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scoray [572]
3 years ago
7

Two speakers hang from racks placed in an open field. When sound of the same frequency comes from both speakers, no sound is hea

rd in some locations. The sound is loud in other places at about the same distance from the speakers. Explain.
Physics
1 answer:
Vanyuwa [196]3 years ago
8 0

Answer:

Explanation:

Wgen the sound is emitting from two speakers, the sound waves interfere each other. the locations at which the destructive interference occurs, we get no sound and the locations where constructive interference occurs, the sound occurs at that locations.

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Using an unmanned rocket to visit the space station requires 85.2 trillion BTU of energy. The best fuel for the mission will hav
Illusion [34]
We shall convert all of the densities to lbs/gal, so the product of
BTU/lbs and lbs/gal gives us the basis of comparison, which was "ratio of energy to volume".
grams / ml x 1 lbs/454 grams → 1 lbs/ 454 ml
1 lbs/454 ml x 3785.41 ml/gal → 3785.41 lbs/454gal
Conversion of g/ml = 8.34 lbs/gal
Looking at each fuel:

Kerosene:
18,500 x (8.34 x 0.82) = 126,517 BTU/gal

Gasoline:
20,900 x (8.34 x 0.737) = 128,463 BTU/gal

Ethanol:
11,500 x (8.34 x 0.789) = 75,673 BTU/gal

Hydrogen:
61,000 x (8.34 x 0.071) = 36,120 BTU/gal

The best fuel in terms of energy to volume ratio is Gasoline.
Gallons required:
BTU needed / BTU per gallon
= 85.2 x 10⁹ / 128,463
= 6.6 x 10⁵ gallons
5 0
3 years ago
The state of matter that has particles in a fixed position, with a definte volume and shape is
Rainbow [258]
That's the description of the SOLID phase of matter.
7 0
3 years ago
What is the theory behind a ticker tape lab?
Lana71 [14]
Purpose: experiments will use it to measure the straight-line accelerated motion of a human hand. The displacement data will be measured and velocity and acceleration will be calculated, run the ticker tape under the guides on the timer and under the carbon circle.Hope this helps! ; )
5 0
3 years ago
Learning Goal:
enot [183]

Answer:

A. U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

C=\dfrac{\epsilon A}{d}

C is the capacitance, A is the common plate area, d is the plate separation and \epsilon is the permittivity of the material between the plates.

For air or free space, \epsilon is \epsilon_0 called the permittivity of free space. In general, \epsilon=\epsilon_r \epsilon_0 where \epsilon_r is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, \epsilon_r=1.

The energy stored in a capacitor is the average of the product of its charge and voltage.

U = \dfrac{QV}{2}

Its charge, Q, is related to its capacitance by Q=CV (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for U,

U = \dfrac{CV^2}{2}

A. Substituting for C in U,

U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. When the distance is 3d,

U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}

U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. When the distance is restored but with a dielectric material of dielectric constant, K, inserted, we have

U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

6 0
3 years ago
A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t
levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

F=m\omega^2 r

where

m is the mass of the object

\omega is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s is the angular velocity

r = 30 cm = 0.30 m is the radius of the circle

Substituting, we find the force:

F=(0.040)(6.28)^2(0.30)=0.47 N

3 0
3 years ago
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