All categories

3 years ago

Edge 2020 waves and diffraction lab report d a t a

Physics

2 answers:

Andreyy893 years ago

8 0

Answer:

Explanation:

Waves and Diffraction: Lab ReportTitle:Lab: Waves and DiffractionPurpose:When a wave encounters a small obstacle or the edge of a barrier, the phenomenon known asdiffraction will occur. The wave theory of light can help explain diffraction, although observingdiffraction of light directly is difficult due to the very short wavelengths of light. In this lab, you willuse a “ripple tank” simulation, which provides a convenient way to study wave diffraction on alarger scale, since the principles of wave diffraction apply to physical waves (such as soundwaves and waves in liquid) as well as electromagnetic waveswaves

Sever21 [200]3 years ago

5 0

Answer:Theoretical Discussion

The diffraction of classical waves refers to the phenomenon wherein the waves encounter an obstacle that fragments the wave into components that interfere with one another. Interference simply means that the wave fronts add together to make a new wave which can be significantly different than the original wave. For example, a pair of sine waves having the same amplitude, but being 180◦ out of phase will sum to zero, since everywhere one is positive, the other is negative by an equal amount.

You might be interested in

A 0.03-kg bullet is fired with a horizontal velocity of 470 m/s and becomes embedded in block B which has a mass of 3 kg. After

Gala2k [10]

Answer with Explanation:

We are given that

Mass of bullet, $m_1=0.03 kg$

$u_1=470 m/s$

$m_2=3 kg$

$\mu_k=0.2$

$m_3=30 kg$

We have to find the velocity of the bullet and block B after the first impact and final velocity of the carrier.

According to law of conservation of momentum

$m_1u_1=(m_1+m_2)v$

$0.03(470)=(0.03+3)v$

$v=\frac{0.03(470)}{(0.03+3)}=4.65 m/s$

Hence, the velocity of the bullet and block B after the first impact=4.65 m/s

According to law of conservation of momentum

$(m_1+m_2)v=(m_1+m_2+m_3)V$

$(0.03+3)\times 4.65=(0.03+3+30)V$

$V=\frac{(0.03+3)\times 4.65}{(0.03+3+30)}$

$V=0.43 m/s$

3 0

4 years ago

Read 2 more answers

a hunter 412.5m from a cliff moves a distance x towards the cliff and fires a gun. he hears the echo from the cliff after 2.2sec

Inessa [10]

Answer: 49.5 m

Explanation:

The speed of sound $s$ is given by a relation between the distance $d$ and the time $t$ :

$s=\frac{d}{t}$ (1)

Where:

$s=330 m/s$ is the speed of sound in air (taking into account this value may vary according to the medium the sound wave travels)

$d=412.5 m-x$ since we are told th hunter was initially 412.5 meters from the cliff and then moves a distance $x$ towards the cliff

$t=\frac{2.2 s}{2}=1.1 s$ Since the time given as data (2.2 s) is the time it takes to the sound wave to travel from the hunter's gun and then go back to the position where the hunter is after being reflected by the cliff

Having this information clarified, let's isolate $d$ and then find $x$ :

$d=st$ (2)

$412.5 m-x=(330 m/s)(1.1 s)$ (3)

Finding $x$ :

$x=49.5 m$ This is the distance at which the hunter is from the cliff.

3 0

3 years ago

If you push a 200 kg box with a horizontal force of 1,172 N and kinetic friction resists the motion with a force of 962 N, what

Lynna [10]

Answer:

a = 1.05m.s²

Explanation:

Fnet = m×a

Fapplied - friction = m×a

1172 - 962 = 200 × a

210 = 200a

a = 1.05

5 0

2 years ago

Does time stop in a black hole

lbvjy [14]

Answer:

In standard GR, nothing exists at the center of a black hole. The center of a black hole is a singularity, and because GR fails at that point it is simply removed from the manifold. That means that the singularity is not part of spacetime.

To answer your question more realistically, we believe that GR is an approximate theory that fails well before you reach the center. Unfortunately, we have no good alternative theory with which to answer the question in the region where GR fails. We simply don’t have any data from that regime and it is very hard to formulate a good theory without data. So there very well could be time at the center, but we simply don’t have a good way to even guess.

4 0

3 years ago

Read 2 more answers

1. 412.9 g of dry ice sublimes at room temperature. a. What’s changing? --- sublimation b. What constant will you use? ----- 25.

hoa [83]

1. 236 kJ

a. The phase (or state of matter) of the substance: from solid state to gas state (sublimation)

b. The enthalphy of sublimation, given by: $\lambda=571 J/g$

c. The equation to use will be $Q=m\lambda$ , where m is the mass of dry ice and $\lambda$ is the enthalpy of sublimation

d. The energy is being absorbed, because the heat is transferred from the environment to the dry ice: as a consequence, the bonds between the molecules of dry ice break and then move faster and faster, and so the substance turns from solid into gas directly.

e. The amount of energy being transferred is

$Q=m\lambda=(412.9 g)(571 J/g)=2.36\cdot 10^5 J=236 kJ$

2. 165 kJ

a. The phase (or state of matter) of the substance: from gas state to liquid state (condensation)

b. The latent heat of vaporisation of water, given by $\lambda=2260 J/g$

c. The equation to use will be $Q=m\lambda$ , where m is the mass of steam that condenses and $\lambda$ is the latent heat of vaporisation

d. The energy is being released, since the substance turns from a gas state (where molecules move faster) into liquid state (where molecules move slower), so the internal energy of the substance has decreased, therefore heat has been released

e. The amount of energy being transferred is

$Q=m\lambda=(72.9 g)(2260 J/g)=1.65\cdot 10^5 J=165 kJ$

3. 3.64 kJ

a. Only the temperature of the substance (which is increasing)

b. The specific heat capacity of silver, which is $C_s = 0.240 J/gC$

c. The equation to use will be $Q=m C_s \Delta T$ , where m is the mass of silver, Cs is the specific heat capacity and $\Delta T$ the increase in temperature

d. The energy is being absorbed by the silver, since its temperature increases, this means that its molecules move faster so energy should be provided to the silver by the surroundings

e. The amount of energy being transferred is

$Q=m C_s \Delta T=(39.2 g)(0.240 J/gC)(412.9^{\circ}C-25.9^{\circ}C)=3641=3.64 kJ$

4. 89 kJ

a. Both the phase of the substance (from solid to liquid) and then the temperature

b. The latent heat of fusion of ice: $\lambda=334 J/g$ and the specific heat capacity of water: $C_s=4.186 J/gC$

c. The equation to use will be $Q=m\lambda + m C_s \Delta T$ , where m is the mass of ice, $\lambda$ the latent heat of fusion of ice, Cs is the specific heat capacity of water and $\Delta T$ the increase in temperature

d. The energy is being absorbed by the ice, at first to break the bonds between the molecules of ice and to cause the melting of ice, and then to increase the temperature of the water

e. The amount of energy being transferred is

$Q=m\lambda +m C_s \Delta T=(156.3 g)(334 J/g)+(156.3 g)(4.186 J/gC)(56.232^{\circ}C-0^{\circ}C)=8.9\cdot 10^4 J=89 kJ$

6 0

3 years ago

Edge 2020 waves and diffraction lab report d a t a ​

Edge 2020 waves and diffraction lab report d a t a