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3 years ago

How many neutrons does protium have

2 answers:

7 0

The mass number of protium is also 1 since there are no neutrons in the nucleus of protium. The atomic mass of protium is about 1.00794 amu. The symbol for protium is 1H. The electron configuration of protium is 1s1.

Sonbull [250]3 years ago

5 0

Answer:

I believe none

Explanation:

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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 $1$ .

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

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

3 years ago

A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.

Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mb*g*H = 0.5*mb*vb^2

vb = √2*g*H

vb = ( 2*9.81*15 ) ^0.5

vb = 17.15517 m/s

- The angular momentum of system before collision is:

M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

M1 = M2

vo = 247.03448 / (5.10*3)

vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

ΔP.E = ΔK.E

mo*g*h = 0.5*mo*vo^2

h = vo^2 / 2*g

h = 16.14604^2 / 2*(9.81)

h = 13.3 m

3 0

3 years ago

Which object will have the most potential energy?

Svetradugi [14.3K]

Answer: the most potential energy == 5 kg book, 2 m from the ground= 98 Joules

Explanation:

potential energy = m g h

m = mass

g = acceleration due gravity = 9.8 m/s²

h = distance above ground

1. Pe₁ = 1 kg x 2 m x g = 2 g

2. Pe₂ = 5 kg x 2 m x g = 10 g = 10 kg m x 9,8 m/s² = 98 Joules

3. Pe₃ = 1 kg x 0,5 m x g = 0,5 g

4. Pe₄ = 5 kg x 0.5 m x g = 2,5 g

10 > 2,5 > 2 >0,5

5 0

3 years ago

A bar magnet is dropped toward a conducting ring lying on the floor. As the magnet falls toward the ring, does it move as a free

REY [17]

Explanation:

According to the Faraday-Lenz law, a conductive ring generates an induced current due to the change in the magnetic flux caused by the motion of the bar magnet. This induced current generates a magnetic field opposite to the magnetic field of the bar, generating an upward force that opposes the weight of the bar magnet, Therefore, it does not move as a freely falling object.

3 0

3 years ago

The atomic mass of nitrogen is 14.01, hydrogen is 1.01, sulfur is 32.07, and oxygen is 16.00. What is the molar mass of ammonium

MArishka [77]

The molar mass of ammonium sulphate [(NH4)2SO4] is 132.17 g (option E). Details about molar mass can be found below.

<h3>How to calculate molar mass?</h3>

The molar mass of a substance can be calculated by adding the atomic masses of the elements in the compound.

According to this question, the atomic mass of nitrogen is given as 14.01, hydrogen is 1.01, sulfur is 32.07, and oxygen is 16.00.

The molar mass of ammonium sulphate is as follows:

[(NH4)2SO4] = [14.01 + 1(4)]2 + 32.07 + 16.00(4)

= 36.02 + 32.07 + 64

= 132.09

Therefore, the molar mass of ammonium sulphate [(NH4)2SO4] is 132.17 g.

Learn more about molar mass at: brainly.com/question/12127540

#SPJ1

4 0

2 years ago