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Kazeer [188]
3 years ago
14

Hans observed properties of four different waves and recorded observations about each one in his chart.

Physics
2 answers:
bogdanovich [222]3 years ago
8 0
The answer to your question is (Wave W is being diffracted, Wave X is being reflected, and waves Y and Z are being refracted.)


HOPE THIS HELPS YOU CAUSE IM DOING TEST AND I HOPE IM RIGHT AS WELL

PLZ GIVE ME THE BRAINLIEST ANSWER HAVE A GOOD DAY




weeeeeb [17]3 years ago
6 0

The correct answer is Wave W is being diffracted, Wave X is being reflected, and waves Y and Z are being refracted.

Wave X is bouncing off the surface thats why it is being reflected. As the waves pass from one medium to another they change speed i.e. when passing through water into air. This causes waves to bend and change color hence waves X and Z. The scattering of waves through a small opening is called diffraction so wave W

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Whic two types of cells are required during the process of osteogenesis
Mkey [24]

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3 0
3 years ago
A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert
liberstina [14]

Answer:

x = 7.62 m

Explanation:

First we need to calculate the weight of the rocket:

W  = mg

we will use the gravity as 9.8 m/s². We have the mass (500 g or 0.5 kg) so the weight is:  

W  = 0.5 * 9.8 = 4.9 N

We know that the rocket exerts a force of 8 N. And from that force, we also know that the Weight is exerting a force of 4.9. From here, we can calculate the acceleration of the rocket:

F - W = m*a

a = F - W/m

Solving for a:

a = (8 - 4.9) / 0.5

a = 6.2 m/s²

As the rocket is accelerating in an upward direction, we can calculate the distance it reached, assuming that the innitial speed of the rocket is 0. so, using the following expression we will calculate the time which the rocket took to blast off:            

y = vo*t + 1/2 at²

y = 1/2at²

Solving for t:

t = √2y/a

t = √2 * 20 / 6.2

t = √6.45 = 2.54 s

Now that we have the time, we can calculate the horizontal distance:

x = V*t

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x = 3 * 2.54 = 7.62 m            

4 0
3 years ago
two vehicles have a head on collision. one vehicle has a mass of 3000 kg and moves at 25 m/s while the second vehicle has a mass
emmasim [6.3K]

Answer:

The speed of the combined vehicles is 6.82m/s

Explanation:

Using the law of conservation of momentum which stayed that the sum of momentum of bodies before collision is equal to their sum of momentum after collision. After collision, both object moves with the same velocity.

Momentum = mass×velocity

Before collision:

Momentum of vehicle or mass 3000kg moving with velocity 25m/s

= 3000×25

= 75000kgm/s

Pa = 75000kgm/s

Momentum of vehicle with mass 2500kg moving with velocity of -15m/s

= 2500×-15

= -37500kgm/s

After collision:

Momentum = (3000+2500)V

Where v is their common velocity

Momentum after collision = 5500V

Based on the law:

75000+(-37500) = 5500V

75000-37500 = 5500V

37500 = 5500V

V = 37500/5500

V = 6.82m/s

7 0
3 years ago
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