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Semenov [28]
3 years ago
5

what is a hypothesis reffered to as after being verified by a large number or independent experiments

Physics
1 answer:
Iteru [2.4K]3 years ago
7 0

Answer:

The hypothesis may or may not be true and needs to be tested. It might be the answer to the problem. Hence, it must be tested thoroughly. When these predictions are tested again and again in independent scientific experiments and gets verified, the hypothesis is converted into a scientific theory.

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A ferry coming into port is sailing at 12 m/s. It takes 2.5km to come to rest in the port. Calculate the deceleration of the fer
pogonyaev

Answer:    -0.0288 m/s^2

Explanation:

Let's suppose that the ferry decelerates at a constant rate A (deceleration is an acceleration in the opposite direction to the original motion of an object)

Then the acceleration equation of the ferry will be:

a(t) = -A

(the negative sign is because this acceleration is in the opposite direction with respect to the movement of the ferry)

To get the velocity equation of the ferry, we need to integrate with respect to the time, t, we will get:

v(t) = -A*t + v0

where v0 is the initial velocity of the ferry, v0 = 12m/s.

v(t) = -A*t + 12m/s

For the position equation of the ferry we need to integrate again over time:

p(t) = (-A/2)*t^2 + (12m/s)*t + p0

Where p0 is the initial position of the ferry, in this case, it can be zero, because it will depend on where we put the origin on our coordinate axis.

then p0 = 0m

P(t) = (-A/2)*t^2 + (12m/s)*t

The ferry will come to rest at the moment when it's velocity is equal to zero, this will happen when:

v(t) = 0m/s = -A*t + 12m/s

We need to find the value of t.

A*t = 12m/s

t = (12m/s)/A

Now we can replace this in the position equation because we know that the ferry needs 2.5 km or 2500 meters to come to rest.

p(  (12m/s)/A) = 2500m =  (-A/2)*( (12m/s)/A)^2 + (12m/s)*((12m/s)/A)

2500m = (-72 m^2/s^2)/A + (144m^2/s^2)/A

2500m = (72 m^2/s^2)/A

2500m*A = (72 m^2/s^2)

A = (72 m^2/s^2)/2500m = 0.0288 m/s^2

and the acceleration of the ferry was -A, then the acceleration of the ferry is:

-0.0288 m/s^2

4 0
3 years ago
What is the name of the hair-like structures on sponge cells that move back and forth to help move water, nutrients, and waste t
Dima020 [189]
I think it's B. Flagella

5 0
3 years ago
Read 2 more answers
La tension que se transmite en la cuerda BD es de 75 lb. Calcula el momento de fuerza generada por la cuerda respecto al punto C
Bas_tet [7]

Answer:

Mc = 1920[lb*in]

Explanation:

Para poder solucionar este problema debemos realizar un análisis estático, por tal motivo lo primero es realizar un diagrama de cuerpo libre con las respectivas fuerzas actuando sobre la barra ABC. DE igual manera calcular la geometría de la configuración mostrada.

El diagrama de cuerpo libre se puede ver en la imagen adjunta, con la solución de este problema.

Lo primero es determinar el angulo t, el cual por medio de las propiedades del triangulo rectángulo se puede determinar.

Con este angulo (t) ya determinado, fijamos la atención en el triangulo BCD, este triangulo no es rectángulo, pero por medio de la ley de senos podemos determinar el angulo omega.

Después de determinar el angulo omega, restamos el angulo (t) para poder determinar el angulo (a).

Seguidamente realizamos una sumatoria de momentos alrededor del punto C, utilizado las respectivas fuerzas con los ángulos descompuestos.

El momento en el punto C es de 1920 [Lb*in].

Nota: ya que no se menciona la fuerza en el punto A, esta se desprecia y no se tiene en cuenta en los calculos. En la imagen adjunta se puede ver el procedimiento desarrollado.

7 0
3 years ago
In a laboratory, you determine that the density of a certain solid is 5.23× 10 −6 kg/m m 3 . convert this density into kilograms
tamaranim1 [39]

Density is given as

\rho = 5.23 * 10^{-6} kg/mm^3

now we have to convert this density into kg/m^3

now we have

1 m = 1000 mm

\rho = 5.23 * 10^{-6} \frac{kg}{(1*10^-3 m)^3}

\rho = 5.23 * 10^{-6} \frac{1*10^9 kg}{m^3}

\rho = 5.23 * 10^3 kg/m^3

\rho = 5230 kg/m^3

8 0
3 years ago
A spring of original length 3.0 cm is extended to a total length of 5.0 cm by a force of 8.0N
hjlf
4848484 5 5 5 5 5  =++++
6 0
3 years ago
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