Answer:
C.) Wave B has about 2.25 times more energy than wave A
Explanation:
Answer:
The range of wavelengths of the sound is 7692.30 m and 3846.15 m
Explanation:
A bat emits pulses of sound at a frequency between 39 kHz and 78 kHz. It is required to find the range of wavelengths of this sound.
Bat uses ultrasonic waves. It moves with the speed of light.
If f = 39 kHz,
If f = 78 kHz,
So, the range of wavelengths of the sound is 7692.30 m and 3846.15 m.
This can be solved using momentum balance, since momentum is conserved, the momentum at point 1 is equal to the momentum of point 2. momentum = mass x velocity
m1v1 = m2v2
(0.03kg x 900 m/s ) = 320(v2)
v2 = 27 / 320
v2 = 0.084 m/s is the speed of the astronaut
(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.
(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.
<h3>
Magnitude of electric field </h3>
The magnitude of electric field is given by the following equation.
F = qE
But F = mg
mg = qE
E = mg/q
where;
- E is the electric field
- m is mass of the particle
- g is acceleration due to gravity
- q is charge of the particle
<h3>For an electron</h3>
E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)
E = 5.57 x 10⁻¹¹ N/C
<h3>For proton</h3>
E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)
E = 1.02 x 10⁻⁷ N/C
Thus, the required vertical electric field is greater when the charge is proton.
Learn more about electric field here: brainly.com/question/14372859
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Explanation:
When one coulomb charge passes through any cross section of the wire per second,the current passing is one ampere. Charge of electron ,e=1.6X10^-19C. n=1/(1.6X10^-19)=6.25X10^18.Sep 17, 2017
