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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

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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Your answer would be B. It forms hydrogen ions (H+)

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2 years ago

Read 2 more answers

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

Mama L [17]

A) Time needed: 6.24 s

B) Time needed: 2.86 s

Explanation:

A)

In this part, we are told that the power if the engine is constant. The power of the engine is given by

$P=\frac{W}{t}$

where

W is the work done

t is the time

This means that the power of the engine is proportional to the work done, and therefore, to the kinetic energy of the car:

$P=\frac{\frac{1}{2}mv^2}{t}=const.$

where m is the mass of the car and v its velocity.

SInce power is constant, we can write:

$\frac{\frac{1}{2}mv_1^2}{t_1^2}=\frac{\frac{1}{2}mv_2^2}{t_2}$

where:

$t_1=1.40 s$ is the time the car needs to accelerates to $v_1=28.0 mph$

$t_2$ is the time the car needs to accelerate to $v_2=57.0 mph$

Therefore, solving for $t_2$ ,

$t_2 = \frac{v^2}{u^2}t_1=\frac{57^2}{28^2}(1.40)=6.24 s$

B)

First of all, we have to calculate the acceleration of the car. We can do it using the following equation:

$a=\frac{v-u}{t}$

where:

u = 0 is the initial velocity

$v=28.0 mph \cdot \frac{1609 m/mi}{3600 s/h}=12.5 m/s$ is the final velocity

t = 1.40 s is the time elapsed

Substituting, we find the acceleration:

$a=\frac{12.5-0}{1.40}=8.9 m/s^2$

In this part, we are told that the force exerted by the engine is constant: according to Newton's second law, acceleration is proportional to the force,

$F=ma$

This means that the acceleration is also constant.

Now we want to find how long the car takes to accelerate to a final velocity of

$v=57.0 mph \cdot \frac{1609}{3600}=25.5 m/s$

From an initial velocity of

u = 0

Using again the same suvat equation, and using the acceleration we found previously, we find:

$t=\frac{v-u}{a}=\frac{25.5-0}{8.9}=2.87 s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

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About power:

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6 0

3 years ago

Multiple Choice

Digiron [165]

Answer:

What was not required by the Massachusetts Education law of 1642 is;

Children were to be sent to designated schoolmaster for their learning

Explanation:

The Law of 1642 required that parents and master see to it that their children knew the principles of religion and the capital laws of the commonwealth.

The important aspects of the 1642 Law includes;

1) The responsibility for the basic education and literacy of a child are those of the parents and masters of child apprentices

2) Reading and writing competency of all children and servants are a requirement

3) It is the duty of the government, where a parent or master are unable to meet their tutoring responsibility, to see that a child is placed where the basic educational requirement will be met.

The role of a schoolmaster or the setting of a formal school were yet to be formed as at that time.

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3 years ago

Why does the moon appear in the daytime

Brut [27]

Answer:

You can see the moon when it is in the perfect spot and it's reflecting the right amount of light. :)

Hope this helps! If you don't mind, please mark as brainliest! Thx!

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2 years ago

The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much

aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

$\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}$

Upon substituting our given values, we will get:

$\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}$

$\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365$

Taking square root of both sides, we will get:

$\frac{T_1}{T_2}=\sqrt{0.0073878361246365}$

$\frac{T_1}{T_2}=0.0859525225030452495$

$\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}$

$\frac{T_2}{T_1}=11.634329870476699\approx 11.634$

$T_2=11.634\cdot T_1$

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0

2 years ago