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3 years ago

As a transverse wave travels through a rope from left to right, the parts of the rope ______?

1 answer:

7 0

A transverse wave always move perpendicularly from its medium. In the problem the medium is the rope. So when the transverse wave movement is left and right, the perpendicular equivalent movement of the rope is up and down.

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Byeeeeeeeeeeeejdiiwidjwdjj2xi2idiwidisididie

Advocard [28]

Answer:

Be serious or else CEO of brainly will kick you out

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3 years ago

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

4 years ago

The following are examples of physical properties except

Ugo [173]

<h2>Answer:</h2><h3>D. ability to react with oxygen</h3><h2>Explanation:</h2>

Im not sure this in your choices but if it is, this is the answer.

I hope I've helped.

6 0

3 years ago

Read 2 more answers

2. A person began running due east and covered 15 kilometers in 2.0 hours. What is the average velocity of the person?

Angelina_Jolie [31]

Answer:

7.5 km/h (2.1 m/s) due east

Explanation:

The average velocity of the person is given by:

$v=\frac{d}{t}$

where

d is the displacement

t is the time taken

In this problem,

d = 15 km is the displacement

t = 2.0 h is the time elapsed

so the average velocity is

$v=\frac{15 km}{2.0 h}=7.5 km/h$

and the direction is the same as the displacement (east).

We can also convert the velocity into SI units (m/s). We have:

d = 15 km = 15,000 m

t = 2.0 h * 3600 s/h = 7200 s

$v=\frac{15,000 m}{7200 s}=2.1 m/s$

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3 years ago

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Why doesn’t she love me

o-na [289]

I bet she does just give her tule work on yourself

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3 years ago

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