Answer:
Explanation:
At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation

Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.
Simplifying
v = 
Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get
v =
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v = 387 x 10⁻⁵ m/s
Terminal velocity = 387 x 10⁻⁵ m/s
Time taken to fall a distance of 100 m
= 
= 2.6 x 10⁴ s.
The kinetic energy K = 0.5 * m * v² must be equal to the potential energy U = m * g * h.
m mass
v velocity
h height
g = 9.81m/s²
The mass m cancels out:
0.5 * v² = g * h
Solve for height h and transform to distance traveled.
(sin (4°) = height / distance)
Answer:
According to the travellers, Alpha Centauri is <em>c) very slightly less than 4 light-years</em>
<em></em>
Explanation:
For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is <em>very slightly less than 4 light-years. </em>The following expression explains why:
v = d / t
where
- v is the speed of the spaceship
- d is the distance
- t is the time
Therefore,
d = v × t
d = (0.999 c)(4 light-years)
d = 3.996 light-years
This distance is<em> very slightly less than 4 light-years. </em>
W=mgh
W=(6)(9.8)(4)
W= 235.2J
Answer:
confocal microscopy
Explanation:
According to my research on different types of microscopes, I can say that based on the information provided within the question the tool being mentioned in this situation is a confocal microscopy. This is an extremely powerful microscope used to develop extremely sharp images of cells and tissues by viewing one plane of the specimen at a given time.
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