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allsm [11]
1 year ago
7

If a piece of ribbon were tied to a stretched string carrying a transverse wave, then how is the ribbon observed to oscillate?a.

)perpendicular to wave directionb.)both perpendicular to and parallel to wave direction
Physics
1 answer:
noname [10]1 year ago
5 0

The ribbon was observed to oscillate perpendicular to the wave direction. <u>Option A.</u>

<u />

In transverse waves, particles move perpendicular to the direction of wave propagation. Examples of transverse waves are string vibrations and water waves. By moving the Slinky vertically up and down, you can create horizontal transverse waves. Her one type of mechanical wave is the transverse wave.

where the motion of the medium is perpendicular to the direction of energy propagation. A transverse wave is a wave whose vibration is perpendicular to the direction of wave propagation. This is in contrast to longitudinal waves, which propagate in the direction of oscillation. Water waves are an example of transverse waves.

Learn more about The ribbon here:-brainly.com/question/11685530

#SPJ9

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Rearrange the momentum equation to solve for speed. Show all work.
nasty-shy [4]

Definition:                   Momentum  =  (mass) x (speed)


OK. Here we go.
Watch closely:


Divide each side        
by  'mass' :                  <span>Momentum / mass  =  Speed </span>


Did you follow that ?

3 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

6 0
3 years ago
SOH-CAH-TOA is used to solve for the ________
Misha Larkins [42]

Answer:

c. initial (x and y)

Explanation:

When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.

Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

Thus, this method resolves the initial x and y velocities.

3 0
3 years ago
As an emergency vehicle approaches Bob and moves away from Jill, how does the actual frequency of the siren change? A) As an eme
Tamiku [17]
The correct answer is: 
<span>C) The actual frequency of the siren does not change despite appearances.

In fact, Bob will observe an increase in the apparent frequency as the emergency vehicle approaches him, while Jill will observe a decrease in the apparent frequency as the emergency vehicle moves away from him, because of the Doppler effect (the relative velocity between the observer and the source of the sound is changing), but this effect involves the apparent frequency, while the real frequency of the siren will remain the same.</span>
7 0
3 years ago
Read 2 more answers
Other quanto
Alex73 [517]

I'm not sure what you were trying to put here

5 0
3 years ago
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