What potential difference is dropped over the 60 ohms resistor

The cat righting reflex is a cat's innate ability to orient itself as it falls in order to land on its feet. The righting reflex

Inessa [10]

Answer:

The solution is shown in the picture attached below

Explanation:

3 0

3 years ago

Communication with submerged submarines via radio waves is difficult because seawater is conductive and absorbs electromagnetic

REY [17]

Answer:

$\lambda=4000\ km$

Explanation:

It is given that,

Frequency for submarine communications, f = 76 Hz

We need to find the wavelength of those extremely low-frequency waves. the relation between the wavelength and the frequency is given by :

$c=f\times \lambda$

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{3\times 10^8\ m/s}{76\ Hz}$

$\lambda=3947368.42\ m$

$\lambda=3947\ km$

or

$\lambda=4000\ km$

So, the wavelength of those extremely low-frequency waves is 4000 km. Hence, this is the required solution.

4 0

3 years ago

HELP ITS DUE IN 4 MINUTES

photoshop1234 [79]

The answer is pangea

5 0

3 years ago

Read 2 more answers

Es muy común que cuando se viaja hacia un río o lago se juegue "ranita", el cual consiste en lanzar una piedra horizontalmente h

shutvik [7]

Answer:

a) La piedra es lanzada desde una altura de 0,785 metros.

b) La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

Explanation:

a) Dado que la piedra es lanzada horizontalmente, tenemos que la piedra experimenta un movimiento horizontal a velocidad constante y uno vertical uniformemente acelerado debido a la gravedad. La altura de la que fue lanzada la piedra se puede determinar mediante la siguiente ecuación cinemática:

$y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Donde:

$y$ - Altura final, medida en metros.

$y_{o}$ - Altura inicial, medida en metros.

$v_{o,y}$ - Componente vertical de la velocidad inicial, medida en metros por segundo.

$t$ - Tiempo, medido en segundos.

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

Si sabemos que $y = 0\,m$ , $v_{o,y} = 0\,\frac{m}{s}$ , $t = 0,4\,s$ y $g = -9,807\,\frac{m}{s^{2}}$ , entonces la altura inicial de la piedra es:

$y_{o} = y-v_{o,y}\cdot t -\frac{1}{2}\cdot g\cdot t^{2}$

$y_{o} = 0\,m-\left(0\,\frac{m}{s} \right)\cdot (0,4\,s)-\frac{1}{2}\cdot \left(-9,807\,\frac{m}{s^{2}} \right) \cdot (0,4\,s)^{2}$

$y_{o} = 0,785\,m$

La piedra es lanzada desde una altura de 0,785 metros.

b) Ahora, obtenemos el componente horizontal de la velocidad inicial a partir de la siguiente ecuación cinemática:

$v_{o,x} = \frac{x-x_{o}}{t}$ (2)

Donde:

$x_{o}$ , $x$ - Posiciones horizontales iniciales y finales, medidas en metros.

$t$ - Tiempo, medido en segundos.

Si tenemos que $x_{o} = 0\,m$ , $x = 2,5\,m$ y $t = 0,4\,s$ , entonces el componente horizontal de la velocidad inicial es:

$v_{o,x} = \frac{2,5\,m-0\,m}{0,4\,s}$

$v_{o,x} = 6,25\,\frac{m}{s}$

La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

4 0

3 years ago

A baseball is hit almost straight up into the air with a speed of 26 m/s . Estimate how high it goes.

Natalka [10]

Answer:

The maximum height of the ball is 34.5 m.

The ball is 5.31 s in the air.

Explanation:

Hi there!

The equations for the height and velocity of the baseball that is hit straight up are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the baseball at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t.

If we place the origin of the frame of reference at the place where the baseball is hit, then, y0 = 0.

To calculate how high it goes, we have to obtain the time at which the ball is at maximum height. At that point, the velocity is 0. Then using the equation of velocity:

v = v0 + g · t

0 = 26 m/s - 9.8 m/s² · t

-26 m/s / -9.8 m/s² = t

t = 2.65 s

The height at that time will be the maximum height:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

y = 26 m/s · 2.65 s - 1/2 · 9.8 m/s² · (2.65 s)²

y = 34.5 m

The maximum height of the ball is 34.5 m

If it takes the ball 2.65 s to reach the maximum height it will take another 2.65 s to return to the initial position. Then, the time it will be in the air is (2.65 s + 2.65 s) 5.30 s. However, let´s calculate the time it takes the ball to reach the initial position using the equation for height.

At the initial position y = 0. Then:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

0 = 26 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (26 m/s - 1/2 · 9.8 m/s² · t) (t = 0 when the ball is hit)

0 = 26 m/s - 1/2 · 9.8 m/s² · t

-26 / -4.9 m/s² = t

t = 5.31 s ( the difference with the 5.30 s obtained above is due to rounding the time to 2.65 s).

The ball is 5.31 s in the air.

Have a nice day!

5 0

4 years ago