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allsm [11]
2 years 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]2 years 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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As light from air to water to glass, it will refract. The best explanation for this would be
diamong [38]
It d I took the test

6 0
4 years ago
Read 2 more answers
A rock has a specific gravity of 2.32 and a volume of 8.64 in3 how much does it weigh
lianna [129]

Answer: 3.21 N

Specific\hspace{1mm} gravity = \frac {Density\hspace{1mm}of\hspace{1mm}substance}{Density\hspace{1mm} of \hspace{1mm}water}

\Rightarrow Density \hspace{1mm}of\hspace{1mm}substance= 2.32\times 1000\hspace{1mm} kg/m^3 = 2320\hspace{1mm}kg/m^3\\ Mass =Density\times volume\\ \Rightarrow 2320 \hspace{1mm} kg/m^3\times 8.64 \hspace{1mm}in^3\times \frac {1.64\times10^{-5} m^3}{1\hspace{1mm}in^3}=0.328 kg

For weight, we will multiply by g=9.8 m/s^{-2}

weight= 0.328\times9.8=3.21\hspace{1mm}N

Hence, the rock would weigh 3.21 N.

3 0
3 years ago
Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8
Kamila [148]

Answer:

34.6 m/s

Explanation:

From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg

Final mass will be 31.5+25.9=57.4 kg

From formula of momentum

M1v1=m2v2

Making v2 the subject of the formula then

V2=\frac {M1v1}{m2}

Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then

V2=\frac {100.2 kg\times 19.8 m/s}{57.4 kg}=34.56376 m/s\approx 34.6 m/s

4 0
3 years ago
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
Advocard [28]

The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as

  • t=0.476v
  • t=1.967v
  • V2=4.323v

<h3>What is the potential across the capacitor?</h3>

Question Parameters:

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

at

  • t = 1.0 seconds
  • 5.0 seconds
  • 20.0 seconds.

Generally, the equation for the Voltage is mathematically given as

v(t)=Vmax=(i-e^{-t/t})

Therefore

For t=1

V=5(i-e^{-1/10})

t=0.476v

For t=5s

V2=5(i-e^{-5/10})

t=1.967

For t=20s

V2=5(i-e^{-20/10})

V2=4.323v

Therefore, the values of voltages at the various times are

  • t=0.476v
  • t=1.967v
  • V2=4.323v

Read more about  Voltage

brainly.com/question/14883923

Complete Question

A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.

Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.

7 0
2 years ago
During an experiment, you take a measurement of 12.9 inches.
svlad2 [7]

Answer:

1 inch = 2.54 cm

12.9 inches= 12.9 x 2.54

= 32.766

= 32.8 cm (approximately)

Hope it helps...

7 0
3 years ago
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