Answer:
r = 4.21 10⁷ m
Explanation:
Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining
T² = (
) r³ (1)
in this case the period of the season is
T₁ = 93 min (60 s / 1 min) = 5580 s
r₁ = 410 + 6370 = 6780 km
r₁ = 6.780 10⁶ m
for the satellite
T₂ = 24 h (3600 s / 1h) = 86 400 s
if we substitute in equation 1
T² = K r³
K = T₁²/r₁³
K =
K = 9.99 10⁻¹⁴ s² / m³
we can replace the satellite values
r³ = T² / K
r³ = 86400² / 9.99 10⁻¹⁴
r = ∛(7.4724 10²²)
r = 4.21 10⁷ m
this distance is from the center of the earth
To calculate the initial velocity of the bike, we use the following equation
.
or

Here, u is initial velocity, v is final velocity, t is the time and d is the distance covered by bike.
Given,
,
and
.
Substituting these values in above equation, we get
.
Thus, the initial velocity of the bike is 1.2 m/s.
The appropriate response is Zero degrees. The beam will leave the two mirrors along a way parallel to the one it came in on. This is the guideline of the corner reflector, which is frequently utilized as a radar target. Take note of that the corner reflector utilizes three reflecting surfaces (that are set up at 90o from each other) rather than the two like are being utilized here. Wikipedia has a truly awesome drawing that shows this two-dimentional issue pleasantly. A moment connection is given to the article on the corner reflector and the 3-D angles.
Answer:
The angle it subtend on the retina is
Explanation:
From the question we are told that
The length of the warbler is 
The distance from the binoculars is 
The magnification of the binoculars is 
Without the 8 X binoculars the angle made with the angular size of the object is mathematically represented as



Now magnification can be represented mathematically as

Where
is the angle the image of the warbler subtend on your retina when the binoculars i.e the binoculars zoom.
So

=> 

Generally the conversion to degrees can be mathematically evaluated as
