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anygoal [31]
2 years ago
12

a coin is dropped from the top of a tall building. determine the coin's (a) velocity and (b) displacement after 1.5 sec.

Physics
1 answer:
Lilit [14]2 years ago
8 0
1) v = gt = 10*1.5 = 15 m/s
2) r = gt^2 /2 = 10*(1.5)^2 / 2 = 11.25 meters 
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How many electrons are depicted in the electron dot diagram of an electrically neutral nitrogen atom? A. two B. six C. eight D.
goldenfox [79]

Answer is D - five.


<em>Explanation;</em>


- Electron dot diagrams show the valence electrons around the element by using dots.


- Valence electrons are the electrons which are in outermost shell of the atom.


-The atomic number of the N atom is 7.

      Atomic number = number of protons = 7

  If the atom is neutral,

      number of protons = number of electrons.


  Hence, N atom has 7 electrons.


- The electron configuration is 1s² 2s² 2p³.


Hence, N atom has 2 + 3 = 5 valence electrons. So, five electrons are represented in electron dot diagram of N.

4 0
3 years ago
Read 2 more answers
An archer defending a castle is on a 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take
vazorg [7]

Answer: 1.907

Explanation:

I did the math

3 0
3 years ago
Anyone know the answer ?
dlinn [17]
Momentum = mass x velocity, so 500kg x 2m/s = 1000 kg m/s
5 0
3 years ago
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How much force is needed to accelerate a 1000Kg car at a rate of 3m/s2?
ikadub [295]

According to Newton (2nd law),   Force = (mass) x (acceleration)

Substitute what we know :           Force = (1,000 kg) x (3 m/s²)

Do the arithmetic:                        Force = 3,000 kg-m/s²  =  3,000 newtons

6 0
3 years ago
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A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
Kipish [7]

Answer:

C. 0.25J

Explanation:

Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;

L is the inductance

I is the current flowing in the inductor

Given parameters

L = 20mH = 20×10^-3H

I = 5A

Required

Energy stored in the magnetic field.

E = 1/2 × 20×10^-3 × 5²

E = 1/2 × 20×10^-3 × 25

E = 10×10^-3 × 25

E = 0.01 × 25

E = 0.25Joules.

Hence the energy stored in the magnetic field of this inductor is 0.25Joules

7 0
3 years ago
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