Answer:
11.25 amps
Explanation:
For transformers, the magnetic flux
Therefore;
Ф = Фmax (cosωt) = 0.21·(cos(5·t))
From Faraday's law of induction, we have;
ε = -N × dΦ/dt
Which gives;
dΦ/dt = -1.05(sin (5t)
)
ε = -N × dΦ/dt = -50× -1.05(sin (5t)
)
ε = 52.5(sin (5t)
)
I = ε/R = 52.5(sin (5t)
)/3.3 = 15.9091(sin (5t)
) amps
The peak current is therefore = 15.9091 amps
The rms current = Peak current /√2 = 15.9091/(√2) = 11.25 amps.
Let's be clear: The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.
Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.
After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.
Answer:
pleading
Explanation:
the first step in a lawsuit where parties pass their claims and their defenses. the plaintiff or the one complaining states the issue while the defendant states his answer on the complain and his defense
Answer:
No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.
Explanation:
A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.
Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:
* When the diameter and length are comparable (i.e have the same measurement)
When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.
Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.
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
