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3 years ago

The wavelength of a wave is the horizontal distance from a to an adjacent .

Physics

2 answers:

goldenfox [79]3 years ago

5 0

<span>A. crest, crest

hope im right!(: </span>

aleksandr82 [10.1K]3 years ago

5 0

D, It's both crest to crest and trough to trough.

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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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8 0

3 years ago

When atoms are organized in repeating patterns, they generally make a _________. select one of the options below as your answer:

valkas [14]

A. Solid

Solid structures are actually made of repeating patterns.

4 0

3 years ago

When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g

Olegator [25]

Answer:

Your question was incomplete so here is the complete question and answer.

Q. When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g., 10K race)

a) plain water

b) 5-7 percent glucose solution

c) Glucose polymer solution of 6-8 percent

d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Ans. d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Explanation:

Temperature Regulation is an important phenomenon for the person exposed to extreme hot conditions or weather. Exercising in hot conditions increase the body temperature. Greater and intense exercise, greater the production of heat. Then the heat dissipation takes place in the form of excessive sweating which results in dehydration. That was just the brief overview of temperature regulation. Above mentioned techniques are equally good hydration techniques so there is no difference at all. You can have a plain water or glucose solutions of above mentioned percentages.

3 0

3 years ago

How many atoms of sodium do you have if you have one mole of sodium?

Yakvenalex [24]

Answer:

multiply the number of moles of Na by the conversion factor 6.02214179×1023 atoms Na/ 1 mol Na, with 6.02214179×1023 atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na.

3 0

3 years ago

Read 2 more answers

A 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in

chubhunter [2.5K]

Answer:

v' = 1.5 m/s

Explanation:

given,

mass of the bullet, m = 10 g

initial speed of the bullet, v = 300 m/s

final speed of the bullet after collision, v' = 300/2 = 150 m/s

Mass of the block, M = 1 Kg

initial speed of the block, u = 0 m/s

velocity of the block after collision, u' = ?

using conservation of momentum

m v + Mu = m v' + M u'

0.01 x 300 + 0 = 0.01 x 150 + 1 x v'

v' = 0.01 x 150

v' = 1.5 m/s

Speed of the block after collision is equal to v' = 1.5 m/s

5 0

3 years ago

The wavelength of a wave is the horizontal distance from a ________ to an adjacent ________.

The wavelength of a wave is the horizontal distance from a to an adjacent .