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Ivan
2 years ago
12

A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe

ed is 8.45 m/s. The rope's breaking strength is 1,000 N. Will the stuntman make it across the moat without falling in? Yes No (b) What If? What is the maximum speed (in m/s) that the stuntman can have at the bottom of the swing on this vine to safely swing across the river? m/s
Physics
1 answer:
goldenfox [79]2 years ago
5 0

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

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) 5 -5 1 2 3 4 5 Other than at t = 0, when is the velocity of the object equal to zero? 1. 5.0 s 2. 4.0 s 3. 3.5 s 4. At no other time on this graph. correct 5. During the interval from 1.0 s to 3.0 s. Explanation: Since vt = Z t 0 a dt, vt is the area between the acceleration curve and the t axis during the time period from 0 to t. If the area is above the horizontal axis, it is positive; otherwise, it is negative. In order for the velocity to be zero at any given time t, there would have to be equal amounts of positive and negative area between 0 and t. According to the graph, this condition is never satisfied. 005 (part 1 of 1) 0 points Identify all of those graphs that represent motion at constant speed (note the axes carefully). a) t x b) t v c) t a d) t v e) t a 
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2 years ago
****PLEASE HELP**** THERE ARE TWO QUESTIONS (ITS EASY)
notka56 [123]

but I think it's a and f

and

b and e

3 0
2 years ago
A car mass of 1.2 x 10 kilograms starts from rest and attains a speed of 20 meters/seconds in 5 seconds. What net force acted on
lorasvet [3.4K]

Net force on the car=F=4.8 x 10³ N

Explanation:

mass of car= 1.2 x 10³ Kg

initial velocity= Vi=0

Final velocity= Vf= 20 m/s

time = t= 5 s

Using kinematic equation,

Vf= Vi + at

20= 0 + a (5)

5 a=20

a= 20/5

a= 4 m/s²

Now force is given by F = ma

F= 1.2 x 10³ (4)

F=4.8 x 10³ N

7 0
2 years ago
Refer to the attached image!!!​
dimaraw [331]

The time of motion of the track star is determined as 0.837 s.

<h3>Time of motion of the track star</h3>

The time of motion of the track star is calculated as follows;

T = (2u sinθ)/g

where;

  • T is time of motion
  • g is acceleration due to gravity
  • θ is angle of projection

T = (2 x 12 x sin20)/9.8

T = 0.837 s

Learn more about time of motion here: brainly.com/question/2364404

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6 0
1 year ago
If velocity is positive, which woul
shusha [124]

Answer:

C. An inital volocity that is faster than the final volocity

Explanation:

.

5 0
2 years ago
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