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Agata [3.3K]
3 years ago
13

How do modern scientist describe the make up of matter

Physics
1 answer:
lbvjy [14]3 years ago
4 0

The matter is a collection or composition of extremely small particles called "Atoms".

<u>Explanation</u>:

  • The matter is anything that has mass and which utilizes space by having volume.
  • Modern scientists describe the makeup of matter as a collection of tiny substances called atoms.
  • Matter can either be in solid, liquid, or gaseous state and even appear in plasma form.
  • Matter can neither be created nor be destroyed but only transferred from one form to the other.
  • Atoms are composed protons, neutrons and electrons. The protons and neutrons are contained in the nucleus, while the electrons orbit around the nucleus of the atom.
  • Protons are positively charged, neutrons are neutral and electrons are negative.
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A 12.70 g bullet has a muzzle velocity (at the moment it leaves the end of a firearm) of 430 m/s when rifle with a weight of 25.
Norma-Jean [14]

Answer:

2.1844 m/s

Explanation:

The principle of conservation of momentum can be applied here.

when two objects interact, the total momentum remains the same  provided no external forces are acting.

Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\

assume the bullet goes to right side and the gravitational acceleration =10 ms^{-2}

so now the weight of the rifle=\frac{25}{10}  

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}

this is a negative velocity to the right side. that means the rifle recoils to the left side

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3 years ago
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It will be a virtual image that appears on the left side of the mirror

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Answer:

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Explanation:

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The moon's pull on Earth's oceans causes water to bulge and creates
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A small object carrying a charge of -2.50 nc is acted upon by a downward force of 18.0 nn when placed at a certain point in an e
Vesna [10]
Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?

B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>

<span>Solution:

A) A charge q </span>under an electric field of intensity E will experience a force F  equal to:

F=qE

In our problem we have q=-2.5 nC=-2.5\cdot 10^{-9} C and F=18 nN = 18 \cdot 10^{-9} N, so we can find the magnitude of the electric field:

E= \frac{F}{q}= \frac{18\cdot 10^{-9}N}{2.5\cdot 10^{-9}C}=7.2 V/m

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.

B) The proton charge is equal to

e=1.6\cdot 10^{-19} C

Therefore, the magnitude of the force acting on the proton will be

F=eE=1.6\cdot 10^{-19} C \cdot 7.2 V/m=1.15 \cdot 10^{-18} N

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.

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3 years ago
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