The power source with the lowest environmental impact is

Suppose we have three hypothetical atomic nuclei: A, B, and C. Nucleus A has 7 protons and 7 neutrons. Nucleus B has 5 protons a

alina1380 [7]

Answer:

Nucleus A with 7 protons and 7 neutrons and Nucleus C with 7 protons and 5 neutrons are isotopes of the same elements

Explanation:

Isotopes are elements that have the same atomic structure but different molecular structure. An atom that has the same atomic number but different mass number are known to be isotopes.

The proton of an atom is the same as its atomic number while the sum of the proton and neutron is equal to its mass number.

According to the question, nuclei that has the same number of proton are isotopes of the same element. Therefore nuclei A and C with 7 protons each are isotopes of the same element since they have the same atomic number i.e number of proton = atomic number.

Their atomic masses of nuclei A and C are 14 and 12 respectively

3 0

2 years ago

Question 5 of 20 Engineers are using computer models to study train collisions to design safer train cars. They start by modelin

aniked [119]

Answer:

35m/s south

Explanation:

6 0

2 years ago

Read 2 more answers

while working out a man performed 2525j of work in 19seconds . what was his power A:132.9w. B:241.5w C 47.975w. D100.5w

Olenka [21]

A: 132.9w because 2525\19 is how much energy transferred per second which is also known as the power

7 0

3 years ago

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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 $1$ .

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

3 years ago

Need help asap pls

Zinaida [17]

B. third

for every action there is a reaction*

6 0

1 year ago