - \mu_{k} mg cos 38 where \mu_{k} is the coefficient of kinetic friction

 a = g ( sin 38 - \mu_{k} cos 38 )

                 = 9.8 ( sin 38 - (0.51) cos 38 )

                = 2.1m/s^{2}

7 0
3 years ago
Vector B~ has x, y, and z components of 2.5, 3,
brilliants [131]

Answer:

magnitude:

formula: v=√(x)^2 + (y)^2 + (z)^2

√(2.5)^2 +(3)^2 + (7.8)^2

√76.09

=8.7

I hope this helped :)

7 0
2 years ago
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