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3 years ago

What is the car's average velocity (in m/s) in the interval between t = 0.5 s to t = 2 s?

Physics

1 answer:

irakobra [83]3 years ago

4 0

Answer:

$1.0\; \rm m \cdot s^{-1}$ .

Explanation:

Consider a time period of duration $\Delta t$ . Let change in the position of an object during that period of time be denoted as $\Delta x$ . The average velocity of that object during that period would be:

$\displaystyle \bar{v} = \frac{\Delta x}{\Delta t}$ .

For the toy car in this question, the time interval has a duration of $2.0 \; \rm s - 0.5\; \rm s = 1.5\; \rm s$ . That is: $\Delta t = 1.5\; \rm s$ . (One decimal place, two significant figures.)

On the other hand, what would be the change in the position of this toy car during that $1.5\; \rm s$ ?

Note, that from readings on the snapshot in the diagram:

The position of the toy car was $0.1\; \rm m$ at $t = 0.5\; \rm s$ (the beginning of this $1.5$ -second time period.)
The position of the toy car was $1.6\; \rm m$ at $t = 2.0\; \rm s$ (the end of this $1.5$ -second time period.)

Therefore, the change to the position of this toy car over that time period would be $\Delta x = 1.6\; \rm m - 0.1\; \rm m = 1.5\; \rm m$ . (One decimal place, two significant figures.)

The average velocity of this car over this period of time would thus be:

$\displaystyle \bar{v} = \frac{\Delta x}{\Delta t} = \frac{1.5\; \rm m}{1.5\; \rm s} = 1.0\; \rm m \cdot s^{-1}$ . (Two significant figures.)

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Answer:

a. $B= 9.45 \times10^{-3} T$

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c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

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J: current density

c: outer radius

b: inner radius

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$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

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