3 years ago

What word can be used to describe the compression of a longitudal wave

1 answer:

8 0

Answer:

Mechanical longitudinal waves are also called compressional or compression waves, because they produce compression and rarefaction when traveling through a medium, and pressure waves, because they produce increases and decreases in pressure.

Explanation:

You might be interested in

A 12.0-g bullet is fired horizontally into a 112-g wooden block that is initially at rest on a frictionless horizontal surface a

sergiy2304 [10]

The speed would be in a decimal? Or do you want it in a fraction?

7 0

3 years ago

A skydiver reaches terminal velocity. Then he opens his parachute.

schepotkina [342]

Answer:

it is 0.9999.10896

Explanation:

4 0

3 years ago

Read 2 more answers

A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. If the radius of the ride is 9.0 meters, what is

NikAS [45]

<span>4.0 m/s2

it's 9 squared divided by 6</span>

8 0

3 years ago

Read 2 more answers

How is velocity different from speed? Based on distance and/or direction

Mnenie [13.5K]

Answer:

Speed is solved with time and distance but has no direction

Average velocity is solved with Δx/Δt and has a direction

6 0

3 years ago

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0

3 years ago