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2 years ago

12

Halle la velocidad de las ondas de una cuerda de violin de 820 g/m y 22cm de longitud, si la frecuencia de la fundamental es de

920 hz. Calcule la tensión de la cuerda

1 answer:

Mice21 [21]2 years ago

7 0

Answer:

your mom B on thezz NUTZ

Explanation:

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Un cuerpo se lanza verticalmente hacia arriba con una velocidad de 13 m/s. ¿Cuánto tiempo tarda en alcanzar la altura máxima? a)

Elena L [17]

Answer:

D. 1.33 segundos.

Explanation:

El cuerpo es experimenta un movimiento en caída libre al modificarse su velocidad por efecto de la gravitación terrestre. Este cuerpo alcanza instantáneamente el reposo cuando se encuentra a su altura máxima, el tiempo puede obtenerse sabiendo la aceleración y las velocidades incial y final a partir de la siguiente ecuación cinemática:

$v = v_{o}+g\cdot t$

Donde:

$v$ - Velocidad final del cuerpo, medida en metros por segundo.

$v_{o}$ - Velocidad inicial del cuerpo, medida en metros por segundo.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$t$ - Tiempo, medido en segundos.

Ahora se despeja el tiempo:

$t = \frac{v-v_{o}}{g}$

Si $v_{o} = 13\,\frac{m}{s}$ , $v=0\,\frac{m}{s}$ y $g = -9.807\,\frac{m}{s^{2}}$ , entonces:

$t = \frac{0\,\frac{m}{s}-13\,\frac{m}{s}}{-9.807\,\frac{m}{s^{2}} }$

$t = 1.326\,s$

Por ende, la respuesta correcta es D.

6 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.6 m/s at ground level.

Dimas [21]

Answer:

The rocket is motion above the ground for 44.7 s.

Explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engine fails:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 = initial height

v0 = initial speed

t= time

a = acceleration due to the engines

g = acceleration due to gravity

First, let´s calculate the time the rocket is being accelerated by the engines:

(The center of the framer of reference is located at the ground, y0 = 0).

y = y0 + v0 · t + 1/2 · a · t²

1190 m = 0 m + 79.6 m/s · t + 1/2 · 4.10 m/s² · t²

0 = 2.05 m/s² · t² + 79.6 m/s · t - 1190 m

Solving the quadratic equation:

t = 11.5 s (the other value of t is discarded because it is negative).

At that time, the engines fail and the rocket starts to fall. The equation of the height will be:

y = y0 + v0 · t + 1/2 · g · t²

The initial velocity (v0) will be the velocity acquired during 11.5 s of acceleration plus the initial velocity of launch:

v = v0 + a · t

v = 79.6 m/s + 4.10 m/s² · 11.5 s

v = 126.8 m/s

Now, we can calculate the time it takes the rocket to reach the ground (y = 0) from a height of 1190 m:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 1190 m + 126.8 m/s · t - 1/2 · 9.80 m/s² · t²

Solving the quadratic equation:

t = 33.2 s

Then, the total time the rocket is in motion is (33.2 s + 11.5 s) 44.7 s

5 0

3 years ago

A ray of laser light travels through air and enters an unknown material. The laser enters the material at an angle of 36 degrees

V125BC [204]

Answer:1.27

Explanation:

Given

incident angle $i=36^{\circ}$

refracted angle $r=27.5^{\circ}$

Suppose $n_2$ is the refractive index of material then using Snell's law we can write

$n_1\sin i=n_2\sin r$

where $n_1$ =refractive index of air

$1\times \sin (36)=n_2\times \sin (27.5)$

$n_2=\dfrac{0.5877}{0.4617}$

$n_2=1.27$

3 0

3 years ago

Four equal masses m are so small they can be treated as points, and they are equally spaced along a long, stiff wire of neglible

anyanavicka [17]

Answer: 5m/L^2

Explanation:

Inertial I = mr^2 where r = distance from axis of rotation, while m is the mass of the object.

I = 2[m(1L/2)^2] + 2[m(3L/2)^2] = 2m×. 25/L^2+ 3m×2. 25/L^2= 0. 5m/l^2 +4. 5m/l^2

= 5m/l^2.

8 0

3 years ago

A force of F = (2.00ˆi + 3.00ˆj) N is applied to an object that is pivoted about a fixed axle aligned along the z coordinate axi

Vladimir [108]

Explanation:

It is given that,

Force applied to object, $F=(2i+3j)\ N$

Position, $r=(4i+5j)\ m$

(b) The cross product of force and position vector is used to find the net torque about the z axis. It is given by :

$\tau=F\times r$

$\tau=(2i+3j) \times (4i+5j)$

$\tau=\begin{pmatrix}0&0&-2\end{pmatrix}$

or

$\tau=(-2k)\ N-m$

The torque is acting in -z axis.

(a) The magnitude of torque is given by :

$|\tau|=\sqrt{(-2)^2}$

$|\tau|=2\ N-m$

Hence, this is the required solution.

5 0

3 years ago