Which statement describes how light behaves when it strikes a translucent material?

A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,

Rus_ich [418]

Answer:

$\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\$

Explanation:

Given that

$Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\$

As both charges are negative so there exist force of repulsion in direction as shown in figure.

$F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N$

Angle at which force F12 is acting is

$\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o$

$F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\$

$\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=$

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

$F_{21}=-5.63\times 10^{-11}N$

$\vec{F}_{21}=$

7 0

4 years ago

Which of these is the best explaination for why two negatively charged balloons move apart without ever touching?

Dennis_Churaev [7]

Hello, sorry this is a little late!

I believe the correct answer to your question would be option D, electric charges have electric fields surrounding them to allow them to exert forces on other objects without touching them.

I just took this test, and can 100% confirm this is the proper answer.

Hope this helps, and have a great day! :)

6 0

3 years ago

Read 2 more answers

What terms are needed to completely describe velocity?

kotegsom [21]

In order to completely describe a velocity,
you need a speed and a direction.

4 0

4 years ago

Which of the following statements is TRUE about updating the exposure control plan?

iren2701 [21]

Statements that are true as regards exposure control plan and its updating are;

Updates must have the reflection of changes in tasks as well in procedures.

Updates must reflect changes in positions that affect occupational exposure.

Updates must have the cost of PPE that is needed and necessary to reduce exposure

An exposure control plan can be regarded as the framework for compliance between the employer and the workers.

This framework give room for the employer to creates a written plan that will help in protecting their workers from bloodborne pathogens.

This plan gives hope to workers in term of protection when working with their Employer.

There are some elements that is associated with Exposure Control Plan, and theses are;

Health hazards as well as risk that is attributed to each product in the worksite.
Statement of purpose.
procedures and practices in a written form
Responsibilities from the Manager, CEO, designated resources and employer.

Therefore, exposure control plan is avenue to protect workers from bloodborne pathogens.

brainly.com/question/1203927?referrer=searchResults

3 0

3 years ago

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

x0 is the initial position.
v0 is the initial velocity.
a is the acceleration.
t is the time interval in which the motion is studied.

In this case:

x0= 0
v0= 0 because the object is initially stationary
a= 6 $\frac{m}{s^{2} }$
t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 $\frac{m}{s^{2} }$ *(15s)²

Solving:

x=½*6 $\frac{m}{s^{2} }$ *(15s)²

x=½*6 $\frac{m}{s^{2} }$ *225 s²

x= 675 m

The object will travel 675 m during that time.

5 0

3 years ago