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Bess [88]
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.
Physics
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

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Anuta_ua [19.1K]

The term Freebody diagram refer to the diagram that could be drawn to show the forces that act on an object.

<h3>What is a freebody diagram?</h3>

The term Freebody diagram refer to the diagram that could be drawn to show the forces that act on an object. We must note that the forces that act on a body determine the direction in which the body moves.

This is because, force is a  vector quantity. As such the magnitude and direction of a force are both responsible when we are trying to determine the force that acts on the object.

Attached is the image of the upward motion of  a vector that shows the forces that act on the vector. This is what we refer to as a Freebody diagram.

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4 0
1 year ago
How long is a football field?
Nana76 [90]

Answer:

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7 0
3 years ago
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A 15.00 kg particle starts from the origin at time zero. Its velocity as a function of time is given by = 8t2î + 5tĵ where is in
Elden [556K]

Answer:

Explanation:

I will assume the equation reads:

v = 8t²î + 5tĵ

The velocity v is the time derivative of the position x.

x = \int\limits^t_0 {v} \, dt = \int\limits^t_0 {8t^{2}\hat i + 5t\hat j} \, dt = \frac{8}{3} t^{3} \hat i + \frac{5}{2}t^{2}\hat j |^t_0 = \frac{8}{3} t^{3} \hat i + \frac{5}{2}t^{2}\hat j - \frac{8}{3} \hat i - \frac{5}{2} \hat j\\ x = \frac{8}{3} (t^{3} - 1 )\hat i + \frac{5}{2} (t^{2} - 1 )\hat j

4 0
4 years ago
Students are given the following information about a 2.0 kg, motorized toy boat on water. what net force is exerted on the boat
Natasha_Volkova [10]

The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.

The motorized toy boat moves at 8 m/s (u) at 4 s and at 8 m/s (v) at 8 s. We can calculate the acceleration (a) in that period using the following kinematic expression.

a = \frac{v-u}{\Delta t} = \frac{8m/s-8m/s}{8s-4s} = 0 m/s^{2}

The object with a mass (m) of 2.0 kg experiences an acceleration of 0 m/s². We can calculate the net force (F) in that period using Newton's second law of motion.

F = m \times a = 2.0 kg \times 0 m/s^{2} = 0 N

The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.

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6 0
3 years ago
Faraday's law states that voltage can be changed by moving the coil out of the magnetic field true or false
NNADVOKAT [17]

As per Faraday's law of induction we know that induced EMF in a conducting closed loop is equal to rate of change in flux in that loop

So here we have

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Now this induced voltage will remain constant if coil is moved out uniformly

But it will not remain constant if coil is moved out with non uniform speed

So this statement is not always true

so answer must be

<u>FALSE</u>

8 0
3 years ago
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