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butalik [34]
3 years ago
7

What is the direction of the magnetic field if an electron moving in the positive x direction experiences a magnetic force in th

e positive z direction
Physics
1 answer:
Alex787 [66]3 years ago
7 0

Given :

An electron moving in the positive x direction experiences a magnetic force in the positive z direction.

To Find :

The direction of the magnetic field.

Solution :

We know, force is given by :

\vec{F}=q(\vec{v}\times \vec{B)}

Here, q = -e.

\vec{F}=(-e)(\vec{v}\times \vec{B)}\\\\\hat{k}=(-e)(\hat{i}\times \vec{B})

Now, for above condition to satisfy :

\hat{i}\times \vec{B}=-\hat{k}

So, \vec{B}=-\hat{j}

Therefore, direction of magnetic field is negative y direction.

Hence, this is the required solution.

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Which type of bone is this <br>covalent bond<br>hydrogen bond <br>ionic bond​
Fynjy0 [20]

Answer:

Explanation:A covalent bond is formed when electrons are shared between non-metal atoms, and the positive nuclei are attracted towards the pair of negative bonded electrons. ... Hence, the hydrogen bond is weaker than ionic and covalent bonds. Example: Water molecules are held to each other by intermolecular forces of attraction.

5 0
3 years ago
At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then
gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

a=\frac{dv}{dt}=1.76

So, the velocity can be obtained by integrating this expression:

v=1.76t

The velocity is, by definition: v=\frac{dx}{dt}, so

dx=1.76tdt\\x=1.76\frac{t^{2}}{2}.

Do x=11 in order to find the time spent.

11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s

At this time the velocity is: v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}

This velocity remains constant in the section 2, so for that section the movement equation is:

x=v*t\\t=\frac{x}{v}

The left distance is 89 meters, and the velocity is 6.2225\frac{m}{s}, so:

t=\frac{89}{6.2225}=14.303s

So, the total time is 14.303+3.5355s=17.838s

7 0
3 years ago
Please i need your help w/h/w you make the following measurements if an object:42kg, and 22m​
Rufina [12.5K]

Answer:

1.9 kg/m^3

Explanation:

The density of an object is given by

d=\frac{m}{V}

where

m is the mass of the object

V is its volume

In this problem,

m = 42 kg

V = 22 m^3

Substituting into the equation, we find the object's density:

d=\frac{42 kg}{22 m^3}=1.9 kg/m^3

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3 years ago
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5 0
3 years ago
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