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2 years ago

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If you stand next to a wall on a frictionless skateboard and push the wall with a force of 40n, how hard does the wall push on y

ou? if your mass is 80 kg, show that your acceleration is 0.5m/s2.

1 answer:

pentagon [3]2 years ago

4 0

The wall pushes back with the same force you exert so 40N.

to show your acceleration is .5m/s^2 use newtons second law F=ma

so plugging in numbers gives 40N=80kg*a knowing that a newton is equal 1kgm/s^2 we could write 40kg*m/s^2= 80kg*a so solving for a gives

40kg*m/s^2/80kg = a we see the kg's cancel and we're left with
(40m/s^2)/80=a which gives .5m/s^2=a

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2 years ago

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Answer:

<em>a) 17.05 mph</em>

<em>b) 54.7° northeast direction</em>

<em>c) 10.71 mph</em>

<em>The direction is -22.58° relative to the east.</em>

<em></em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

Explanation:

The question is a little confusing but, I guess the correct question should be;

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.

a) What is your airspeed?

b) What angle (direction) are you flying?

c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?

NB: The difference in the question and my suggestion is highlighted boldly.

Your speed = 14 mph

direction is 45° northeast

Th wind speed = 4 mph

direction is north

We resolve the your speed and the wind speed into the horizontal and vertical components

For vertical the component component

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For the horizontal speed component

$V_{x}$ = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph

Resultant speed = $\sqrt{V^{2} _{y}+V^{2} _{x} }$

==> $\sqrt{13.89^{2} +9.89^{2} }$ = <em>17.05 mph This is your airspeed</em>

b) To get your direction, we use

tan ∅ = $V_{y}$ / $V_{x}$

tan ∅ = 13.89/9.89 = 1.413

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$V_{y}$ = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph

Resultant speed = $\sqrt{V^{2} _{y}+V^{2} _{x} }$

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Your direction will be,

tan ∅ = $V_{y}$ / $V_{x}$

tan ∅ = -4.11/9.89 = -0.416

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2 years ago

Read 2 more answers

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Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

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2 years ago