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Masteriza [31]
3 years ago
6

A node is a point on a standing wave that has no displacement from the rest position. At the nodes, _____.

Physics
2 answers:
Kitty [74]3 years ago
8 0

Answer: (B) There is complete destructive interference between the incoming and reflected waves

Explanation:

For example, if you pluck a guitar the waves will travel back and forth. They consist of nodes and anti-nodes. It is created, when the wave traveling to one side and bounces of the other end and comes back. As it travels to the other side, it is reflected thus, comes back. So standing waves occurs when there is interference.

When the wave is produced, the points where the string is not moving are called nodes and where they are moving are called anti-nodes. The positions where nodes are produced, destructive interference occurs and where anti-nodes are produced, constructive interference occurs

Debora [2.8K]3 years ago
5 0

Answer:

There is complete destructive interference between the incoming and reflected waves

Explanation:

I did gradpoint

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A freight train has a mass of [02] kg. The wheels of the locomotive push back on the tracks with a constant net force of 7.50 ×
otez555 [7]

Answer:

t = 300.3 seconds

Explanation:

Given that,

The mass of a freight train, m=1.01\times 10^7\ kg

Force applied on the tracks, F=7.5\times 10^5\ N

Initial speed, u = 0

Final speed, v = 80 km/h = 22.3 m/s

We need to find the time taken by it to increase the speed of the train from rest.

The force acting on it is given by :

F = ma

or

F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s

So, the required time is 300.3 seconds.

4 0
3 years ago
A one-dimensional plane wall of thickness 2l= 100 mm experiences uniform thermal energy generation of q˙= 800 w/m3 and is convec
slega [8]

Answer:

The thermal conductivity of the wall = 40W/m.C

h = 10 W/m^2.C

Explanation:

The heat conduction equation is given by:

d^2T/ dx^2 + egen/ K = 0

The thermal conductivity of the wall can be calculated using:

K = egen/ 2a = 800/2×10

K = 800/20 = 40W/m.C

Applying energy balance at the wall surface

"qL = "qconv

-K = (dT/dx)L = h (TL - Tinfinity)

The convention heat transfer coefficient will be:

h = -k × (-2aL)/ (TL - Tinfinty)

h = ( 2× 40 × 10 × 0.05) / (30-26)

h = 40/4 = 10W/m^2.C

From the given temperature distribution

t(x) = 10 (L^2-X^2) + 30 = 30°

T(L) = ( L^2- L^2) + 30 = 30°

dT/ dx = -2aL

d^2T/ dx^2 = - 2a

4 0
3 years ago
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