Answer:
Following are the answer to this question:
Explanation:
In option (a):
- The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.
- Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.
In option (b):
- Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.
- Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.
The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N.
<span>Fx = [(233 + 840)/g]*v²/7.5 </span>
<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>
<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>
<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>
<span>233 + 840 = Ti*cos40º </span>
<span>solve for Ti. (This is the answer to the part b) </span>
<span>Horizontally </span>
<span>[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º </span>
<span>Solve for Th </span>
<span>Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º </span>
<span>using v and Ti computed above.</span>
Answer:
a) 0.162 V/m
b) 0.54 nT
c) 22 kW
Explanation:
Given
Distance of flight, d = 20 km
Intensity of signal, I = 35.1 μW/m²
Magnitude of the electric component is gotten using the formula
E(m) = √(2Iμc), where
E(m) = √(2 * 35*10^-6 * 4*3.142*10^-7 * 3*10^8)
E(m) = √(2 * 35*10^-6 * 1.257*10^-6 * 3*10^8)
E(m) = √0.0264
E(m) = 0.162 V/m
Magnitude of magnetic component can be gotten by using the relation
B(m) = E(m) / c
B(m) = 0.162 / 3*10^8
B(m) = 5.4*10^-10
B(m) = 0.54 nT
Transmission power, P = IA
where A = 1/2 * 4πr²
P = 2Iπr²
P = 2 * 35*10^-6 * 3.142 * (10000)²
P = 2 * 35*10^-6 * 3.142 * 1*10^8
P = 21994 W
Thus, the transmission power is 22 kW
Answer:
Yes
Explanation:
sun rays towards the earth is an example of velocity
