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natita [175]
2 years ago
7

How much water is displaced by a solid object with a volume of 9 cm3 that is completely submerged in the water?

Physics
1 answer:
Sati [7]2 years ago
6 0

Given :

An object with a volume of 9 cm³ is completely submerged in the water.

To Find :

How much water is displaced by the object.

Solution :

According to Archimedes' principle, the amount of volume displaced of fluid when an object is fully and partially submerged into that is equal to the volume of the object below the surface of liquid.

Now, it is given that object is fully submerged into water.

Therefore, volume of water displaced by the object is 9 cm³.

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Describa con sus palabras que fue lo que descubrio galileo en su legendario experimento en la torre inclinada de pisa
german

Answer:

En 1589 Galileo realizó un experimento lanzando dos bolas de diferentes masas desde la famosa Torre Inclinada de Pisa para demostrar que el tiempo de caída es independiente de la masa de la bola. A través de este experimento, Galileo descubrió que los cuerpos caían casi simultáneamente, refutando la teoría de Aristóteles de que la tasa de caída era proporcional a la masa del cuerpo.  

Debido a la imperfección de los equipo de medición de esa época, la caída libre de los cuerpos era casi imposible de estudiar. En busca de una forma de reducir la velocidad de movimiento, Galileo reemplazó la caída libre por rodar sobre una superficie inclinada, donde había velocidades y resistencia del aire significativamente más bajas. Se notó que con el tiempo, la velocidad del movimiento aumenta: los cuerpos se mueven con aceleración. Se concluyó que la velocidad y la aceleración no dependen ni de la masa ni del material de la pelota.

 

5 0
3 years ago
A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp
Anna71 [15]

\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}

Explanation:

       Natural length of a spring is 0.5\text{ }m. The spring is streched by 0.02\text{ }m. The resultant energy of the spring is 0.5\text{ }J.

       The potential energy of an ideal spring with spring constant k and elongation x is given by \dfrac{1}{2}kx^{2}.

       So, in the current problem, the natural length of the spring is not required to find the spring constant k.

       \text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}

∴ The spring constant of the spring = 2500\text{ }\frac{kg}{s^{2}}

4 0
3 years ago
Help me pls I WILL GIVE BRAINIEST I DONT CARE PLS HELP
PilotLPTM [1.2K]

Answer:

A) was reusable

Explanation:

Check this website out for more information about the space shuttle: https://www.nasa.gov/audience/forstudents/k-4/stories/nasa-knows/what-is-the-space-shuttle-k4.html

3 0
2 years ago
Read 2 more answers
Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

starting point. Higher

         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

6 0
2 years ago
What type of electromagnetic radiation is being used in the picture?
pashok25 [27]

Answer: b i think

Explanation:

6 0
3 years ago
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