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3 years ago

What real world measurable property of waves is the A value in the equation relating frequency and wavelength?

Physics

1 answer:

navik [9.2K]3 years ago

3 0

Answer:

Hello There!!

Explanation:

The frequency of a wave is the number of times per second that the wave cycles. Frequency is measured in Hertz or cycles per second.The period and frequency are closely related to each other. The period equals 1 over the frequency and the frequency is equal to one over the period.The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.

hope it helps,have a great day!!

~Pinky~

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3 physical adaptations of a frog

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Water, land. breath using skin and lungs

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4 years ago

A steel ball and a wooden ball of the same diameter are released together from the top of a tower. In the absence of air resista

ella [17]

Answer:

False

Explanation:

The steel ball and the wooden ball do not have the same force acting on them because their masses are different. But, they have the same acceleration which is the acceleration due to gravity g = 9.8 m/s².

Using the equation of motion under freefall, s = ut +1/2gt². Since u = 0,

s = 1/2gt² ⇒ t = √(2s/g)

Since. s = height is the same for both objects, they land at the same time neglecting air resistance.

8 0

3 years ago

A marshmallow is dropped from a 5.71 meter high pedestrian bridge, and 0.921 seconds later, it lands right on the head of an uns

Natalka [10]

let the height of the person with marshmallow on her head be "h"

consider the motion of the marshmallow after it is dropped from bridge.

Y₀ = initial position of the marshmallow above the ground = 5.71 m

Y = final position of marshmallow on head of person = h

v₀ = initial velocity of the marshmallow = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

t = time of travel for marshmallow = 0.921 sec

Using the kinematics equation

Y = Y₀ + v₀ t + (0.5) a t²

inserting the values

h = 5.71 + 0 (0.921) + (0.5) (-9.8) (0.921)²

h = 5.71 - 4.16

h = 1.55 m

5 0

3 years ago

A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch

ANTONII [103]

Answer:

A. A = 0.913 m

B. amax = 132.24m/s^2

C. Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

$v_{max}=A\omega$ (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

$\omega=\sqrt{\frac{k}{m}}$ (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

$\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}$

Next, you solve the equation (1) for A and replace the values of vmax and w:

$A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m$

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

$a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}$

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

$F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N$

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0

3 years ago

What frequency is received by the ambulance after reflecting from a wall near the person watching the oncoming ambulance (as in

Zepler [3.9K]

Answer:

900.48925 Hz

979.9785 Hz

Explanation:

$v_a$ = Relative velocity of ambulance = $109\ km/h=\dfrac{109}{3.6}$

$v_w$ = Velocity of wall = 0

v = Velocity of sound in air = 343 m/s

From doppler effect we have

$f=f'\dfrac{v+v_w}{v-v_a}\\\Rightarrow f=821\dfrac{343+0}{343-\dfrac{109}{3.6}}\\\Rightarrow f=900.48925\ Hz$

The frequency of sound is 900.48925 Hz

When the wall acts like a source

$f=f'\dfrac{v+v_a}{v-v_w}\\\Rightarrow f=900.48925\dfrac{343+\dfrac{109}{3.6}}{343-0}\\\Rightarrow f=979.9785\ Hz$

The frequency of sound is 979.9785 Hz

4 0

3 years ago