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3 years ago

Four students made a graphic organizer describing the parts of the atom.

Physics

2 answers:

Nikitich [7]3 years ago

8 0

Answer:

I believe the answer is D.

Explanation:

Protons are found inside the nucleus so are neutrons. Electrons are found outside the nucleus.

vodomira [7]3 years ago

5 0

Answer : The correct option is,

Particles Charge Location

Proton + inside nucleus

Electron - outside nucleus

Neutron 0 inside nucleus

Explanation :

As we know that an atom is the smallest unit of a matter that consist of three subatomic particles which are electrons, protons and neutrons.

The protons and the neutrons are located inside the nucleus or center of the nucleus where the mass of the an atom is concentrated and the electrons are located around the nucleus.

The protons are positively charged, the electrons are negatively charged and the neutrons are neutral that means it has no charge.

From the given table, we conclude that

The particle proton has positive charge and location is inside the nucleus.

The particle electron has negative charge and location is outside the nucleus.

The particle neutrons has no charge and location is inside the nucleus.

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A grindstone increases in angular speed from 6.00 rad/s to 12.20 rad/s in 16.00 s. Through what angle does it turn during that t

Akimi4 [234]

Answer:

Explanation:

Given that the grand stone has initial angular velocity of

w(ini)= 6rad/

And it has a final angular velocity of

w(fin)=12.20rad/sec

Time taken is t=16s

Using equation of angular motion

To get angular acceleration (α)

w(fin)=w(ini)+αt

12.20=6+16α

16α=12.20-6

16α=6.2

α=6.2/16

α=0.3875rad/sec²

The angular acceleration is 0.39rad/s²

Angle that he turn using

w(fin)²=w(ini)²+2αθ

12.2²=6²+2×0.3875θ

12.2²-6²=0.775θ

0.775θ=112.84

Then, θ=112.84/0.775

θ=145.6radian

The angular displacement is 145.6rad

6 0

3 years ago

As a train enters the station it slows down from 40 m/s to a 10 m/s in 5 seconds

lora16 [44]

Answer:

See the answers below.

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}-a*t$

where:

Vf = final velocity = 10 [m/s]

Vo = initial velocity = 40 [m/s]

t = time = 5 [s]

a = acceleration [m/s²]

Now replacing:

$10=40-a*5\\40-10=a*5\\30=5*a\\a=6[m/s^{2}]$

Note: The negative sign in the above equation means that the velecity is decreasing.

To solve this second part we must use the following equation of kinematics.

$v_{f}^{2} =v_{o}^{2} -2*a*x\\$

where:

x = distance [m]

$(10)^{2} =(40)^{2} -2*6*x\\100=1600-12*x\\12*x=1600-100\\12*x=1500\\x=125[m]$

7 0

3 years ago

An object is dropped from rest from a height of 4.1 107 m above the surface of the Earth. If there is no air resistance, what is

Valentin [98]

Answer:

2.83 x 10^4m/s

Explanation:

First, let us calculate the time taken by the object to hit the surface of the earth.

H = 4.1 x 10^7m

g = 9.8m/s2

t = √(2H/g)

t = √((2x 4.1 x 10^7) /9.8)

t = 2892.64secs

Now, we can find the velocity with which the object strikes the earth as follows:

V = gt

V = 9.8 x 2892.64

V = 2.83 x 10^4m/s

3 0

3 years ago

A blue-green photon (λ = 488 nm ) is absorbed by a free hydrogen atom, initially at rest. What is the recoil speed of the hydrog

Natalka [10]

Answer:

The recoil speed is $2.207\times 10^{4} m/s$

Solution:

Wavelength of a blue-green photon, $\lambda_{BG} = 488 nm = 488\times 10^{- 9} m$

Now, the energy associated with the blue-green photon:

$E_{BG} = \frac{hc}{\lambda_{BG}}$

where

h = Planck's constant

C = speed of light ion vacuum

$E_{BG} = \frac{6.626\times 10^{- 34}\times 3\times 10^{8}}{488\times 10^{- 9}}$

$E_{BG} = 4.07\times 10^{- 19} J$

Also, we know that the recoil speed can be calculated by the KInetic energy which is equal to the Energy of the blue-green photon:

$KE_{H} =\frac{1}{2}m_{p}v_{H}$

where

$v_{H}$ = velocity of Hydrogen atom

$m_{p} = 1.67\times 10^{- 27} kg$ = mass of H-atom

Now,

$KE_{H} =\frac{1}{2}m_{p}(v_{H})^{2}$

$4.07\times 10^{- 19} =\frac{1}{2}\times 1.67\times 10^{- 27}\times (v_{H})^{2}$

$v_{H} = \sqrt(4.87\times 10^{8}) = 2.207\times 10^{4} m/s$

7 0

3 years ago

If an astronaut could travel a v 0.980c for at time oft 12.0 hr as measured on Earth, how much time would the astronauts measure

bonufazy [111]

Answer:

Time observed by the astronaut = 2.388 hours

Distance traveled be according to an observer on Earth = 1.27 × 10¹¹ m

Distance traveled according to the astronaut = 2.52 × 10¹² m

Explanation:

Given:

Speed of the astronaut, v = 0.980c

time = 12.0 hr

now, from the time dilation formula we have

$t'=t\times\sqrt{1-\frac{v^2}{c^2}}$

here

t' is the time observed by the astronaut

c is the speed of the light = 3 × 10⁸ m/s

thus,

$t'=12\times\sqrt{1-\frac{(0.980c)^2}{c^2}}$

t' = 2.388 hours

Now,

Distance = Velocity × time

The distance traveled be according to an observer on Earth will be

Distance = 0.980c × ( 12 × 60 × 60 )

D = 0.980 × 3 × 10⁸ × ( 12 × 60 × 60 )

Distance = 1.27 × 10¹¹ m

And, The distance traveled according to the astronaut will be

Distance = velocity × t'

D = 0.980 × 3 × 10⁸ × ( 2.388 × 60 × 60 )

Distance = 2.52 × 10¹² m

5 0

3 years ago