Answer: Velocity of the ball just after the collision is -1.414 m/s.
Explanation:
As energy is conserved in a reaction so here, energy before collision will be equal to the energy after collision.
[tex]E_{before} = mgh = E_{after} = \frac{1}{2}mv_{o}^{2}[/tex]
where,
m = mass
g = gravitational energy = [tex]9.8 m/s^{2}[/tex]
h = height or length
[tex]v_{o}[/tex] = initial velocity
Also here, height is the length of wire. Let the height be denoted by 'L'. Therefore,
[tex]\frac{1}{2}mv_{o}^{2} = mgL\\v_{o}^{2} = 2gL\\v_{o} = \sqrt{2gL}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.11 m}\\= 4.66 m/s[/tex]
Formula used to calculate velocity after the collision is as follows.
[tex]v_{f ball} = v_{o} [\frac{m_{ball} - m_{block}}{m_{ball} + m_{block}}][/tex]
where,
[tex]v_{f ball}[/tex] = final velocity of ball after collision
[tex]m_{ball}[/tex] = masses of ball
[tex]m_{block}[/tex] = masses of block
Substitute the values into above formula as follows.
[tex]v_{f ball} = v_{o} [\frac{m_{ball} - m_{block}}{m_{ball} + m_{block}}]\\= 4.66 m/s [\frac{1.48 kg - 2.77 kg}{1.48 kg + 2.77 kg}]\\= 4.66 m/s \times (-0.303)\\= -1.414 m/s[/tex]
Thus, we can conclude that velocity of the ball just after the collision is -1.414 m/s.
Which of the following statements is true concerning the induced current in a coil of wire for a magnet approaching or moving away from the coil?
a. The induced magnetic field repels an approaching magnet.
b. The induced magnetic field attracts a departing magnet.
c. The induced magnetic field always opposes the motion of the magnet.
d. All of the above statements are correct.
Answer:
c
Explanation:
it is c because I hade got it right
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 167 cm , but its circumference is decreasing at a constant rate of 13.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 1.00 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
(a) Find the emf induced in the loop at the instant when 9.0 s have passed.
(b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Answer:
a) fem = - 0.0103 V, b) the applied field is in a vertical upward direction, the induced current is clockwise.
Explanation:
a) For this exercise we use Faraday's law
fem = [tex]- \frac{d \phi}{dty}[/tex]
the magnetic flux is
Ф = B. A = B A cos θ
The bold letters indicate vectors, in this case the direction of the magnetic field and the normal to the circumference is parallel therefore the angle is zero and the cos 0 = 1
fem = - B dA / dt
the area of a circle is
A = π r²
l
et's perform the derivative
dA / dt =π 2r [tex]\frac{dr}{dt}[/tex]
we substitute
fem = - B 2π r [tex]\frac{dr}{dt}[/tex]
the circumference of a circle is
L = 2π r
we substitute
fem = - B L ( L )
fem =
Let's find the circumference for the 9 s, let's use a direct rule of proportions
If the circumference changes 13cm at t = 1. how much does it change at t=9s
ΔL = 13cm (9s / 1s) = 117cm
the circumference that is
L = Lo - ΔL
L = 167 - 117
L = 50 cm
let's reduce all magnitudes to the SI system
L = 0.50 m
= 0.130 m / s
calculate us
fem = - 1.00 0.50 0.130
fem = - 0.0103 V
b) the electromotive force induced in the opposite direction to the change of the radius and is decreasing with time, the current follows the direction of the decreased voltage therefore the current is induced in the opposite direction to the change of the magnetic flux.
If the applied field is in a vertical upward direction, the induced current is clockwise.
How much heat is needed to warm 365 mL of water in a baby bottle from 240C to 38C
Answer:98.6 degrees Fahrenheit
Explanation:
What are the similarities between a resultant force equilibrant force?
Answer:
Explanation:
Resultant is a single force that can replace the effect of a number of forces. "Equilibrant" is a force that is exactly opposite to a resultant. Equilibrant and resultant have equal magnitudes but opposite directions.
What is the magnitude of the gravitational force acting on a
1.0 kg object which is 1.0 m from another 1.0 kg object?
Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!
bit.[tex]^{}[/tex]ly/3a8Nt8n
The coefficient of linear expansion of lead is 29 x 10 K What change in temperature will cause a 5-m long lead bar to change in length by 3.0 mm?
Answer:
These linear thermal expansion coefficients are room temperature values of metals. Linear thermal expansion coefficient is defined as material's fractional change in length divided by the change in temperature.
Explanation:
The change in temperature caused is 0.0000207 K.
To calculate the change in temperature we use the formula of linear expansivity below.
⇒ Formula:
∝ = ΔL/(LΔT).................. Equation 1⇒ Where:
∝ = Coefficient of linear expansionΔL = Change in lengthL = Original lengthΔT = Change in Temperature.⇒ Make ΔT the subject of the equation
ΔT = ΔL/(∝L)................ Equation 2From the question,
⇒ Given:
∝ = 2.9×10 K⁻¹ΔL = 3 mm = 3×10⁻³ mL = 5 m⇒ Substitute these values into equation 2
ΔT = (3×10⁻³)/(29×5)ΔT = 0.003/145ΔT = 0.0000207 KHence, the change in temperature caused is 0.0000207 K
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In a little league baseball game, the 145 grams ball enters the strike zone with a speed of 11.0 meters per second. The batter hits the ball and it leaves his bat with a speed of 25.0 meters per second in exactly the opposite direction. If the bat is in contact with the ball for 1.0 m/s, what is the magnitude of the average force exerted by the bat on the ball?
Answer:
Force = 5.22 N
Explanation:
According to Newton's Second Law of motion:
[tex]Force = Rate\ of\ Change\ of\ Momentum\\\\Force = \frac{mv_f-mv_i}{t}\\[/tex]
where,
m = mass of ball = 145 g = 0.145 kg
vf = final speed of ball after hit = 25 m/s
vi = initial speed of ball before hit = - 11 m/s (negative sign due to opposite direction)
t = time of contact = 1 s
Therefore,
[tex]Force = \frac{(0.145\ kg)(25\ m/s)-(0.145\ kg)(-11\ m/s)}{1\ s} \\\\[/tex]
Force = 5.22 N
How does the wave period relate to the frequency of a wave?
Answer:
its in the picture hope it helps make brainlliest ty
Which two waves combine to form a surface wave?
a transverse wave and a light wave
two transverse waves
two longitudinal waves
a transverse wave and a longitudinal wave
Answer:
i beilieve that the answer is a transverse wave and a longitudinal wave
Explanation:
i dont really have an explanation
The two waves combine to form the surface wave will be the transverse wave and the longitudinal waves.
What are surface waves?Surface waves are generated from the point of the earthquake inside the earth. There are generally three types of waves are generated primary, secondary, and surface waves.
Now this surface wave is the slowest wave which reach to the surface of earth very slowly and is the most destructive wave.
The surface waves travel both in the vertical as the well horizontal direction along their direction of flow.
An example of surface waves is the car park in the parking that goes up down just like the waves in the oceans.
Thus the two waves combine to form the surface wave will be the transverse wave and the longitudinal waves.
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An ideal horizontal spring-mass system has a mass of 1.0 kg and a spring with constant 78 N/m. It oscillates with a period of 0.71 seconds. When this same spring-mass system oscillates vertically instead, the period is _______ seconds. Enter 2 significant figures (a total of three digits) and use g = 10.0 m/s2 if necessary.
Answer:
T = 0.71 seconds
Explanation:
Given data:
mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.
We have to calculate time period when this same spring-mass system oscillates vertically.
As we know
[tex]T = 2\pi \sqrt{\frac{m}{K} }[/tex]
This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.
Therefore, T = 0.71 seconds
When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of J. Determine the wavelength of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface
This question is incomplete, the complete question is;
When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.28 × 10⁻¹⁹ J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.
Answer:
the required wavelength of light is 161.9 nm
Explanation:
Given the data in the question;
Let us represent work function of the metal by W.
