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4 years ago

Which wave is the result of amplification and produces a louder sound

1 answer:

4 0

https://www.khanacademy.org/science/ap-physics-1/ap-mechanical-waves-and-sound/introduction-to-sound-waves-ap/v/production-of-sound

BTW- i am a teacher and recently used this video to explain

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a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially

CaHeK987 [17]

Answer:

Velocity of the helium nuleus = 1.44x10⁴m/s

Velocity of the proton = 2.16x10⁴m/s

Explanation:

From the conservation of linear momentum of the proton collision with the He nucleus:

$P_{1i} + P_{2i} = P_{1f} + P_{2f]$ (1)

where $P_{1i}$ : is the proton linear momentum initial, $P_{2i}$ : is the helium nucleus linear momentum initial, $P_{1f}$ : is the proton linear momentum final, $P_{2f}$ : is the helium nucleus linear momentum final

From (1):

$m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f}$ (2)

where m₁ and m₂: are the proton and helium mass, respectively, $v_{1i}$ and $v_{2i}$ : are the proton and helium nucleus velocities, respectively, before the collision, and $v_{1f}$ and $v_{2f}$ : are the proton and helium nucleus velocities, respectively, after the collision

By conservation of energy, we have:

$K_{1i} + K_{2i} = K_{1f} + K_{2f}$ (3)

where $K_{1i}$ and $K_{2i}$ : are the kinetic energy for the proton and helium, respectively, before the colission, and $K_{1f}$ and $K_{2f}$ : are the kinetic energy for the proton and helium, respectively, after the colission

From (3):

$\frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2}$ (4)

Now we have two equations: (2) ad (4), and two incognits: $v_{1f}$ and $v_{2f}$ .

Solving equation (2) for $v_{1f}$ , we have:

$v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f}$ (5)

From getting (5) into (4) we can obtain the $v_{2f}$ :

$v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0$

$v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0$

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

$v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s}$ (6)

Now, by introducing (6) into (5) we get the proton velocity after the collision:

$v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4}$

$v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s}$

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!

7 0

4 years ago

What do protoplanets eventually grow into?

kow [346]

Answer:

a planet

Explanation:

If planetesimals, objects that combined to form protoplanets, were to collide head on in the early solar nebula, they would shatter to the point where they would be unable to grow into a larger size, a protoplanet. A protoplanet is a massive object that will eventually become a planet.

7 0

2 years ago

2N2 Hy 03 2N₂²_0₂ +42 H₂O

Sholpan [36]

Yfytctucctiycbjj itvbyobiouvuviybhv

4 0

3 years ago

write a good scientific question that can be answered by Guadalupe based on what is described above. Label the independent and d

kakasveta [241]

To do that, we would need to know what is "described above".

4 0

3 years ago

A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei

Digiron [165]

Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

$F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\$

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

$F = (150\ N)Sin\ 20^o$

F = 51.3°

5 0

3 years ago