What is the medium through which the wave is moving?
What is the medium through which the wave is moving? - 1

Answers

Answer 1
Answer:

Answer:

STRING held by the man

Explanation:

Wave is defined as a disturbance that travels through a MEDIUM and transfer energy from one point to another without causing any permanent displacement of the medium itself.

The medium through which wave travels varies e.g water, string, air etc

According to the diagram, wave travels through the string held by the man. This string is referred to as the MEDIUM through which the wave moves. The wave generated produces both crest (D) and trough (C) when displaced from its initial position (A)

Answer 2
Answer:

Rope or one could consider it to be C to B or A to D


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Nicki rides her bike at a constant speed for 6 km. That part of her ride takes her 1 h. She then rides her bike at a constant speed for another 9km. That part of her trip takes her 2 h. What is her average speed?

Answers

Answer:

24km/h

Explanation:

go it right on ingenuity 2020

An athlete can exercise by making mechanical waves in ropes. What is themedium of these waves?
A. Energy
B. The rope
C. The athlete
D. Air

Answers

Answer:

It would be B. The Rope

Explanation:

I say this because the rope is transferring energy from one location to another. Now, I could be totally wrong on this but I think this is right lol.

Answer:

The answer is B the rope.

Hope this helps <3 ;)

A student lifts their 75 N backpack 0.50 m onto their chair. How much work is done?

Answers

Answer:

37.5 J

Explanation:

With work done equation: W=Fs

W=75*0.50=37.5 J

or use mgh=(75)(0.5) which is the same

A 50 kg woman and an 80 kg man stand 12.0 m apart on frictionless ice.(a) How far from the woman is their CM?
m

(b) If each holds one end of a rope, and the man pulls on the rope so that he moves 1.3 m, how far from the woman will he be now?
m

(c) How far will the man have moved when he collides with the woman?
m

Answers

Answer:

Given that

m₁ = 50 kg

m₂=80 kg

d= 12 m

a)

We know that center of mass given as

X = (x₁m₁+x₂m₂)/(m₁+m₂)

Lets take distance of CM from woman is X

So now by putting the value

X = (0 x 50+12 x 80)/(50+80)

x=7.38 m

b)

There is no any external force so the CM  will not move.

So we can say that

x₁m₁+x₂m₂ = 0

50(x) - 80(1.3)=0

x=2.08

So the distance move by woman d=12-2.08-1.3=8.62 m

d=8.62 m

c) lets take distance move by man is x

50 (x) - 80 (12-x) =0

x=7.38

So the distance move by woman d=12-7.38

d=4.62 m

What allows two or more atoms to be held together? shared protons shared electrons shared energy shared neutrons

Answers

It is shared electrons.

The following information should be considered:

  • In the case when two or more atoms can be together at the time when they share electrons with each other.
  • By sharing, they  create a covalent bond and that way the atoms can be stable.

learn more: brainly.com/question/2514933?referrer=searchResults

Answer:

try electrons i hope this helps!!

Explanation:

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

Answers

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

q_(1) +q_(2) =-q_(3) -----eqution 1

where,

q_(1) is the heat absorbed by the solid at 0⁰C

q_(2) is the heat absorbed by the liquid at 0⁰C

q_(3) the heat lost by the warmer water sample

Important equations to be used in solving this problem

q=m *c*\delta {T}, where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

\delta {T} is change in temperature

Again,

q=n*\delta {_f_u_s} -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

=237g *(1 mole H_(2) O)/(18g) = 13.167 moles of H_(2)O

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

q_(1) = 13.167 moles *6.01(KJ)/(mole) = 79.13KJ

This means that equation (1) becomes

79.13 KJ + q_(2) = -q_(3)

Step 4: calculate the final temperature of the water

79.13KJ+M_(sample) *C*\delta {T_(sample)} =-M_(water) *C*\delta {T_(water)

Substitute in the values; we will have,

79.13KJ + 237*4.18(J)/(g^(o)C)*(T_(f)-218}) = -350*4.18(J)/(g^(o)C)*(T_(f)-100})

79.13 kJ + 990.66J* (T_(f)-218}) = -1463J*(T_(f)-100})

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* (T_(f)-218}) = -1.463KJ*(T_(f)-100})

79.13 + 0.99066T_(f) -215.96388= -1.463T_(f)+146.3

collect like terms,

2.45366T_(f) = 283.133

T_(f) = = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C