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irinina [24]
2 years ago
5

If the volume on your TV is low, turning the volume up one click of the remote control will make the TV seem louder than if the

volume were already relatively high. This ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the __________ fraction.
A. Fechner
B. Young
C. Weber
D. Helmholtz
Physics
1 answer:
marusya05 [52]2 years ago
7 0

From what we know, we can confirm that this ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the Weber fraction.

<h3>What is the Weber fraction?</h3>

This fraction describes the ratio needed for change to a stimulus in which the change is just barely noticeable. This question is a prime example in that it seeks to find out just how low of a difference is needed in TV volume in order for the difference to be noticeable.

Therefore, we can confirm that this ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the Weber fraction.

To learn more about Weber visit:

brainly.com/question/5004433?referrer=searchResults

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A wave travels at 330m/s^-1. the wavelength is found to be 2.4m.
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Answer: frequency 137.5

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A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w
Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, a=3\ m/s^2

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.  

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

m=\dfrac{W}{g}

m=\dfrac{150}{9.8}

m=15.3\ kg

Frictional force is given by :

F=ma

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So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0
3 years ago
A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subj
KATRIN_1 [288]

Answer:

\Delta t = 8 s

Explanation:

As we know that the angular acceleration of the wheel due to friction is constant

so we can use kinematics

\theta = \omega_i t + \frac{1}{2}\alpha t^2

so we have

(65 \times 2\pi) = (2\pi \times 9)(10) + \frac{1}{2}(\alpha)(10^2)

130\pi = 180\pi + 50 \alpha

\alpha = -\pi rad/s^2

now time required to completely stop the wheel is given as

\omega_f = \omega_i + \alpha t

0 = (2\pi \times 9) + (-\pi) t

t = 18 s

now time required to stop the wheel is given as

\Delta t = 18 - 10

\Delta t = 8 s

6 0
3 years ago
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