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ryzh [129]
3 years ago
13

An electron, moving south, enters a uniform magnetic field. Because of this uniform magnetic field, the electron curves upward.

We can conclude that the magnetic field must have a component A. upward; B. downward; C. toward the north; D. toward the east; E. toward the south; F. toward the west.
Physics
1 answer:
Flura [38]3 years ago
3 0

Answer:

Towards the west

Explanation:

Magnetic force is the interaction between a moving charged particle and a magnetic field.

Magnetic force is given as

F = q (V × B)

Where F is the magnetic force

q is the charge

V is the velocity

B is the magnetic field

V×B means the cross product of the velocity and the magnetic field

NOTE:

i×i=j×j×k×k=0

i×j=k.  j×i=-k

j×k=i.  k×j=-i

k×i=j.  i×k=-j

So, if the electron is moving southward, then, it implies that the velocity of it motion is southward, so the electron is in the positive z-direction

Also, the electron is curved upward due to the magnetic field, this implies that the force field is directed up in the positive y direction.

Then,

V = V•k

F = F•j

Then, apply the theorem

F •j = q ( V•k × B•x)

Let x be the unknown

From vector k×i =j.

This shows that x = i

Then, the magnetic field point in the direction of positive x axis, which is towards the west

You can as well use the Fleming right hand rule

The thumb represent force

The index finger represent velocity

The middle finger represent field

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SOVA2 [1]

Answer:

\omega=0.12\frac{rad}{s}

Explanation:

In a uniform circular motion, since a complete revolution represents 2π radians, the angular velocity, which is defined as the angle rotated by a unit of time, is given by:

\omega=\frac{2\pi}{T}(1)

Here T is the period, that is, the time taken to complete onee revolution:

T=\frac{2\pi r}{v}(2)

Replacing (2) in (1):

\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}

3 0
3 years ago
A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t
HACTEHA [7]

Answer:

(a) 2.85 m

(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

7 0
3 years ago
The x vector component of a displacement vector has a magnitude of 86.2 m and points along the negative x axis. The y vector com
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Answer:

(a) Magnitude of Vector = 207.73 m

(b) Direction = 65.48°

Explanation:

(a)

The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:

Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\

<u>Magnitude of Vector = 207.73 m</u>

(b)

For the direction of the vector we have the formula:

Direction = tan^{-1}(\frac{y}{x})\\\\Direction = tan^{-1}(\frac{189\ m}{86.2\ m})\\\\

<u>Direction = 65.48°</u>

4 0
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Alisiya [41]
<h3><u>Answer;</u></h3>

<u>  = 55.2 Coulombs </u>

<h3><u>Explanation</u>;</h3>

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7 0
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OverLord2011 [107]

Answer:

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Explanation:

In a longitudinal wave, the particles vibrate at right angles in reference to the wave motion.

In a transverse wave, the particles vibrate parallel to the wave motion

Electromagnetic waves occur as a result of the interaction between two  waves and are normally transverse in nature.  

8 0
3 years ago
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