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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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Two models of the same compound are shown. In what way is Model A better than Model B?
astra-53 [7]

The correct answer is Model A shows the three-dimensional shape of the molecule, but Model B does not.

Explanation:

Model A and B show the structure of a molecule. In the case of model A, the structure is represented through the use of three-dimensional shapes, while in model B the structure is represented using the letters of each element and showing how each element is connected to others.

In this context, one feature that makes model A better is that this represents the molecule using a 3D model, which is better to understand how the molecule looks like and what is its structure. Moreover, both models are alike because they show the number of atoms of each element, although model A does not show the types of elements.

7 0
3 years ago
Read 2 more answers
How long does it take the students’ kettle to come to the boil if 3 kg of ice at 0 °C is put in it? The latent heat of fusion of
Alborosie

Answer:

Explanation:

111.333

3 0
3 years ago
Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s
alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

  • Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
  • Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
  • vₓ₀ = v * cos θ (1)
  • where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
  • Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        x_{f} = v_{ox} * t = v_{o} * cos \theta * t  (2)

  • Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)

  • ⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

  • At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
  • Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
  • vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

  • At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

  • Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       \Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)

  • Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
  • v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
  • Replacing (7) in (6), we get:

       \Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)

e)

  • When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
  • The horizontal component, since it keeps constant, is just v₀x:
  • v₀ₓ = 13.7 m/s
  • The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       v_{fy} = v_{oy} - g*t  (9)

  • Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s  (10)

  • Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)

3 0
3 years ago
What would the final temperature be if 8.94 X 10 3 joules of heat were added to 454 grams of copper at 23.0 o C?
ozzi

Answer: 74.1^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=8.94\times 10^3 Joules

m= mass of copper = 454 g

c = specific heat capacity = 0.385J/g^0C

Initial temperature of the copper = T_i = 23.0°C

Final temperature of the water = T_f  = ?

Change in temperature ,\Delta T=T_f-T_i

Putting in the values, we get:

8.94\times 10^3=454\times 0.385\times (T_f-23.0)^0C

T_f=74.1^0C

The final temperature of copper will be 74.1^0C

8 0
3 years ago
A hydraulic lift has two connected pistons with cross-sectional areas 15 cm2 and 670 cm2. It is filled with oil of density 510 k
BigorU [14]

Answer:

Explanation:

Given

Cross-sectional area of two areas is

A_1=15\ cm^2

A_2=670\ cm^2

It is filled with oil of density \rho _0=510\ kg/m^3

mass of car place on Large area M=1100\ kg

Suppose a mass of m kg is placed on smaller area

According to pascal law's intensity of pressure is same at every point on Liquid

P_1=P_2

\frac{F_1}{A_1}=\frac{F_2}{A_2}

\frac{mg}{15}=\frac{Mg}{670}

m=1100\times \frac{15}{670}

m=24.62\ kg                            

4 0
3 years ago
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