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TEA [102]
3 years ago
8

Which of the following describes sound waves?

Physics
2 answers:
lianna [129]3 years ago
8 0
The answer is c) electromagnetic sawed in which the vibrations are perpendicular to the motion of the sound
hoa [83]3 years ago
3 0

Answer:

D. Mechanical waves in which the vibrations are parallel to the

motion of the sound

Explanation:

Ap3x

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When placed near another charge, a 20 microcoulomb charge experiences an attractive force of 0.080 N. What is the electric field
prohojiy [21]

Answer:

E = 4000 N/C

Explanation:

Given the following data;

Force = 0.080 N.

Charge, q = 20 microcoulomb = 20 * 10^-6 = 2 * 10^-5 Coulombs

To find the electric field strength;

Mathematically, the electric field strength is given by the formula;

Electric field strength = force/charge

Substituting into the formula, we have;

E = 0.080/0.00002

E = 4000 N/C

8 0
3 years ago
An 97 kg climber climbs to the top of Mount Everest, which has a peak height of 8850 m above sea level. What is the climber’s po
Kisachek [45]
Eg=mgh
Eg=(97)(9.8)(8850)
Eg=8412810J
4 0
3 years ago
GP A sinusoidal wave traveling in the negative x direction (to the left) has an amplitude of 20.0 cm, a wavelength of 35.0 cm, a
SIZIF [17.4K]

The period of the wave is determined as 0.083 seconds.

<h3>What is period of a wave?</h3>

The period of a wave is the time taken by a particle of the medium to complete one vibration.

<h3>Period of the wave</h3>

The period of the wave is calculated as follows;

T = 1/f

where;

  • T is the period of the wave
  • f is frequency of the wave

T = 1/12

T = 0.083 seconds

Thus, the period of the wave is determined as 0.083 seconds.

Learn more about period of a wave here: brainly.com/question/18818486

#SPJ4

8 0
1 year ago
What is the magnitude of the electric field at a point midway between a −5.0μC and a +5.8μC charge 8.4cm apart? Assume no other
Alex73 [517]

Answer:

Electric Field = E = 36.848 N/C

Explanation:

In accordance with Columb's law

E = k Q1 Q2 / r.r = 8.99 x 10^9 x 5.0 x 10^-6 x 5.8 x 10^-6 / 0.084 x 0.084

= 36948.6961 x 10^-3 = 36.848 N/C

4 0
3 years ago
A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.30 m/s2. At 20.0 s after bla
lesya [120]

Answer:

Explanation:

v = u +at

u = 0

a = 2.3 m /s²

t = 20 s

v = 2.3 x 20

= 46 m /s

Distance covered under acceleration of 2.3 m/s²

s = ut + 1/2 at²

= 0 + .5 x 2.3 x 20²

= 460 m

After that it moves under free fall ie g acts on it downwards .

v² = u² - 2gh , h is height moved by it under free fall

0 = 46² - 2 x 9.8 h

h = 107.96 m

Total height attained

= 460 + 107.96

= 567.96 m

b ) At its highest point ,it stops so  its velocity = 0

c ) rocket's acceleration at its highest point = g = 9.8 downwards .

At highest  point , it is undergoing free fall so its acceleration  = g

6 0
3 years ago
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