Answer:
t = 4.17 hours
Explanation:
given,
The distance between Sun and Neptune, d = 4.5 billion Km
= 4.5 x 10⁹ Km
= 4.5 x 10¹¹ m
The velocity of light, c = 3 x 10⁸ m/s
The velocity is always equal to displacement by the time.
<em>V = d / t m/s</em>
∴ t = d / V
= 4.5 x 10¹¹ m / 3 x 10⁸ m/s
= 15,000 s
= 4.17 h
Hence, the time taken by the light rays to reach the Neptune is, t = 4.17 h
I think you can just sub the values in? unless the qn is asking for smth else?
Answer:
I dont see a picture where is it?
Explanation:
I cannot see anything L question Luestion
Answer:
v = 10 m/s
Explanation:
Let's assume the wheel does not slip as it accelerates.
Energy theory is more straightforward than kinematics in my opinion.
Work done on the wheel
W = Fd = 45(12) = 540 J
Some is converted to potential energy
PE = mgh = 4(9.8)12sin30 = 235.2 J
As there is no friction mentioned, the remainder is kinetic energy
KE = 540 - 235.2 = 304.8 J
KE = ½mv² + ½Iω²
ω = v/R
KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²
v = √(2KE / (m + I/R²))
v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6
v = 10.07968...
Answer:
7.78x10^-8T
Explanation:
The Pointing Vector S is
S = (1/μ0) E × B
at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,
S = (1/μ0) E B
where S, E and B are magnitudes. The average value of the Pointing Vector is
<S> = [1/(2 μ0)] E0 B0
where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)
Also at any instant,
E = c B
where E and B are magnitudes, so it must also be true at the instant of peak values
E0 = c B0
Substituting for E0,
<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²
Solve for B0.
Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)
= 7.79 x10 ^-8 T