All categories

1 year ago

Identify the names, charges, and location of three types of subatomic particles that make up an atom.

2 answers:

3 0

Answer:Proton (charge of +e, in the nucleus), Neutron (0 charge, in the nucleus), and Electron (charge of –e, outside the nucleus).Nov 13, 2015

Explanation: YA

vredina [299]1 year ago

3 0

Protons are positive and in the center ( nucleus)

Neutrons have no charge and are also in the center (nucleus)

Electrons are negative and orbit around the nucleus

You might be interested in

Help me it’s for a test I have??

Volgvan

1. C
2. A
3. E
4. D
5. B
6. F
i might have 2 and 6 mixed up, not completely sure tho

4 0

2 years ago

Read 2 more answers

What is 1.2 kg converted into mg.<br> I need to know for step by step please?

astra-53 [7]

Here’s a good photo to reference when converting in the metric system.

Each time you move down a step you move the decimal to the right, each time you move up a step you move the decimal to the left.

We are going from 1.2 kg or kilograms, which is at the very top left of the ladder. To get to mg or milligrams, we would have to make six jumps, so we’d move the decimal over six times.

1.2 > 12. > 120. > 1200. > 12000. > 120000. > 1200000.

So our final answer would be 1,200,000mg.

4 0

3 years ago

Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a

wel

Answer:

a) $v=7.32m/s$

b) $\alpha =-35$ º

c) $ΔK=-1094.62J$

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

$p_{1x}=p_{2x}$

To simplify the problem, lets name D for Daniel and R for Rebecca

a) $p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}$

Since Daniel's initial velocity is 0

$m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}$

$v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}$

$v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s$

Now, lets analyze the movement in the vertical direction

$p_{1y}=p_{2y}$

Since $p_{1y}=0$

$0=m_{D}v_{D2y}+m_{R}v_{R2y}$

$v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s$

Now, we can find the magnitude of Daniel's velocity after de collision

$v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s$

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

$\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35$ º

c) The change in the total kinetic energy is:

ΔK= $K_{2}-K_{1}$

ΔK= $\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J$

That means that the kinetic energy decreases

5 0

2 years ago

Too much skepticism can

Schach [20]

Too much skepticism will lead one to doubt everything while too little will lead to gullibility.

6 0

2 years ago

Read 2 more answers

A 2 kg picture frame sits on a shelf at a height of 0.5 m. how much gravitational potential energy is added to the picture frame

AysviL [449]

The change in gravitational potential energy is A) 15.68 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in the gravitational field.

The change in gravitational potential energy of an object is given by the equation:

$\Delta U = mg\Delta h$

where

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration due to gravity

$\Delta h$ is the change in height of the object

In this problem, we have

m = 2 kg is the mass of the frame

$\Delta h = 1.3 - 0.5 = 0.8 m$ is the change in height

Substituting, we find:

$\Delta U = (2)(9.8)(0.8)=15.68 J$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

7 0

3 years ago