Question 9:<br> All parts thank you !

You need to get to class, which is 282 meters away. You can only walk in the hallway at about 2.0 m/s (if you go any faster, you

worty [1.4K]

Answer:

Time, t = 141 seconds

Explanation:

Given that,

Distance between the class and the person, d = 282 m

Speed of the person, v = 2 m/s

We need to find the time taken by the person to get to your class. Let the time taken is t. The speed of the person is given by :

$v=\dfrac{d}{t}$

$t=\dfrac{d}{v}$

$t=\dfrac{282\ m}{2\ m/s}$

t = 141 seconds

So, the time taken by the person to get to your class is 141 seconds. Hence, this is the required solution.

6 0

2 years ago

A man seeking to set a world record wants to tow a 101,000-kg airplane along a runway by pulling horizontally on a cable attache

NemiM [27]

Answer:

6.19 x $10^{-3}$ m/ $s^{2}$

Explanation:

For this exercise we need to sum the forces on the y-axis and x-axis as follows:

∑ $F_{y}$ = N - mg = m. $a_{y}$ = 0

From the exercise, we deduce there is no motion in y-axis, so:

N = mg

Then for x-axis we have:

∑ $F_{x}$ = H - $f^{s}$ = m. $a_{x}$ = 0

Now, from the exercise we deduce that we are looking for the greatest static friction which means to have the maximun static friction to start moving, so at this point the acceleration is zero, so we can find horizontal force (H), which then will act in the airplane to move it. Therefore we have:

H = $f^{s}$ = $f^{sma_{x} }$ = $u_{s}$ N = $u_{s}$ mg