A SPIDER CRAWLS VERTICALLY ON A WALL Y VS TIME T WHAT IS THE AVERAGE VELOCITY OF THE SPIDER BETWEEN THE TIME T=4s and t=10s

Physics

1 answer:

GarryVolchara [31]3 years ago

3 0

Answer:

The question is incomplete,

the displacement time graph is missing. Check attachment for the graph.

Explanation:

Average velocity can be find by using

Average velocity = change in displacement / time taken

A.V = ∆y / ∆t

So, at time t1 = 4s, the position of the body is at 2m

y1 = 2m

Also, at time t2 = 10s, the position of the body is at 5m

y2 = 5m.

So,

A.V = ∆y / ∆t

A.V = (y2 - y1) / (t2 - t1)

A.V = (5 - 2) / (10 - 4)

A.V = 3 / 6

A.V = 0.5m/s

So, the average velocity is 0.5m/s

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The length of a cylindrical axon is 8 cm and its radius of 8μm,and the thickness of the membrane is 0.01μm,dielectric constant (

GuDViN [60]

Answer:

9.965 nF

Explanation:

The capacitance of the axon C = εA/d where ε = dielectric constant = 24.78 × 10⁻¹² F/m, A = surface area of axon = 2πrL where r = radius of axon = 8 μm = 8 × 10⁻⁶ m and L = length of axon = 8 cm = 8 × 10⁻² m and d = thickness of membrane = 0.01 μm = 0.01 × 10⁻⁶ m

So, C = εA/d

C = ε2πrL/d

Substituting the of the values variables into the equation, we have

C = ε2πrL/d

C = 24.78 × 10⁻¹² F/m × 2π × 8 × 10⁻⁶ m × 8 × 10⁻² m/0.01 × 10⁻⁶ m

C = 9964.63 × 10⁻²⁰ Fm/0.01 × 10⁻⁶ m

C = 996463 × 10⁻¹⁴ F

C = 9.96463 × 10⁻⁹ F

C = 9.96463 nF

C ≅ 9.965 nF

6 0

3 years ago