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Natalka [10]
3 years ago
14

State examples of a transverse wave. ​

Physics
2 answers:
blondinia [14]3 years ago
7 0

Answer

Light and other types of electromagnetic radiation are examples of transverse waves. Some other examples of transverse waves include a ripple on a pond and a wave in a string.

laiz [17]3 years ago
6 0

ripples on the surface of water.

vibrations in a guitar string.

a Mexican wave in a sports stadium.

electromagnetic waves – eg light waves, microwaves, radio waves.

seismic S-waves.

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A motorcycle and a police car are moving toward one another. The police car emits sound with a frequency of 523 Hz and has a spe
Firdavs [7]

To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.

It can be described as

f = \frac{c\pm v_r}{c\pm v_s}f_0

c = Propagation speed of waves in the medium

v_r= Speed of the receiver relative to the medium

v_s= Speed of the source relative to the medium

f_0 =Frequency emited by the source

The sign depends on whether the receiver or the source approach or move away from each other.

Our values are given by,

v_s = 32.2m/s \rightarrow Velocity of car

v_r = 14.8 m/s \rightarrow velocity of motor

c = 343m/s \rightarrow Velocity of sound

f_0 = 523Hz \rightarrowFrequency emited by the source

Replacing we have that

f = \frac{c + v_r}{c - v_s}f_0

f = \frac{343 + 14.8}{343 - 32}(523)

f = 601.7Hz

Therefore the frequency that hear the motorcyclist is 601.7Hz

8 0
4 years ago
If Abdul pedals his 20 kg bike with a force of 100 N, what will the bike’s acceleration be? The formulas above may be helpful in
Scorpion4ik [409]

Answer:

5 m/s/s

Explanation:

F = ma

100 = 20a

100/20 = a

a = 5 m/s/s

4 0
3 years ago
Brainliest if correct
Bad White [126]

Answer:

D: Increase the distance between the objects.

E: Decrease the mass of one of the objects.

6 0
2 years ago
How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00
laiz [17]

Answer:

Work= -7.68×10⁻¹⁴J

Explanation:

Given data

q₁=q₂=1.6×10⁻¹⁹C

r₁=2.00×10⁻¹⁰m

r₂=3.00×10⁻¹⁵m

To find

Work

Solution

The work done on the charge is equal to difference in potential energy

W=ΔU

Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\  Work=-7.68*10^{-14}J

4 0
4 years ago
2. What type of motion objects that fall under the pull of gravity?
vivado [14]
<u><em>Kinetic Energy Pulls Any object to the ground.This Energy is a part of Gravity.</em></u>
<u><em>Wish you happy timez!</em></u>
4 0
3 years ago
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