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egoroff_w [7]
3 years ago
14

A 35.1 kg wagon is towed up a hill inclined at 18.3 ◦ with respect to the horizontal. The tow rope is parallel to the incline an

d has a tension of 125 N in it. Assume that the wagon starts from rest at the bottom of the hill, and neglect friction. The acceleration of gravity is 9.8 m/s 2 . How fast is the wagon going after moving 75.4 m up the hill? Answer in units of m/s.
Physics
1 answer:
allsm [11]3 years ago
3 0

Answer:

4.933m/s

Explanation:

the wagon has a weight of <u>35.1kg*9.81m/s2 = 343.98N</u>

of that weight <u>343.98N*sin(18.3)=108N </u>are parallel to the hill and oposit tothe tension of the rope.

then, the force that is moving the wagon is <u>125N-108N=17N</u>

<u>F=m*a then 17N=35.1kg*a</u>

<u>a=0.4843 m/s2</u>

we have two equations

v=a.t\\x=v.t+\frac{1}{2} . a . t^{2}

then

x=a.t^{2} +\frac{1}{2}. a t^{2} \\x=\frac{3}{2}. a t^{2}

t=\sqrt{\frac{2x}{3a} }

t=10.188s

v=a.t

<em><u>v=4.933 m/s</u></em>

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A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches
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Answer:

(a) The parachutist spent 24.84 secs in air

(b) The height the fall begins is 472 m

Explanation:

Here is the complete question:

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches the ground with a speed of 3.3 m/s. (a) How long is the parachutist in the air  (b) At what height does the fall begin?

Explanation:

From one of the equations of kinematics for free fall

H = ut - \frac{1}{2}gt^{2}

Where H is the height

u is the initial velocity

t is the time

and g is the acceleration due to gravity (Take g = 9.8 m/s2)

Now, we can find the time spent before the parachute opens.

u = 0 m/s (we assume the parachutist starts from rest)

H = - 63 m

∴-63 = 0(t) - \frac{1}{2}(0.98) t^{2}  \\-63 = -4.9 t\\t^{2} = \frac{63}{4.9} \\t^{2} = 12.86\\t = \sqrt{12.86} \\t = 3.59 s

This the time spent before the parachute opens

Also from one of the equations of kinematics for free fall

v = u - gt

where v is the final velocity

We can determine the final velocity before the parachute opens and she starts to decelerate

∴v = - 9.8(3.59)\\v = - 35.18m/s

Now, we will calculate the time spent after the parachute opens

From one of the equation of kinematics for linear motion,

v = u + at

Here, the initial velocity will be the final velocity just before the parachute opens, that is

u = - 35.18 m/s

From the question,

v = - 3.3 m/s

a = 1.5 m/s^{2}

We then get

-3.3 = - 35.18 + (1.5)t

-3.3 + 35.18 =  1.5t\\31.88 = 1.5t

t = \frac{31.88}{1.5}

t = 21.25 secs

(a) To determine how long the parachutist is in the air,

That is sum of the time used when falling freely and the time used after the parachute opens

= 3.59 secs + 21.25 secs

24.84 secs

Hence, the parachutist spent 24.84 secs in air

(b) To determine what height the fall begins

First, we will calculate the height from which the parachute opens

From one of the equation of kinematics for linear motion,

x = ut + \frac{1}{2}at^{2}  \\

x = -35.18(21.25) + \frac{1}{2}(1.5)(21.25)^{2}  \\x = -747.58 + 338.67\\x = -408.91m\\

x ≅ - 409m

Hence, the height the fall begins is 63m + 409m

= 472 m

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