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charle [14.2K]
3 years ago
7

At a location near the equator, the earth’s magnetic field is horizontal and points north. An electron is moving vertically upwa

rd from the ground. What is the direction of the magnetic force that acts on the electron?(a) North (b) East (c) South (d) West (e) The magnetic force is zero.

Physics
2 answers:
ivann1987 [24]3 years ago
3 0

Answer:

(b) EAST

Explanation:

you can assume that the magnetic field points rightward, that is, in the positive x direction (NORTH). Furthermore, you can assume that the direction of the motion of the electron is in the positive y direction. Hence, you have:

\vec{B}=B_o\hat{i}\\\\\vec{v}=v_o\hat{j}

You use the Lorentz formula to known which is the direction of the magnetic force over the electron:

F=qv\ X\ B

which implies the cross product between the unitary vecors j and i, that is

\hat{i} \ X\ \hat{j} = -\hat{k}  (WEST)

However, the minus sign of the charge of the electron changes the direction 180°. Hence, the direction is k. That is, to the EAST

Rina8888 [55]3 years ago
3 0

Answer:

The electron will move east.

Explanation:

According to fleming's left hand rule for when the field and current are at right angles, the first finger represents the field's direction ( north to south), the second finger represents the current direction ( + to -) and the thumb represents the direction of motion..

See image below.

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Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo
Naily [24]

Answer:

t= 1,56 s ,   x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

        y = y₀ + v_{oy} t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

       0 = y₀ – ½ g t²

       t= √ (2 y₀/g)

calculemos

       t= √ ( 2 12 / 9,8)

       t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida  

        x = v₀ₓ t

        x = 80 1,56

        x= 124,8 m

La velocidad de impacto cuando toca el suelo

        vx = v₀ₓ = 80 ms

        v_{y} = v_{oy} – gt

        v_{y} = - 9,8 1,56

        v_{y} = - 15,288 m/s

la velocidad es

       v = (80 i^ - 15,288 j ) m/s

Traducttion  

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

        y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero (v_{oy} = 0)

       0 =y₀I - ½ g t²

       t = √ (2y₀ / g)

let's calculate

       t = √ (2 12 / 9.8)

       t = 1.56 s

Projectile range is the horizontal distance traveled

        x = v₀ₓ t

        x = 80 1.56

        x = 124.8 m

Impact speed when it hits the ground

        vₓ = v₀ₓ = 80 ms

        v_{y} = v_{oy} - gt

        v_{y} = - 9.8 1.56

        v_{y} = - 15,288 m / s

the speed is

       v = (80 i ^ - 15,288 j) m / s

7 0
3 years ago
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