100 POINTS 100 POINTS 100 POINTS!!!!!<br> HELP PLEASE I DON'T KNOW WHAT TO DO!!!!!

A positive charge of 18nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a rad

sergey [27]

Explanation:

Formula for angle subtended at the center of the circular arc is as follows.

$\theta = \frac{S}{r}$

where, S = length of the rod

r = radius

Putting the given values into the above formula as follows.

$\theta = \frac{S}{r}$

= $\frac{4}{2}$

= $2 radians (\frac{180^{o}}{\pi})$

= $114.64^{o}$

Now, we will calculate the charge density as follows.

$\lambda = \frac{Q}{L}$

= $\frac{18 \times 10^{-9} C}{4 m}$

= $4.5 \times 10^{-9} C/m$

Now, at the center of arc we will calculate the electric field as follows.

E = $\frac{2k \lambda Sin (\frac{\theta}{2})}{r}$

= $\frac{2(9 \times 10^{9} Nm^{2}/C^{2})(4.5 \times 10^{-9}) Sin (\frac{114.64^{o}}{2})}{2 m}$

= 34.08 N/C

Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.

5 0

3 years ago

Why do the lighter isotopes disappear first from the atmosphere? Where do those isotopes go?

denpristay [2]

Lighter molecules move fast and escape from the upper atmosphere relatively quickly.

To find the answer, we have to know more about the lighter isotopes.

<h3>What are lighter isotopes?</h3>

Lighter molecules are mobile and soon leave the higher atmosphere.
A particular element's stable isotopes have slightly different atomic masses and quantum mechanical energies.
The lighter isotope of an element's chemical bonds are more easily broken than the heavier isotope's.
As a result, the light isotope typically benefits from chemical reactions.

Thus, we can conclude that, lighter molecules move fast and escape from the upper atmosphere relatively quickly.

Learn more about the isotopes here:

brainly.com/question/364529

#SPJ4

3 0

1 year ago

A block of mass 200g is oscillating on the end of a horizontal spring of spring constant 100 N/m and natural length 12 cm. When

malfutka [58]

In order to determine the acceleration of the block, use the following formula:

$F=ma$

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

$F=kx$

Then, you have:

$ma=kx$

by solving for a, you obtain:

$a=\frac{kx}{m}$

In this case, you have:

k: spring constant = 100N/m

m: mass of the block = 200g = 0.2kg

x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m

Replace the previous values of the parameters into the expression for a:

$a=\frac{(\frac{100N}{m})(0.02m)}{0.2\operatorname{kg}}=10\frac{m}{s^2}$

Hence, the acceleration of the block is 10 m/s^2

8 0

1 year ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

Does friction always oppose relative motion?

Ilya [14]

Answer:

Yes

Explanation:

Friction is a force that opposes relative motion between systems in contact. One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction that opposes motion or attempted motion of the systems relative to each other.

3 0

3 years ago

100 POINTS 100 POINTS 100 POINTS!!!!! HELP PLEASE I DON'T KNOW WHAT TO DO!!!!!