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zhannawk [14.2K]
3 years ago
6

A jogger runs 4.0 km [W] in 0.50 h, then turns and runs 1.0 km [E] in 0.20 h, then 1.5 km [N] in 0.25 h, then 3.0 km [E] in 0.75

h and finally 1.5 km [S] in 0.30 h.  What is the average velocity?
Physics
2 answers:
Kisachek [45]3 years ago
6 0
After you draw the displacement vectors you will find the gogger returned to the start location
velocity = (displacement) /  ( t )      >>>>>>>>>>displacement = 0 ( jogger returns to its started point) , t = 0.5 h + 0.2h + 0.25h + 0.75h = 1.7 h
So v = 0/ 1.7 h = 0 
GrogVix [38]3 years ago
3 0
This is a sneaky trick question, to help you discover whether you know
one of the differences between velocity and speed.
=======================================
If you make a list of the distances and directions, and ignore the times,
you find these:

4 - west,  (3 + 1) - east . . . . .  zero in the east/west direction

1.5 - north,  1.5 - south . . . . . zero in the north/south direction

This jogger went out, had a nice jog around the neighborhood,and ended up exactly where he started.

Average velocity = (distance between start point and end point) / (time)

IF the question asked for average SPEED, then you would need the total distance, and divide it by the total time.  But it asks for VELOCITY, and <u>that</u> only involves the straight distance between the start point and the end point, regardless of the route taken in between.

The jogger ended up exactly where he started.  The distance between start and end points was zero.  Average velocity is  (zero) / (time) .  And that fraction is going to be <em><u>Zero</u></em>, no matter how long or how short the trip was, and no matter how much time it took.


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560.06714 Nm

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\omega_f = Final angular velocity

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R = Radius of swing = 0.984 m

\omega_f=\dfrac{v}{r}\\\Rightarrow \omega_f=\dfrac{29.8}{0.984}\\\Rightarrow \omega_f=30.28455\ rad/s

From equation of rotatational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{30.28455^2-0^2}{2\times \pi}\\\Rightarrow \alpha=145.96958\ rad/s^2

Moment of inertia is given by

I=\dfrac{1}{3}MR^2+mR^2\\\Rightarrow I=\dfrac{1}{3}11.3\times 0.984^2+0.196\times 0.984^2\\\Rightarrow I=3.83687577\ kgm^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=3.836875776\times 145.96958\\\Rightarrow \tau=560.06714\ Nm

The torque the pitcher applies is 560.06714 Nm

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MA_775_DIABLO [31]

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The density of the object does not change when it is broken in more pieces, therefore Maria will be able to find the density of the material.

In fact, by breaking the object, then she can take only one piece, and by measuring the mass of this piece and its volume she will be able to calculate the density of the material by using the formula:

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8 0
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A vehicle moving with a uniform acceleration of 2 m/s2 has
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Answer:

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b) a = Δv / Δt

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