Now, using Einstein photoelectric effect equation;
[tex]E_{proton[/tex] = W + [tex]K_{max[/tex]
hc/λ = W + [tex]K_{max[/tex] ------- let this be equation 1
we solve for W
W = hc/λ - [tex]K_{max[/tex]
given that; λ = 221 nm = 2.21 × 10⁻⁷ m, [tex]K_{max[/tex]= 3.28 × 10⁻¹⁹ J
we know that speed of light c = 3 × 10⁸ m/s and Planck's constant h = 6.626 × 10⁻³⁴ Js.
so we substitute
W = [( (6.626 × 10⁻³⁴)(3 × 10⁸) )/2.21 × 10⁻⁷ ] - 3.28 × 10⁻¹⁹
W = 8.99457 × 10⁻¹⁹ - 3.28 × 10⁻¹⁹
W = 5.71457 × 10⁻¹⁹ J
Now, to determine λ for which maximum kinetic energy is double
so;
[tex]K'_{max[/tex] = double = 2( 3.28 × 10⁻¹⁹ J ) = 6.56 × 10⁻¹⁹ J
from from equation 1
we solve for λ'
λ' = hc / W + [tex]K'_{max[/tex]
we substitute
λ' = ( (6.626 × 10⁻³⁴)(3 × 10⁸) ) / ( (5.71457 × 10⁻¹⁹ J) + ( 6.56 × 10⁻¹⁹ J ))
λ' = 1.9878 × 10⁻²⁴ / 1.227457 × 10⁻¹⁸
λ' = 1.619 × 10⁻⁷ m
λ' = 161.9 nm
Therefore, the required wavelength of light is 161.9 nm
500 J of work is used to decrease the angular velocity of a disk from 65 rad/s to 52 rad/s.What is the rotational inertia of the disk?
Answer:
The correct answer is "0.66 kg.m²".
Explanation:
The given values are:
K.E = 500 J
w₁ = 65 rad/s
w₂ = 52 rad/s
As we know,
⇒ [tex]K.E=\frac{1}{2}I[w_1^2-w_2^2][/tex]
then,
⇒ [tex]I=\frac{2(K.E)}{w_1^2-w_2^2}[/tex]
On putting the given values, we get
⇒ [tex]=\frac{2\times 500}{(65)^2-(52)^2}[/tex]
⇒ [tex]=\frac{1000}{4225-2704}[/tex]
⇒ [tex]=\frac{1000}{1521}[/tex]
⇒ [tex]=0.66 \ kg.m^2[/tex]
Which of the following lists the elements in order, from those having the least protons to those having the most protons in the atoms?
A. O, N, B, Li
B. Na, S, Al, Cl
C. O, S, Se, Te
D. Rb, K, Na, Li
Answer:
C
Explanation:
O has 8 protones,S tiene 16, Se 34 y Te 52.
Determine the direction of the force on a charge.
a. along the line between the charge and the center of the square outward of the center
b. along the side of the square outward of the other charge that lies on the side
c. along the line between the charge and the center of the square toward the center
d. along the side of the square toward the other charge that lies on the side
Answer:
hi your question is incomplete below is the complete question
A charge of 2.15 mC is placed at each corner of a square 0.500 m on a side. Determine the direction of the force on a charge.
answer : Along the line between the charge and the center of the square outward of the center ( A )
Explanation:
The direction of the force on a charge is along the line between the charge and the center of the square outward of the center
Given that; Fnet = 3.17 * 10^-1 N ( calculated value )
The nature of the Force is repulsive in nature
attached below is a Pictorial representation of the direction of the force on a charge
Particles q1, 92, and q3 are in a straight line.
Particles q1 = -1.60 x 10-19 C, 92 = +1.60 x 10-19 C,
and q3 = -1.60 x 10-19 C. Particles qi and q2 are
separated by 0.001 m. Particles q2 and q3 are
separated by 0.001 m. What is the net force on q2?
Answer: 0
Explanation: Acellus
The charge q2 is kept in between the charges q1 and q3. The net force on charge q2 is zero.
What is force?The force is the action of push or pull in order to move or stop the object.
Given, q1 = -1.60 x 10⁻¹⁹ C, q2 = +1.60 x 10⁻¹⁹ C, and q3 = -1.60 x 10⁻¹⁹ C. Particles q1 ,q2 and q2 , q3 are separated by d =0.001 m.
