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3 years ago

How do mechanical waves travel through a medium?

2 answers:

8 0

Particles of the medium move perpendicular to the direction of energy transport. ... Mechanical waves require a medium in order to transport energy. Sound, like any mechanical wave, cannot travel through a vacuum. the particles

EXPLANATION:

uo

max2010maxim [7]3 years ago

7 0

Answer:

sound waves need to travel through a medium such as solids,liquids and gases. hoped this helped

Explanation:

sound waves

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How large a band of frequencies does each television broadcasting channel get ?

k0ka [10]

Answer:

Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.)

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2 years ago

Read 2 more answers

A yo-yo has a string that is 0.95 m in length. What is the period of oscillation if the yo-yo is allowed to swing back and forth

yulyashka [42]

Answer:

Explanation:

As we know the , equation of time period for simple pendulum ,

T = 2*pi* $\sqrt{l/g}$

hence putting values we get ,

the solution is in picture ,

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3 years ago

A 6 m long, uniform ladder leans against a frictionless wall and makes an angle of 74.3 ◦ with the floor. The ladder has a mass

olganol [36]

Answer: µ=0.205

Explanation:

The horizontal forces acting on the ladder are the friction(f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,

f=Fw

The sum of the moments about the base of the ladder Is 0

ΣM = 0 = Fw*L*sin74.3º - (25.8kg*(L/2) + 67.08kg*0.82L)*cos74.3º*9.8m/s²

Note that it doesn't matter WHAT the length of the ladder is -- it cancels.

Solve this for Fw.

0= 0.9637FwL - (67.91L)2.652

Fw=180.1/0.9637

Fw=186.87N

f=186.81N

Since Fw=f

We know Fw, so we know f.

But f = µ*Fn

where Fn is the normal force at the floor --

Fn = (25.8 + 67.08)kg * 9.8m/s² =

910.22N

µ = f / Fn

186.81/910.22

µ= 0.205

4 0

3 years ago

Please help!! I will give brainliest!

grandymaker [24]

Answer:

Depends.

Explanation:

Whether the object is going left or right, the speed will stay the same until friction eventually stops it. However, if, for example, we're talking about an object going straight before veering right, then yes, speed does matter. An object will normally have to speed up or slow down momentarily when changing direction to keep itself sustained on the ground.

So, honestly? It really depends on what we're talking about!

Hope this helped!

Source(s) used: None.

7 0

3 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

Friction Force

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$