<span>move away from one another
(convergent is the opposite way)</span>
Answer:
24445.85 J/s
Explanation:
Area, A = 300 m^2
T = 33° C = 33 + 273 = 306 k
To = 18° C = 18 + 273 = 291 k
emissivity, e = 0.9
Use the Stefan's Boltzman law
Where, e be the energy radiated per unit time, σ be the Stefan's constant, e be the emissivity, T be the temperature of the body and To be the absolute temperature of surroundings.
The value of Stefan's constant, σ = 5.67 x 10^-8 W/m^2k^4
By substituting the values
E = 24445.85 J/s
Answer:
Mirage is an optical illusion due to the refraction of light as it travels through different layers of air at different temperature.
Explanation:
On a clear hot afternoon, the sun heats up the ground, and the ground heats up the air immediately above it. The result is that the air directly above the ground is hotter than the air above it, leaving the air layers with different refractive indies. Since refractive index decreases with increase in temperature, the air immediately above the ground bends the light that travels through it from the sky, directly into our eyes. The image seen is perceived to be reflected from the ground, but in an actual sense is a direct image of the sky, giving one the impression of seeing a pool of water on the ground.
Answer:
B meet A 0.01 km east of flagpole
Explanation:
given data
distance A = 5.7 km west
velocity V1 = 8.9 km/h
distance B = 4.5 km east
velocity V2 = 7 km/h
to find out
How far runners from the flagpole, when paths cross
solution
we know A and B are 5.7 + 4.5 = 10.2 km apart
and we consider here B will run distance x km for meet
so time will be for B is
time B = distance / velocity
time B = x / 7 ...................1
and
for A distance for meet = ( 10.2 - x ) km
so time A = distance / velocity
time A = ( 10.2 - x ) / 8.9 .............2
now equating equation 1 and 2
time A = time B
x / 7 = ( 10.2 - x ) / 8.9
x = 4.490
so distance of B run for meet is 4.490 km
so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km
so B meet A 0.01 km east of flagpole
Initially its moving with tail wind so here the speed of wind will support the motion of the plane
so we can say
now when its moving with head wind we can say that wind is opposite to the motion of the plane
now by using above two equations we can find speed of palne as well as speed of wind
