The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensity if the frequency is increased to 2.20 kHz while a constant displacement amplitude is maintained.(b) Calculate the intensity if the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled.
Answers
Final answer:
The intensity of sounds is dependent on the square of the amplitude, not the frequency. Therefore, the intensity of sound remains the same when frequency is altered but the amplitude is constant. When the amplitude is quadrupled, the intensity of the sound becomes sixteen times greater.
Explanation:
In the field of physics, the intensity of a sound wave is defined as the power per unit area carried by the wave. This question involves calculating the change in sound wave intensity when the frequency and displacement amplitude of the source are altered.
(a) When the frequency is increased to 2.20 kHz while keeping the displacement amplitude constant, the intensity does not change, as the intensity in this case is not dependent on the frequency but on the square of the amplitude. Therefore, the intensity remains 0.750 W/m2.
(b) When the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled, the intensity changes. Since the intensity of a sound wave is proportional to the square of the amplitude, by quadrupling the amplitude, the intensity will become 16 times greater (since 4 squared is 16). Hence, the new intensity will be 16 * 0.750 = 12 W/m2.
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The focal length of a concave mirror is 17.5 cm. An object is located 38.5 cm in front of this mirror. How far in front of the mirror is the image located?
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Answer:
Air pockets.
Explanation:
Air pockets in the cooling system are bubbles of air trapped within the lines (hoses and pipes) of the cooling system. This air bubbles enter the cooling system usually during the process of filling the radiator coolant fluid (usually water), or replacing the water pump or the radiator hose during repairs or servicing of the cooling system. The trapped air prevent pressure movement that is needed by the coolant to move the heat generated from the engine cylinder, resulting in heat build up. The solution is to "bleed" the engine through the radiator lid or some air release valves.
Answers
The acceleration of the box is 0.81 m/sec².
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
According to Newton's second law, the resultant of the forces acting on the box is equal to the product between its mass and its acceleration:
(1)
we are only concerned about the horizontal direction, so there are only two forces acting on the box in this direction:
- the horizontal component of the force exerted by the rope, which is equal to
the frictional force, acting in the opposite direction, which is equal to
By applying Newton's law (1), we can calculate the acceleration of the
box,
The acceleration of the box is 0.81 m/sec².
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Answers
Answer:
L > 0.08944 m or L > 8.9 cm
Explanation:
Given:
- Flux intercepted by antenna Ф = 0.04 N.m^2 / C
- The uniform electric field E = 5.0 N/C
Find:
- What is the minimum side length of the antenna L ?
Solution:
- We can apply Gauss Law on the antenna surface as follows:
Ф =
- Since electric field is constant we can pull it out of integral. The surface at hand is a square. Hence,
Ф = E.(L)^2
L = sqrt (Ф / E)
L > sqrt (0.04 / 5.0)
L > 0.08944 m
Final answer:
The area of a square antenna needed to intercept a flux of 0.040 N⋅m2/C in a uniform electric field of magnitude 5.0 N/C is 0.008 m². Consequently, each side of the antenna must be about 0.089 meters (or 8.9 cm) long.
Explanation:
The question pertains to the relationship between electric field and flux. The electric flux through an area is defined as the electric field multiplied by the area through which it passes, oriented perpendicularly to the field.
We are given that the electric field (E) is 5.0 N/C and the flux Φ must be 0.040 N⋅m2/C.
Hence, to intercept this amount of flux, the antenna must have an area (A) such that A = Φ / E.
That is, A = 0.040 N⋅m2/C / 5.0 N/C = 0.008 m².
Since the antenna is square, each side will have a length of √(0.008) ≈ 0.089 meters (or 8.9 cm).
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B. 1.65 m/s
C. 10.4 m/s
D. 1040 m/s
Answers
Answer:
Maximum speed ( v ) = 10.4 m/s (Approx)
Explanation:
Given:
Amplitude A = 15.0 cm = 0.15 m
Frequency f = 11.0 cycles/s (Hz)
Find:
Maximum speed ( v )
Computation:
Angular frequency = 2πf
Angular frequency = 2π(11)
Angular frequency = 69.14
Maximum speed ( v ) = WA
Maximum speed ( v ) = 69.14 x 0.15
Maximum speed ( v ) = 10.371
Maximum speed ( v ) = 10.4 m/s (Approx)
Answers
The movement of a positively charged particle from point A to point B. the motion-induced electrostatic work done on the positively charged particle.
Whether positively or negatively charged, an object that is neutral will interact with it in a pleasing way. Both positively charged and neutral items attract one another, as do negatively charged and neutral objects. These electrons gather on the further surface of sphere B, depleting the electron supply in sphere A. Therefore, sphere A (which is closer to the rod) obtains a positive charge and sphere B acquires a negative charge when the two spheres separate in the presence of the rod. The change in the particle's electrostatic potential energy in the external field equals the work done by the external force. When a charge is pushed from point A to point B, its potential energy changes, representing the labor of an outside force.
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Answers
Answer: 38.5rad/s
Explanation: The calculations can be viewed on the image attached below. Thanks