Net electrostatic force on q2 will be
F12 =F23
k q1 q2 / d² = kq2 q3 /d²
q1 q2 = q2 q3
Substitute the values of forces, we get
1.60 x 10⁻¹⁹ x 1.60 x 10⁻¹⁹ = 1.60 x 10⁻¹⁹ x 1.60 x 10⁻¹⁹
1 =1
The net force F12 - F23 =0
Thus, the net force on q2 is zero.
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A compact disc (CD) is played by a cd player, which uses a laser to read the tracks on the disc. The disc spins initially at approximately 200 rotations per minute (rpm), and increases to a maximum of approximately 500 rpm as the laser spirals inward towards the center of the disc. This ensures that the laser covers equal distance in an equal amount of time during playback. During the time that the cd is played, which of the following statements is true?
a. The laser tracking mechanism experiences a changing tangential velocity
b. The laser tracking mechanism experiences a constant angular velocity
c. The laser tracking mechanism experiences a non-zero angular acceleration
d. The laser tracking mechanism experiences a non-zero tangential acceleration
Answer:
a. The laser tracking mechanism experiences a changing tangential velocity
c. The laser tracking mechanism experiences a non-zero angular acceleration
d. The laser tracking mechanism experiences a non-zero tangential acceleration
Explanation:
a. The laser tracking mechanism experiences a changing tangential velocity
This is because the tangential velocity v = rω where r = radius of disc and ω = angular speed of discs. Since r is constant, v ∝ ω.
Since the angular speed changes from 200 rpm to 500 rpm, thus, the tangential velocity would also change.
So, the laser tracking mechanism experiences a changing tangential velocity
c. The laser tracking mechanism experiences a non-zero angular acceleration
Since angular acceleration, α = Δω/Δt where Δω = change in angular speed and Δt = change in time.
Since there is a change in angular speed from 200 rpm to 500 rpm in time Δt, there is thus a non-zero angular acceleration.
So, The laser tracking mechanism experiences a non-zero angular acceleration
d. The laser tracking mechanism experiences a non-zero tangential acceleration
Since tangential acceleration, a = rα where r = radius of disc and α = angular acceleration.
Since there is an angular acceleration of the disc, there is thus going to be a tangential acceleration given by a = rα.
So, the laser tracking mechanism experiences a non-zero tangential acceleration
Statement b is false because, the disc experiences a changing angular speed from 200 rpm to 500 rpm.
Active listening includes all of the following EXCEPT: A. paraphrasing B. clarifying C. ignoring D. empathizing Please select the best answer from the choices provided. A B C D
Answer:
C: Ignoring
Explanation:
On edge 2021
Answer:
It's C.
Explanation:
Ways if improving soil fertility
Answer:
Soil fertility can be further improved by incorporating cover crops that add organic matter to the soil, which leads to improved soil structure and promotes a healthy, fertile soil; by using green manure or growing legumes to fix nitrogen from the air through the process of biological nitrogen fixation; by micro-dose ........
Answer:
Continuous cultivation of water intensive crops like sugarcane and rice has resulted in their water tables running low at 149 percent. Alternating crops with millet, which consumes less water and has a high nutritional value, can increase the soil fertility and balance nutrients present in it.Apr 19, 2019
Explanation:
pls mark me as a brainlist
hope my answer is helpful
A lead fishing weight of a mass of 0.20 kg is tied to a fishing line that is 0.50 m long. The weight is then whirled in a vertical circle. The finishing line will break if its tension exceeds 100.0 N.
Required:
a. If the weight is whirled at higher and higher speeds, at what point in the vertical circle will the string break (top, bottom, or random position)?
b. At what speed will the string break?
Answer:
The solution of the given question is summarized in the below section.
Explanation:
The given values are:
Tension,
T = 100 N
mass,
m = 0.2 kg
length,
l = 0.5 m
Now,
(a)
Somewhere at bottom, string or thread breaks since string voltage seems to be the strongest around this stage.
then,
⇒ [tex]T-mg=\frac{mv^2}{l}[/tex]
or,
⇒ [tex]T=mg+\frac{mv^2}{l}[/tex]
(b)
As we know,
⇒ [tex]\frac{mv^2}{l}=T-mg[/tex]
or,
⇒ [tex]v^2=\frac{(T-mg)l}{m}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{(100-0.2\times 10)0.5}{0.2}[/tex]
⇒ [tex]=\frac{49}{0.2}[/tex]
⇒ [tex]v =\sqrt{245}[/tex]
⇒ [tex]=15.65 \ m/s[/tex]
A student must analyze data collected from an experiment in which a block of mass 2M traveling with a speed v0 collides with a block of mass M that is initially at rest. After the collision, the two blocks stick together. What applications of the equation for the conservation of momentum represent the initial and final momentum of the system for a completely inelastic collision between the blocks?
