Answer:
f = 1.8 10² Hz
Explanation:
With the readings of the oscilloscope screen we can calculate the period of the wave
T = #_divisions time_base
T = 5.5 1 10⁻³
T = 5.5 10⁻³ s
the period and frequency are related
f = 1 / T
f = 1 / 5.5 10⁻³
f = 1.8 10² Hz
Simple harmonic oscillations can be modeled by the projection of circular motion at constant angular velocity onto the diameter of a circle. When this is done, the analog along the diameter of the acceleration of the particle executing simple harmonic motion is
Answer:
the analog along the diameter of the acceleration of the particle executing simple harmonic motion is the projection along the diameter of the centripetal acceleration of the particle in the circle
A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated
We know
[tex]\boxed{\sf m=-\dfrac{v}{u}}[/tex]
[tex]\\ \sf\longmapsto 3=-\dfrac{-40}{u}[/tex]
[tex]\\ \sf\longmapsto 3=\dfrac{40}{u}[/tex]
[tex]\\ \sf\longmapsto u=\dfrac{40}{3}[/tex]
[tex]\\ \sf\longmapsto u=13.3cm[/tex]
Which of the following frequencies could NOT be present as a standing wave in a 2m long organ pipe open at both ends? The fundamental frequency is 85 Hz.
Answer:
382Hz
Explanation:
The question lacks the required option. Find the complete question in the attachment.
The long organ pipe open at both ends is called an open pipe. The fundamental frequency for an open pipe is expressed as F0 = V/2L
Harmonics are integral multiples of the fundamental frequency. For open pipes its harmonics are 2fo, 3fo, 4fo, 5fo...
Given fundamental frequency f0 to be 85 Hz, the following frequencies will be present as a standing wave;
First overtone f1 = 2fo = 2(85) = 170Hz
Second overtone f2 = 3fo = 3(85) = 255Hz
Third overtone = 4fo = 4(85) = 340Hz
Based on the option it can be seen that the only frequency that is not present as a standing wave is 382Hz
Two objects, one of mass m and the other of mass 2m, are dropped from the top of a building. When they hit the ground:_______.
a) the heavier one will have four times the kinetic energy of the lighter one.
b) the heavier one will have twice the kinetic energy of the lighter one.
c) the heavier one will have times the kinetic energy of the lighter one.
d) both of them will have the same kinetic energy.
Answer:
b) the heavier one will have twice the kinetic energy of the lighter one.
Explanation:
The kinetic energy of object with mass, m
K.E₁ = ¹/₂mv²
where;
m is mass of the object
v is the velocity of the object
Since, the two objects are falling under same acceleration due to gravity, their velocity will be increasing at the same rate
The kinetic energy of object with mass, 2m
K.E₂ = ¹/₂(2m)v²
K.E₂ = 2(¹/₂mv²)
BUT K.E₁ = ¹/₂mv²
K.E₂ = 2(K.E₁)
Therefore, the heavier one will have twice the kinetic energy of the lighter one.
b) the heavier one will have twice the kinetic energy of the lighter one.
A car moving at 36 m/s passes a stationary police car whose siren has a frequency of 500 Hz. What is the change in the frequency (in Hz) heard by an observer in the moving car as he passes the police car? (The speed of sound in air is 343 m/s.)
Answer:
Change in the frequency (in Hz) = 104.96 Hz
Explanation:
Given:
Speed of sound in air (v) = 343 m/s
Speed of car (v1) 36 m/s
Frequency(f) = 500 Hz
Find:
Change in the frequency (in Hz)
Computation:
Frequency hear by the observer(before)(f1) = [f(v+v1)] / v
Frequency hear by the observer(f1) = [500(343+36)] / 343
Frequency hear by the observer(f1) = 552.48 Hz
Frequency hear by the observer(after)(f2) = [f(v-v1)] / v
Frequency hear by the observer(f2) = [500(343-36)] / 343
Frequency hear by the observer(f2) = 447.52 Hz
Change in the frequency (in Hz) = f1 - f2
Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz
Change in the frequency (in Hz) = 104.96 Hz
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by in SI units. What is the frequency of the wave
Complete Question
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by
[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex] in SI units.
Answer:
The value is [tex]f = 1.98918*10^{5}\ Hz[/tex]
Explanation:
From the question we are told that
The magnetic field is [tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex]
This above equation can be modeled as
[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ] \equiv A sin ( kz -wt )[/tex]
So
[tex]w = \frac{10^7}{8}[/tex]
Generally the frequency is mathematically represented as
[tex]f = \frac{w}{2 \pi}[/tex]
=> [tex]f = \frac{ \frac{10^7}{8} }{2 \pi}[/tex]
=> [tex]f = 1.98918*10^{5}\ Hz[/tex]