Answer:
6.7 × 
Explanation:
Initial P = Final P
(1.8 × 0) + (.0075 × 0) = (1.8 × x) + (.0075 × 16)
-.12 + 1.8x
x = .0067
The period of the wave is 4.35 ms. The sound waves are called longitudinal waves
Explanation:
The period of a wave is related to its frequency by the equation:

where
T is the period
f is the frequency
For the bee in this problem, the frequency of the sound wave emitted by it is

Therefore, the period of the sound wave is

The sound wave is a type of wave called longitudinal wave. In longitudinal waves, the oscillation of the medium occurs in a direction parallel to the direction of motion of the wave: therefore in a sound wave, the particle of the medium (air, in this case) oscillate back and forth along the direction of propagation of the wave, forming alternating areas of higher density of particles (called compressions) and of lower density of particle (called rarefactions).
The other type of wave, instead, is called transverse wave. In a transverse wave, the oscillation of the wave occurs in a direction perpendicular to the direction of motion of the wave. An example of transverse waves are the electromagnetic waves, which consists of electric field and magnetic fields that vibrate in a plane perpendicular to the direction of motion of the wave itself.
Learn more about waves:
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Answer:
False
Explanation:
The formula of force that exists between two charges is expressed as;
F = kq1q2/r²
If two charges separated by one meter exert a 9 N force on each other, the;
9 = kq1q2/1²
9 = kq1q2 ..... 1
If the charges are pushed to a 3 meter separation, then;
F = kq1q2/3²
F = kq1q2/9 .... 2
Divide both equations;
9/F = (kq1q2)/ kq1q2/9
9/F = kq1q2 * 9/ kq1q2
9/F = 9
F = 9/9
F = 1N
Hence if the charges are pushed to a 3 meter separation, then the force on EACH charge will be 1N. Hence the answer is False
Answer:
2.59 T
Explanation:
Parameters given:
Current flowing through the wire, I = 29 A
Angle between the magnetic field and wire, θ = 90°
Magnetic force, F = 2.25 N
Length of wire, L = 3 cm = 0.03 m
The magnetic force, F, is related to the magnetic field, B, by the equation below:
F = I * L * B * sinθ
Inputting the given parameters:
2.25 = 29 * 0.03 * B * sin90
2.25 = 0.87 * B
=> B = 2.25/0.87
B = 2.59 T
The magnetic field strength between the poles is 2.59 T