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Nana76 [90]
3 years ago
11

A horn on a boat sounds a warning, and the sound penetrates the water. How does the frequency of the sound in the air compare to

its frequency in the water? How does the wavelength in the air compare to the wavelength in the water?
Physics
1 answer:
Illusion [34]3 years ago
7 0
<span>Sound travels faster in water than in air because water has closer particles than in air. This means that the wavelength of sound in water is larger and longer. and its frequency is lower than the sound as heard in the air.</span>
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Which, if any, of the following statements about electric field lines is/are true? A.The electric field is always perpendicular
jonny [76]
The electric field is always perpendicular to the surface outside of a conductor. TRUE

<span> If an electron were placed on an electric field line, it would move in a direction perpendicular to the field. FALSE, it would move in an anti-parallel direction because its charge is negative </span>
 
<span>Electric field lines originate on positive charge and terminate on negative charge. TRUE ; but they can also go to infinity </span>
 
It is possible for two electric field lines to cross each other.
<span> Usually FALSE; though technically possible at special points where field is zero. </span>
 
If an electron and a positron were in the presence of a very strong electric field, they would move away from each other.
<span> TRUE; one is positive, and one is negative. If the field is strong enough, the action of the field will overcome the mutual attraction between them </span>
 
It is not possible for the electric field to ever be zero. FALSE: it IS possible, inside a conductor for instance
   
If a proton were placed on an electric field line, it would move in a direction anti-parallel to the field.
<span> FALSE: being positive, it would move in the SAME direction as the field</span>ic 
8 0
3 years ago
The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizont
Elza [17]

Answer:

1923 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 65 Kg

Radius (r) = 2.5 m

Velocity (v) = 8.6 m/s

Centripetal force (F) =?

The centripetal force, F, can be obtained by using the following formula:

F = mv²/r

F = 65 × 8.6² / 2.5

F = 65 × 73.96 / 2.5

F = 4807.4 / 2.5

F = 1922.96 ≈ 1923 N

Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N

3 0
3 years ago
A heat engine exhausts 3 000 J of heat while performing 1 500 J of useful work. What is the efficiency of the engine
amid [387]

efficiency=work output/work input×100

since it exhausts(use up)3000j of heat that's the work input and the 1500j is the work input

efficiency=1500/3000×100

=50%

3 0
3 years ago
An oven mitt is made of a material that is a poor thermal conductor.
ddd [48]

Answer:

D

Explanation:

5 0
3 years ago
A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em
Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

P = \ro g h = 1000 * 10 * 20 = 200000 Pa

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

P_1V_1 = P_2V_2

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

V_1 = Ah

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

P_1Ah_1 = P_2Ah_2

h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m

b) If the temperatures changes, we can still reuse the ideal gas equation above:

\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}

h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m

3 0
3 years ago
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