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Gemiola [76]
2 years ago
14

How u do this atomic mass

Physics
2 answers:
bezimeni [28]2 years ago
6 0
Vas happenin!!


To find atomic mass you you add the protons and the neutrons


Hydrogen-
Atomic number: 1
Atomic mass: 1.008
Number of protons: 1
Number of electrons: 1
Number of neutrons: 1



Helium-
Atomic number: 2
Atomic mass: 4.0026
Number of protons: 2
Number of neutrons: 2
Number of electrons: 2


Hope this helps

-Zayn Malik
Eddi Din [679]2 years ago
5 0

Explanation:

Hydrogen

Atomic Number=1

Atomic Mass= 1

number of protons= 1

number of electrons= 1

number of neutrons= 0

Helium

atomic number=2

atomic mass=2

number of protons=2

number of electrons=2

number of neutrons=2

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If a string completes 500 vibrations in one second, what is its frequency?
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True or false<br> Hydrogen fuel cells generate electricity by combining hydrogen with oxygen.
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The answer is: True.

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A child walks due east on the deck of a ship at 2 miles per hour. the ship is moving north at a speed of 18 miles per hour. find
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Refer to the figure shown below.

The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.

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8 0
3 years ago
A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti
LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision = 7.668\ ms^-1

Their common speed after the collision = 2.56\ ms^-1

Height achieved by the package of mass m when it rebounds = 0.33\ m

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

where K is Kinetic energy and U is Potential energy.

K= \frac{mv^2}{2} and U= mgh

Considering the fact  K_{initial} = 0\ and U_{final} =0 we will plug out he values of the given terms.

So V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1

Keypoints:

  • Sum of energies and momentum are conserved in all collisions.
  • Sum of KE and PE is also known as Mechanical energy.
  • Only KE is conserved for elastic collision.
  • for elastic collison we have e=1 that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

m_1V_1(i) = (m_1+m_2)V_f where V_1{i} =7.668\ ms^-1.

Plugging the values we have

m\times 7.668 = (3m)\times V_{f}

Cancelling m from both sides and dividing 3 on both sides.

V_f = 2.56\ ms^-1

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic e=1

We have to find the value of V_{f} for m mass.

As here V_{f}=-2.56\ ms^-1 we can use that if both are moving in right ward with 2.56 then there is a  -2.56 velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object 1.

So

V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1

Now using law of conservation of energy.

K_{initial} + U_{initial} = K_{final}+U_{final}

\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh

\frac{v(f1)^2}{2g} = h

h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed =7.668\ ms^-1 for part b we have their common speed =2.56\ ms^-1 and for part c we have the rebound height =0.33\ m.

3 0
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