Solution :
In the question, it is given that the collision is inelastic and the blocks stick together.
In an inelastic collision, the linear momentum is conserved but the kinetic energy is not conserved.
The linear momentum is given by :
[tex]$\vec p = m \vec v$[/tex] (mass x velocity)
So according to the conservation of linear momentum,
[tex]$\vec p_{(\text{before collision})}=\vec p_{(\text{after collision})}$[/tex]
Let the velocity after the collision is [tex]$v_F$[/tex]
[tex]$m_1v_0+m_2 \times 0 = m_1v_F+m_2v_F$[/tex]
Putting the values of [tex]$m_1 \text{ and}\ m_2$[/tex]
[tex]$m_1=2M \text{ and}\ m_2=M$[/tex]
∴ [tex]$2Mv_0=2Mv_F+Mv_F$[/tex], as the blocks stick together after the collision.
and [tex]$2MV_0=3Mv_F$[/tex], as the blocks stick together after the collision.
Which is a property of bases?
A.
highly metal reactive
B.
sour to the taste
C.
slippery feel
D.
low pH
Answer:
C. Slippery feel
Explanation:
A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?
Answer:
the force of attraction between the two charges is 4.2 x 10⁹ N.
Explanation:
Given;
the magnitude of first charge, q₁ = 0.06 C
the magnitude of the second charge, q₂ = 0.07 C
distance between the two charges, r = 3 m
The force of attraction between the two charges is calculated as ;
[tex]F = \frac{Kq_1q_2}{r^2}[/tex]
where;
k is Coulomb's constant = 9 x 10⁹ Nm²/C²
[tex]F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(0.06)(0.07)}{3^2} \\\\F = 4.2 \times 10^{6} \ N[/tex]
Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.
A battery is connected to a light bulb in a circuit. There is a current of 2 A in the light bulb.
The voltage of the battery is 1.5 V. What is the resistance of the light bulb?
V
Use R = to solve this problem.
Answer:
Resistance of the light bulb = 1.33 ohm (Approx.)
Explanation:
Given:
Current in electric bulb = 2 Amp
Voltage in battery = 1.5 Volt
Find:
Resistance of the light bulb
Computation:
Resistance = Voltage / Current
Resistance of the light bulb = Voltage in battery / Current in electric bulb
Resistance of the light bulb = 2 / 1.5
Resistance of the light bulb = 1.33 ohm (Approx.)
When antimatter interacts with an equal mass of ordinary matter, both matter and antimatter are converted completely into energy, in the form of photons. In an antimatter-fueled spaceship, a staple of science fiction, the newly created photons are shot from the back of the ship, propelling it forward. Suppose such a ship has a mass of 2.00×10^6kg, and carries a mass of fuel equal to 4% of its mass, or 4.00×10^4kg of matter and an equal mass of antimatter.
Required:
What is the final speed of the ship, assuming it starts from rest, if all energy released in the matter-antimatter annihilation is transformed into the kinetic energy of the ship?
Answer:
v = 5.88 10⁷ m / s
Explanation:
For this exercise we use the relation
E = m c²
also indicate that all energy is converted into kinetic energy
E = K = ½ (M-2m) v²
where m is the mass of antimatter and M is the mass of the ship's mass. Factor two is due to the fact that equal amounts of matter and antimatter must be combined
we substitute
m c² = ½ (M-2m) v²
v² = [tex]2 \frac{m}{M+2m} \ c^2[/tex]
let's calculate
v = [tex]\sqrt{2 \ \frac{4 \ 10^4 }{2 \ 10^6 + 2 \ 4 \ 10^4} \ (3 \ 10^8)^2}[/tex]
v = [tex]\sqrt{ 34.615 \ 10^{14}}[/tex]
v = 5.88 10⁷ m / s
what is force,momentum,and velocity.
