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deff fn [24]
3 years ago
6

What is the difference between a neap and a spring tide in terms of size?

Physics
1 answer:
Mamont248 [21]3 years ago
7 0
Spring tides occur when the moon is either new or full, and the sun, the moon, and the Earth are aligned. ... neap tide- A tide in which the difference between high and low tide is the least. Neap tides occur twice a month when the sun and moon are at right angles to the Earth.
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Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de
marshall27 [118]
Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos: 

Altura o distancia recorrida: 40 m
Vo: 6 m/s 
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes. 

Entonces tenemos que:

Vf ^{2} -Vo ^{2} =2 x g x h

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2}  \\  \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2}  }  \\ Vf=28,64 m/s

Que tengas un buen día!
6 0
3 years ago
Consider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir.
Marina CMI [18]

Answer:

(P_1-P_2)=1913.31 N/m^2

Explanation:

given:

\frac{A_t}{A_1}=0.85

V_1=90 m/s

γ∞=1.23 kg/m^3

solution:

since outside pressure is atm pressure vaccum can be defined by (P_1-P_2)

V_1=√2(P_1-P_2)/γ∞[\frac{A_t}{A_1}^2-1]

(P_1-P_2)=1913.31 N/m^2

6 0
3 years ago
A wave has an amplitude of 0.2 cm.
Anettt [7]
The distance between the resting point and maximum height of the wave is 0.2 cm.

The amplitude is measured from the resting point up to the highest point of the wave.
3 0
3 years ago
How are the direction of the net force on an object and the direction of the object’s acceleration related?
Oksanka [162]

They're the same direction.

8 0
3 years ago
A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th
svp [43]

a) x(t)=2.0 sin (10 t) [m]

The equation which gives the position of a simple harmonic oscillator is:

x(t)= A sin (\omega t)

where

A is the amplitude

\omega=\sqrt{\frac{k}{m}} is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s

The amplitude, A, can be found from the maximum velocity of the spring:

v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m

So, the equation of motion is

x(t)= 2.0 sin (10 t) [m]

b)  t=0.10 s, t=0.52 s

The potential energy is given by:

U(x)=\frac{1}{2}kx^2

While the kinetic energy is given by:

K=\frac{1}{2}mv^2

The velocity as a function of time t is:

v(t)=v_{max} cos(\omega t)

The problem asks as the time t at which U=3K, so we have:

\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}

However, \frac{m}{k}=\frac{1}{\omega^2}, so we have

(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\

with two solutions:

\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s

\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s

c) 3 seconds.

When x=0, the equation of motion is:

0=A sin (\omega t)

so, t=0.

When x=1.00 m, the equation of motion is:

1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

T=2 \pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum. By using T=0.628 s, we find

L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






5 0
3 years ago
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