Answer:
i) Telescopes can be used to view far distant objects but the human eye can't view far distant objects.
ii) Telescopes uses two convex lenses producing a magnified image while the human eye only possesses one convex lens (image seen are smaller than that viewed under telescopes)
Explanation:
The telescopes can be used to view far distant objects due to their presence of two convex lenses. The two convex lenses are the objective lens (lens closer to object) and the eye piece lens (lens closer to eye). The object to be viewed forms an intermediate image first before the final image is seen using the eye piece lens.
The human eye only possess one convex lens and as such cannot view far ranged objects.
There's no way to tell. Without seeing a diagram of the circuit,
I'll need to know much more about it than you've told me.
I don't know anything about the components or power supply
that are in the circuit, and I don't know where point ' f ' is in it.
Right now, even with the copious volume of all the available
information, no answer to your question is possible.
The farthest position the mouse reaches inside the tunnel is 4 meters into the tunnel.
From the graph,
The positions reached after,
5 s = 4 m
10 s = 2 m
20 s = 2 m
35 s = 3 m
40 s = 0 m
So the farthest position here is 4 m into the tunnel.
The rate of change of positions is displacement. So displacement will be change in initial and final positions divided by change in time.
s = Δx / Δt
Therefore, the farthest position the mouse reaches inside the tunnel is 4 meters into the tunnel.
To knw more about displacement
brainly.com/question/28609499
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Answer:
Explanation:
an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
Answer:
The time rate of change in air density during expiration is 0.01003kg/m³-s
Explanation:
Given that,
Lung total capacity V = 6000mL = 6 × 10⁻³m³
Air density p = 1.225kg/m³
diameter of the trachea is 18mm = 0.018m
Velocity v = 20cm/s = 0.20m/s
dv /dt = -100mL/s (volume rate decrease)
= 10⁻⁴m³/s
Area for trachea =
0 - p × Area for trachea =
⇒
ds/dt = 0.01003kg/m³-s
Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s
