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3 years ago

According to the periodic table, the average atomic mass of helium is A) 2 amu. B) 2.0026 amu. C) 4.0026 amu. D) 6.0026 amu.

Physics

1 answer:

RUDIKE [14]3 years ago

3 0

Answer:

C) 4.0026 amu

Explanation:

Helium (He) is the 2nd element on the Periodic Table.

It's neutral atom has 2 protons and 2 electrons.

It is in the 1st period and the 18th row.

It is also a Noble gas.

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What happens when an atoms losses an electron-

inysia [295]

It becomes positive

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3 years ago

Help me plz on the one i just put i neeed it uhg plz anter

pychu [463]

Answer:

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Explanation:

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4 years ago

Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s

anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of $-5.2^{\circ}$

D) The impulse exerted on C is 4.29 kg m/s at a direction of $-5.2^{\circ}$

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

$p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B$

$m_A = 0.600 kg$ is the mass of sphere A

$u_A = 4.00 m/s$ is the initial velocity of the sphere A (taking the right as positive direction)

$v_A$ is the final velocity of sphere A

$m_B = 1.80 kg$ is the mass of sphere B

$u_B = 2.00 m/s$ is the initial velocity of the sphere B

$v_B = 3.00 m/s$ is the final velocity of the sphere B

Solving for vA:

$v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s$

The sign is positive, so the direction is to the right.

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

$K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J$

After the collision:

$K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J$

The total kinetic energy is conserved: therefore, the collision is elastic.

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

$p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s$

While the total momentum along the y-axis is zero:

$p_y = 0$

We can now write the equations of conservation of momentum along the two directions as follows:

$p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy}$ (1)

We also know the components of the momentum of B after the collision:

$p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s$

So substituting into (1), we find the components of the momentum of C after the collision:

$p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s$

So the magnitude of the momentum of C is

$p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s$

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

$v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s$

And the direction is

$\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}$

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

$p_C = 0$

While the final momentum is:

$p'_C = 4.29 kg m/s$

So the magnitude of the impulse exerted on C is

$I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s$

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore $-5.2^{\circ}$ .

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

$K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J$

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

$K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J$

Since the total kinetic energy is not conserved, the collision is inelastic.

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

$F=Ma$

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

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#LearnwithBrainly

8 0

3 years ago

A car travels initially at 24 m/s, until it enters the highway. If the car accelerates at 4 m/s^2 for a 96 meters, what is the c

marishachu [46]

initial velocity=u=24m/s
Acceleration=a=4m/s^2
Distance=s=96m
Final velocity=v

Using 3rd equation of kinematics

$\boxed{\Large{\sf v^2-u^2=2as}}$

$\\ \Large\sf\longmapsto v^2=u^2+2as$

$\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)$

$\\ \Large\sf\longmapsto v^2=576+768$

$\\ \Large\sf\longmapsto v^2=1344$

$\\ \Large\sf\longmapsto v=\sqrt{1344}$

$\\ \Large\sf\longmapsto v=36.6m/s$

3 0

3 years ago

Which of the following statements is/are true? Check all that apply. A nonconservative force permits a two-way conversion betwee

saul85 [17]

Answer:

A conservative force permits a two-way conversion between kinetic and potential energies.

The work done by a nonconservative force depends on the path taken.

A potential energy function can be specified for a conservative force.

Explanation:

A conservative force is defined as a force whose work done does not depend on the path taken, but only on the initial and final position of motion.

This means that for a conservative force, it is possible to defined a potential energy function U which depends only on the position of the object. An example of conservative force is gravity: the gravitational potential energy of an object, in fact, depends only on its position in the field, not on the path taken.

This behaviour also implies that when an object moves from A to B and then back from B to A, the potential energy gained (or lost) moving from A to B is lost (or re-gained) when moving from B to A. This means that the total mechanical energy (sum of kinetic energy and potential energy) of the object is conserved, and therefore there is a constant conversion between potential and kinetic energy during the motion.

A non-conservative force instead does not show this properties, as the work done by it depends on the path taken, and therefore it is not possible to define a potential energy function. An example of non-conservative force is friction.

According to what we wrote above, therefore, the only correct statements are:

A conservative force permits a two-way conversion between kinetic and potential energies.

The work done by a nonconservative force depends on the path taken.

A potential energy function can be specified for a conservative force.

3 0

3 years ago