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2 years ago

Con que nombre se conocieron los primeros gimnasio que se crearon

Physics

1 answer:

notsponge [240]2 years ago

7 0

Answer:

Los primeros gimnasios registrados datan de hace más de 3000 años en la antigua Persia, donde se los conocía como zurkhaneh, áreas que fomentaban la aptitud física.

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Suppose that during any period of ¼ second there is one instant at which the crests or troughs of component waves are exactly in

kolezko [41]

<h3>Answer;</h3>

A. 4

<h3>Explanation;</h3>

The period of a wave or periodic time is the time taken for a complete oscillation to occur. For example its is the time taken between two successive crests or troughs.
The beats or oscillation that occur in one second represents the frequency. Frequency is the number of complete oscillations or beats in one second in a wave.
Frequency, measured in Hertz is given by the reciprocal of the periodic time.
Thus; Frequency or beats per second = 1/(1/4) = 4
Hence , 4 beats per second

6 0

3 years ago

A generator makes electricity from _____ A. Chemical reactions B. Friction C. Kinetic energy D. Heat

coldgirl [10]

Answer:

The answer is C. Kinetic energy

Explanation:

Generators don't actually create electricity. Instead, they convert mechanical or chemical energy into electrical energy. They do this by capturing the power of motion and turning it into electrical energy by forcing electrons from the external source through an electrical circuit.

7 0

3 years ago

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The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago

(1 point) find the half-life (in hours) of a radioactive substance that is reduced by 1010 percent in 9595 hours.

Bad White [126]

The decay function is of the form
$N(t) = N_{0} \, e^{-kt}$
where
N₀ = initial amount
k = decay constant
t = hours

The material decays by 10% in 95 hours. Therefore
$0.9N_{0} = N_{0} \, e^{-95k} \\\\ -95k = ln(0.9) \\\\ k= \frac{ln(0.9)}{-95}=0.001109$

The time for the half life is given by
$0.5N_{0} = N_{0} \, e^{-0.001109t} \\\\ -0.0001109t = ln(0.5) \\\\ t = \frac{ln(0.5)}{-0.001109} = 625 \, h$

Answer: The half life is 625 hours

8 0

3 years ago

Explain the phases of matter.

Ludmilka [50]

Answer:

liquid, gas, solid, and plasma.

Explanation:

8 0

3 years ago

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Con que nombre se conocieron los primeros gimnasio que se crearon​

Con que nombre se conocieron los primeros gimnasio que se crearon