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Mars2501 [29]
3 years ago
6

Where is each subatomic particle is found in an atom

Physics
1 answer:
Ronch [10]3 years ago
7 0
Protons and neutrons are located in the nucleus, a dense central core in the middle of the atom, while the electrons are located outside the nucleus.
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PLEASE ANSWER! WILL MARK BRAINLIEST!
Angelina_Jolie [31]

The first one would be thermal energy

8 0
3 years ago
Draw a picture of an atom.include protons,neutrons, and electrons.
gogolik [260]
Hey, unfortunately it’s impossible for me to draw one for you here, but you should be able to draw it yourself pretty easily!

Answer:


An atom is built with a combination of three distinct particles: electrons, protons, and neutrons. Each atom has a center nucleus, where the protons and neutrons are densely packed together. Surrounding the nucleus are a group of orbiting electrons. And here is a picture that may assist your further :

Hope this helps
6 0
4 years ago
Two objects of the same radius start from rest and roll without slipping down a ramp. The objects are a hollow cylinder and a so
Westkost [7]

Answer:

Solid cylinder

Explanation:

because hollow may blow and just bounce

8 0
3 years ago
How to find final kinetic energy given height, mass, distance, and velocity.
Orlov [11]

Answer:

KE = 1/2(m)(v^2)

Explanation:

You only need mass and velocity to find kinetic energy

4 0
3 years ago
2000, the Millennium Bridge, a new footbridge over the River Thames in London, England, was opened to the public. However, after
sweet [91]

Answer:

n = 1810

A = 25 mm

Explanation:

Given:

Lateral force amplitude, F = 25 N

Frequency, f = 1 Hz

mass of the bridge, m = 2000 kg/m

Span, L = 144 m

Amplitude of the oscillation, A = 75 mm = 0.075 m

time, t = 6T

now,

Amplitude as a function of time is given as:

A(t)=A_oe^{\frac{-bt}{2m}}

or amplitude for unforce oscillation

\frac{A_o}{e}=A_oe^{\frac{-b(6T)}{2m}}

or

\frac{6bt}{2m}=1

or

b=\frac{m}{3T}

Now, provided in the question Amplitude of the driven oscillation

A=\frac{F_{max}}{\sqrt{(k-m\omega_d^2)+(b\omega_d^2)}}

the value of the maximum amplitude is obtained (k=m\omega_d^2)

thus,

A=\frac{F_{max}}{(b\omega_d}

Now, for n people on the bridge

Fmax = nF

thus,

max amplitude

0.075=\frac{nF}{((\frac{m}{3T})2\pi}

or

n = 1810

hence, there were 1810 people on the bridge

b)A=\frac{F_{max}}{(b\omega_d}

since the effect of damping in the millenium bridge is 3 times

thus,

b=3b

therefore,

A=\frac{F_{max}}{(3b\omega_d}

or

A=\frac{1}{(3}A_o

or

A=\frac{1}{(3}0.075

or

A = 0.025 m = 25 mm

6 0
3 years ago
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