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Arada [10]
3 years ago
7

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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V125BC [204]
A) The acceleration is due to gravity at any given point if you look at it vertically, so -10 m/s^2.

b) sin(25) = V_y/V, so V_y = V*sin(25). We use V = V_0 + a t and then the final speed must be 0 because it stops at the highest point. So 0 = V_y - 10t. Solve for t and you get t = 32sin(25)/10 = 16sin(25)/5

c) Y = Y_0 + V_0t + (1/2)at^2, and then we plug the values: Y_m_a_x = 32sin(25)*t - (1/2)*10*t^2 and we already have the time from "b)", so Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2; then we just rearrange it Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100] and finally Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20
6 0
3 years ago
We are sending a 30 Mbit file from source host A to destination host B. All links in the path between source and destination hav
Dmitrij [34]

Answer:

t=1.5\times 10^{-4}\ s

Explanation:

Given:

  • file size to be transmitted, D=30\ Mb
  • transmission rate of data, \dot D=10\ Mb.s^{-1}
  • propagation speed, v=2\times 10^8\ m.s^{-1}
  • distance of data transfer, s=10000\ km=10^4\ m

<u>Now the delay in data transfer from source to destination for each 10 Mb:</u>

t'=\frac{s}{v}

t'=\frac{10^4}{2\times 10^8}

t'=5\times 10^{-5}\ s

<u>Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:</u>

So,

t=3\times t'

t=3\times 5\times 10^{-5}

t=1.5\times 10^{-4}\ s

3 0
4 years ago
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