Answer:
A force is a push or pull upon an object resulting from the object's interaction with another object.
Momentum is force or speed of movement.
Velocity defines the path of the motion of the frame or the object
How does the presence of a nucleus provide a method of basic cell classification?
Answer:D.
Explanation:
Pause
He
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be:
OA. 40 m/s
OB. 9.8 m/s2
OC 2.5 m/s2
OD. 0.40 m/s2
Answer:
C
Explanation:
a=f/m
10Kgm/s2/4kg
2.5m/s2
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity, or g. It is a vector quantity whose strength or magnitude is determined by the norm and whose direction correlates with a plumb bob.
Given in the question, force 10 N and mass 4.0 Kg the acceleration is,
a = 10/4 = 2.5 m/sec²
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².
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Question 5 of 10
The graph below shows the downloads of two songs over time.
70
Song 1
60
50
40
Number of downloads
(spopunu)
Song 2
30
20
10
O
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Time (minutes)
Which term describes the slope of the graph of song 2 between minute 6 and
minute 7?
A. Positive
B. Zero
C. It is impossible to calculate
D. Negative
SUBMIT
Answer:
b
Explanation:
A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During a 5.0-s interval the angular velocity increases to 40 rad/s. Assume that the angular acceleration was constant during the 5.0-s interval. How many revolutions does the wheel turn through during the 5.0-s interval ( in revelations)? Hint: You need to use one of the equations of constant angular acceleration that independent of angular acceleration.
Answer:
The appropriate solution is "23.87 rev".
Explanation:
The given values are:
Initial angular velocity,
[tex]\omega_i=20 \ rad/s[/tex]
Final angular velocity,
[tex]\omega_f=40 \ rad/s[/tex]
Time taken,
[tex]t = 5.0 \ s[/tex]
If, α be the angular acceleration, then
⇒ [tex]\omega_f=\omega_i+\alpha t[/tex]
or,
⇒ [tex]\alpha t=\omega_f-\omega_i[/tex]
⇒ [tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{40-20}{5.0}[/tex]
⇒ [tex]=\frac{20}{5.0}[/tex]
⇒ [tex]=4 \ rad/s^2[/tex]
If, ΔФ be the angular displacement, then
⇒ [tex]\Delta \theta=\omega_i t+\frac{1}{2} \alpha t^2[/tex]
On substituting the values, we get
⇒ [tex]=[(20\times 5.0)+(\frac{1}{2})\times 4\times (5.0)^2][/tex]
⇒ [tex]=100+50[/tex]
⇒ [tex]=150[/tex]
On converting it into "rev", we get
⇒ [tex]\Delta \theta=(\frac{150}{2 \pi} )[/tex]
⇒ [tex]=23.87 \ rev[/tex]
Two loudspeakers S1 and S2, 2.20 m apart, emit the same single-frequency tone in phase at the speakers. A listener L is located directly in front of speaker S1, in other words, the lines LS1 and S1S2 are perpendicular. L notices that the intensity is at a minimum when L is 5.50 m from speaker S1. What is the lowest possible frequency of the emitted tone
Answer:
the lowest possible frequency of the emitted tone is 404.79 Hz
Explanation:
Given the data in the question;
S₁ ← 5.50 m → L
↑
2.20 m
↓
S₂
We know that, the condition for destructive interference is;
Δr = ( 2m + [tex]\frac{1}{2}[/tex] ) × λ
where m = 0, 1, 2, 3 .......
Path difference between the two sound waves from the two speakers is;
Δr = √( 5.50² + 2.20² ) - 5.50
Δr = 5.92368 - 5.50
Δr = 0.42368 m
v = f × λ
f = ( 2m + [tex]\frac{1}{2}[/tex])v / Δr
m = 0, 1, 2, 3, ....
Now, for the lowest possible frequency, let m be 0
so
f = ( 0 + [tex]\frac{1}{2}[/tex])v / Δr
f = [tex]\frac{1}{2}[/tex](v) / Δr
we know that speed of sound in air v = 343 m/s
so we substitute
f = [tex]\frac{1}{2}[/tex](343) / 0.42368
f = 171.5 / 0.42368
f = 404.79 Hz
Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz