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2 years ago

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When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (

C) 4 (D) 8 (E) 258

1 answer:

Degger [83]2 years ago

3 0

Answer:

(C) 4 beats per second.

Explanation:

As we know that the no of beats can be calculated as.

No. of beats is equal to difference in the tuning forks frequencies.

So,

$n= \nu _{1}- \nu _{2}$ .

Substitute the values of frequencies of 2 tuning forks in the above equation.

$n=(260 Hz-256 Hz)\\n=4$

Therefore the number of beats per second will be hear by the observer is 4 beats per second.

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Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^

ioda

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

$T \approx e^{-2CL}$

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

$0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)$

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

$ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)$

Next, you divide the equation (3) into (4), and finally, you solve for L':

$\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L$

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

8 0

2 years ago

A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the

Sedbober [7]

Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is

$f = \sqrt{ \frac{T}{4mL} }$ <span> -- (A)</span>

<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>

Plug in the values in Equation (A)

<span>so </span> $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ <span> = 197.97Hz </span>

<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>

3 0

2 years ago

Read 2 more answers

How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle abov

sveta [45]

Answer:

x = v₀ cos θ t , y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This is a projectile launch exercise, in this case we will write the equations for the x and y axes

Let's use trigonometry to find the components of the initial velocity

sin θ = $v_{oy}$ / v₀

cos θ = v₀ₓ / v₀

v_{y} = v_{oy} sin θ

v₀ₓ = vo cos θ

now let's write the equations of motion

X axis

x = v₀ₓ t

x = v₀ cos θ t

vₓ = v₀ cos θ

Y axis

y = y₀ + $v_{oy}$ t - ½ g t2

y = y₀ + v₀ sin θ t - ½ g t2

v_{y} = v₀ - g t

v_{y} = v₀ sin θ - gt

$v_{y}^{2}$ = v_{oy}^2 sin² θ - 2 g y

As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components

7 0

2 years ago

A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume

dalvyx [7]

Explanation:

It is given that,

Mass of the car 1, $m_1=900\ kg$

Initial speed of car 1, $u_1=15i\ m/s$ (east)

Mass of the car 2, $m_2=750\ kg$

Initial speed of car 2, $u_1=20j\ m/s$ (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$900\times 15i +750\times 20j=(900+750)V$

$13500i+15000j=1650V$

$V=(8.18i+9.09j)\ m/s$

The magnitude of speed,

$|V|=\sqrt{8.18^2+9.09^2}$

V = 12.22 m/s

(b) Let $\theta$ is the direction the wreckage move just after the collision. It is given by :

$tan\theta=\dfrac{v_y}{v_x}$

$tan\theta=\dfrac{9.09}{8.18}$

$\theta=48.01^{\circ}$

Hence, this is the required solution.

4 0

2 years ago

A stationary particle of charge q = 2.1 × 10-8 c is placed in a laser beam (an electromagnetic wave) whose intensity is 2.9 × 10

alisha [4.7K]

(a) The intensity of the electromagnetic wave is related to the amplitude of the electric field by
$I= \frac{1}{2} c \epsilon_0 E^2$
where
I is the intensity
c is the speed of light
$\epsilon_0$ is the electric permittivity
E is the amplitude of the electric field

By substituting the numbers of the problem and re-arranging the equation, we can find E:
$E= \frac{2 I}{c \epsilon_0} = \frac{2 ( 2.9 \cdot 10^3 Wm^{-2})}{(3 \cdot 10^8 m/s)(8.85 \cdot 10^{-12} Fm^{-1})} =2.2 \cdot 10^6 N/C$

Now that we have the intensity of the electric field, we can calculate the electric force on the charge:
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(b) We can calculate the amplitude of the magnetic field starting from the amplitude of the electric field:
$B= \frac{E}{c}= \frac{2.2 \cdot 10^6 N/C}{3 \cdot 10^8 m/s}=7.3 \cdot 10^{-3} T$

The magnetic force is given by
$F=qvB \sin \theta$
where v is the particle's speed, B the magnetic field intensity and $\theta$ the angle between B and v.
In this case the charge is stationary, so v=0, and so the magnetic force is zero: F=0.

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(d) This time, the particle is moving with speed $v=3.7 \cdot 10^4 m/s$ , in a direction perpendicular to the magnetic field (so, the angle $\theta$ is $90^{\circ}$ ), and so by using the intensity of the magnetic field we found in point (b), we can calculate the magnetic force on the particle:
$F=qvB \sin \theta = (2.1 \cdot 10^{-8}C)(3.7 \cdot 10^4 m/s)(7.3 \cdot 10^{-3} T)(\sin 90^{\circ} )=$
$=5.7 \cdot 10^{-6} N$

5 0

2 years ago