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Kitty [74]
2 years ago
6

To compare the hearing capacities of 5 of his friends, Ravi designs a simple experiment. He places a CD player at the end of a l

ong hall containing a CD with 5 well-known songs.
One by one each person hears the same 5 songs he plays softly but at a pre-determined volume.

Each person starts at a distance of 1 metre from the player and moves away from it. The distance at which a person says that he cannot hear the song any more, is noted. Others are not in the room when 1 person is hearing the songs.
Which of these steps may increase the reliability of the results obtained?

A)having each person listen to just 1 song instead of 5
B)allowing all people to stay in the room for the entire experiment
C)having each person start from far and walk towards the C
D )player Dvarying the volume for the different songs being played
Physics
1 answer:
aliya0001 [1]2 years ago
8 0

Answer:

A) having each person listen to just 1 song instead of 5.

Explanation:

Ravi could accurately use the outcome to compare the hearing capacities of his friends by this experiment if appropriate precautions are observed.

Playing 5 well known songs simultaneously could result to the interference of sounds when each participant moves away from the player. Which could naturally cause the variation in volume of the songs played with respect to the increasing distance.

Therefore, the reliability of the result from this experiment can be increased if each person listen to just 1 song instead of 5 at a predetermined volume. Each participant would be able to focus on hearing a song during the experiment.

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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
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

6 0
2 years ago
On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max
sesenic [268]

Work done by a given force is given by

W = F.d

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = W_1 = Fdcos\theta

W_1 = 12*5cos15

W_1 = 57.96 J

Now similarly work done by frictional force

W_2 = Fdcos\theta

W_2 = 4*5cos180

W_2= -20 J

Now total work done on sled

W_{net}= W_1 + W2

W_{net} = 57.96 - 20 = 37.96 J

7 0
2 years ago
The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s2. Neptune has mass 1.02×1026 kg and radiu
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Answer: (a) The gravitational force on the object at the North Pole of Neptune is 51.7N

(b) The apparent weight of the object at Neptune's equator is 50.4N

Explanation: Please see the attachments below

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2 years ago
Fill in the blanks to correctly complete the statement. The motion of an object moving with uniform circular motion is always to
Citrus2011 [14]
The sentence can be completed as follows:

<span>The motion of an object moving with uniform circular motion is always tangential to the circle, so the speed of an object moving in a circle is known as tangential speed.

The object moves by uniform circular motion due to the presence of a force (called centripetal force) pointing toward the center of the circle. Due to the presence of this force, the object experiences an acceleration (called centripetal acceleration) that makes the object turning in a circle. This centripetal acceleration changes only the direction of the velocity of the object, not its magnitude, which is called tangential speed and it is constant.</span>
8 0
2 years ago
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