Ask question

Ask question

All categories

3 years ago

12

Un movil pasa por el punto A en direccion hacia B (350cm más adelante) y, luego, sigue hasta el punto C. Sabiendo que pasa por B

a las 11:42:38 y por C a las 11:43:16, completando un tiempo total de recorrido de 3min 47s, calule la distancia entre B y C y a la hora a la que paso por el punto A.

1 answer:

xeze [42]3 years ago

6 0

Explanation:

PRIMERO HACES EL RECUENTO DEL TIEMPO Y LO CONVIERTES EN

SEGUNDOS Y ENTONCES

<em>t</em> = 227 s $t_{AB}$ = 227 S - 38 s = 189 s

$t_{BC}$ = 38 s

LUEGO USANDO LA ECUACIÓN DE GALILEO GALILEI SSUPONIENDO

QUE EL MOVIL VIAJA A VELOCIDAD CONSTANTE

<em>v</em> = 3.50 m/189 s = 0.0185 m/s

PARA LA DISTANCIA NTRE B Y C

$x_{BC}$ = 0.0185 m/S( 38 s) = 0.703 m

LA HORA EN QUE EL MOVIL PASA POR A ES

11:43:15 - 38 s - 189 s = 11:39:29

You might be interested in

If a proton and an electron are released when they are 2.50×10^-10m apart (typical atomic distances), find the initial accelerat

katrin [286]

To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

$F = k \frac{q_1q_2}{r^2}$

Here,

k = Coulomb's constant

q = Charge of proton and electron

r = Distance

Replacing we have that,

$F = (9*10^9)(\frac{(1.602*10^{-19})^2}{2.5*10^{-10}})$

$F = 3.6956*10^{-9}N$

The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.

The acceleration of the electron is given as

$a_e = \frac{F}{m_e}$

$a_e = \frac{3.6956*10^{-9}}{9.11*10^{-31}}$

$a_e = 4.0566*10^{21}m/s^2$

The acceleration of the proton is given as,

$a_p = \frac{F}{m_p}$

$a_p = \frac{3.6956*10^{-9}}{1.672*10^{-27}}$

$a_p = 2.21*10^{18}m/s^2$

3 0

3 years ago

Which of the following consists of photovoltaic cells?

Charra [1.4K]

The answer is number 1 solar panels

7 0

2 years ago

Why are citizens obligated to respond to such documents?

Zina [86]

To protect the constitutional right to confront ones accused

5 0

3 years ago

A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

3 years ago

Charles' law explains which of these phenomena?

Luden [163]

I think the correct answer would be that Charles' law explains why <span>a balloon deflates when the air around it cools. Charles' law is a simplification of the ideal gas law. At constant pressure, volume and temperature have a direct relationship. Hope this helps.</span>

5 0

3 years ago