I need someone that has a course hero subscription to help me I need all the pages

astraxan [27]

Answer:

Explanation:

Commonly area is a scalar quantity because there is no need of direction to define and also follow the algebrac summation. When we talk about vector there exist a frame of reference with a certain orogin

8 0

2 years ago

An electron moving with a velocity = 5.0 × 10 7 m/s enters a region of space where perpendicular electric and a magnetic fields

inna [77]

Answer:

The magnetic field is $2 \times 10^{-4}$ T

Explanation:

Given:

Velocity of electron $v = 5 \times 10^{7} \frac{m}{s}$

Electric field $E = 10^{4} \frac{V}{m}$

The force on electron in magnetic field is given by,

$F = qvB \sin \theta$ ......(1)

The force on electron in electric field is given by,

$F = qE$ ......(2)

Compare both equation,

$qE = qvB \sin \theta$

Here $\sin \theta = 1$

$E = vB$

$B= \frac{E}{v}$

$B = \frac{10^{4} }{5 \times 10^{7} }$

$B = 2 \times 10^{-4}$ T

Therefore, the magnetic field is $2 \times 10^{-4}$ T

5 0

3 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

3 years ago

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A city of punjab has 15 percemt chance of wet weather on any given day. What is probability that it will take a week for it thre

sesenic [268]

Answer:

The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

Explanation:

We are given that A city of Punjab has 15 percent chance of wet weather on any given day.

So, Probability of wet weather = 0.15

Probability of not being a wet weather = 1-0.15 =0.85

We are supposed to find probability that it will take a week for it three wet weather on 3 separate days

Total number of days in a week = 7

We will use binomial over here

n = 7

p =probability of failure = 0.15

q = probability of success=0.85

r=3

Formula : $P(r=3)=^nC_r p^r q ^{n-r}$

$P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166$

Standard deviation = $\sqrt{n \times p \times q}$

Standard deviation = $\sqrt{7 \times 0.15 \times 0.85}$

Standard deviation =0.9447

Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

4 0

3 years ago

A ball is fired at an angle of 45 degrees, the angle that yields the maximum range in the absence of air resistance. What is the

Nesterboy [21]

Look at the picture for the answer

7 0

3 years ago

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