Answer:
E ’= E / 8
therefore the correct answer is A
Explanation:
Let's calculate the electric field in an insulating sphere with a radius r <R, let's use Gauus's law, with a spherical Gaussian surface
Фi = ∫ E. dA = /ε₀
E (4πr²) = q_{int} / ε₀
density is
ρ = q_{int} / V
q_{int} = ρ V = ρ 4/3 π r³
we substitute
E (4π r²) = ρ 4/3 π r³ /ε₀
E = 1 /3ε₀ ρ r
let's change the density by
ρ = Q / V = Q / (4/3 π R³)
E = 1 / 4πε₀ Q r / R³
if we now distribute the same charge on a sphere of radius R' = 2R
E ’= 1 / 4pieo Q r / (2R)³
E ’= 1 / 4ft Qr / R³ ⅛
E ’= E / 8
therefore the correct answer is A
<span>The
heavier the body is, the stronger its gravitational pull. Example, the Milky Way
Galaxy has a gravitational pull because of the heavenly bodies such as stars and planets are surrounding it. A strong force is exerted if the mass of another body is bigger than the other body.</span>
<span>The period of a wave can be calculate according T = 1 / f
where T is wave period and f is frequency,
then in our problem we have a wave of 100 Hz so:
T = 1 / 100 = 0.01 seconds
The wave have a period of 0.01 seconds.
Optinaly we could also calculate the speed: V=wavelenght/period= 2/0.01= 200 m/s</span>
Answer:
a) 361.7 Hz
b) 318.1 Hz
c) 29.3 m/s
Explanation:
a) Speed of sound in 20 degrees celcius is 343 m/s.
Use v=wavelength*frequency to find wavelength.
343=wavelength*340
wavelegth=1.009 meters
Relative speed of waves to observer A = 343+22 = 365 m/s
Plug it back in.
365=1.009*F
F=361.7 Hz
b) Relative speed of waves to observer B = 343-22 = 321 m/s
Plug it back in.
321=1.009*F
F=318.1 Hz
c) STATIONARY CAR, so 369Hz=v/1.009. v=372.3, 372.3-343=29.3m/s!
Answer:
<h2><em>
0.165Tesla</em></h2>
Explanation:
The Force experienced by the wire in the uniform magnetic field is expressed as F = BILsin∝ where;
B is the magnetic field (in Tesla)
I is the current (in amperes)
L is the length of the wire (in meters)
∝ is the angle that the conductor makes with the magnetic field.
Given parameters
L = 0.56 m
I = 2.6A
F = 0.24N
∝ = 90°
Required
magnitude of the magnetic field (B)
Substituting the given values into the formula given above we will have;
F = BILsin∝
0.24 = B * 2.6 * 0.56 sin90°
0.24 = B * 2.6 * 0.56 (1)
0.24 = 1.456B
1.456B = 0.24
Dividing both sides by 1.456 will give;
1.456B/1.456 = 0.24/1.456
B ≈ 0.165Tesla
<em>Hence the magnitude of the magnetic field is approximately 0.165Tesla</em>