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Arturiano [62]
3 years ago
12

To determine a waves frequency you must know the??

Physics
2 answers:
saul85 [17]3 years ago
8 0

Answer:

To determine a waves frequency you must know the distance it travels!

Explanation:

Frequency=speed. Something must move for you to determine how fast it  moves!

lutik1710 [3]3 years ago
7 0

Answer: I think, the number of oscillations in a given period of time.

Explanation: Well I guess because in a period time is known as the rate of occurrence of the wave. Hope this helps!

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Five wheels are connected as shown in the figure. Find the velocity of the block “Q”, if it is known that: RA= 5 [m], RB= 10 [m]
Tanzania [10]

Answer:

-5 m/s

Explanation:

The linear velocity of B is equal and opposite the linear velocity of E.

vB = -vE

vB = -ωE rE

10 m/s = -ωE (12 m)

ωE = -0.833 rad/s

The angular velocity of E is the same as the angular velocity of D.

ωE = ωD

ωD = -0.833 rad/s

The linear velocity of Q is the same as the linear velocity of D.

vQ = vD

vQ = ωD rD

vQ = (-0.833 rad/s) (6 m)

vQ = -5 m/s

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3 years ago
What do scientists call water
pickupchik [31]

Answer:

h20

Explanation:

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Which technology is most efficient in mapping large areas of the ocean floor?
qaws [65]
''Sonar'' is the answer.
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What is the angle of reflection of the incident angle is 35°?<br> 35<br> 25°<br> 55<br> 90
Anestetic [448]

Answer:

i think its the third one or second one

4 0
3 years ago
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0
3 years ago
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