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Aliun [14]
2 years ago
14

A pontoon boat travels across a lake in a straight line and increases in speed uniformly from vi = 18.5 m/s to vf = 36.0 m/s in

a displacement Δx of 250 m. We wish to find the time interval required for the boat to move through this displacement. g
(a) Draw a coordinate system for this situation.

(b) What

analysis model is most appropriate for describing this situation?

(c) From the analysis model, what equation is most

appropriate for finding the acceleration of the speedboat?

(d) Solve the equation selected in part (c) symbolically for

the boat’s acceleration in terms of vi , vf , and ?x. (e) Substitute

numerical values to obtain the acceleration numerically.

(f) Find the time interval mentioned above.
Physics
1 answer:
omeli [17]2 years ago
8 0

Answer:

Explanation:

a ) The motion is one dimensional , so motion is along x - axis , starting from origin ( 0 , 0 )

b ) Initial velocity is 18.5 m /s when boat is situated at origin . When he displaces by 250 m along x axis and his position is ( 250 , 0 ) along x axis , his velocity becomes 36 m /s . Both his velocity and acceleration is along x - axis.

c ) Initial velocity vi = 18.5 m /s

final velocity vf = 36 m/s

Displacement x  = 250 m

Acceleration a = ?

Most appropriate formula is given below .

vf² = vi² + 2 a x

2ax = vf² - vi²

x =  ( vf² - vi² ) / 2 a

d )

Putting the given values

36² - 18.5² / 2 x 250

= 1296 - 342.25 / 500

= 1.9 m /s².

f ) Time interval t = ?

Required formula

vf = vi + at

t = (vf - vi ) / a

Putting the values

t = (30 - 18.5) / 1.9

= 6.05 second .

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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
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Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

    gm = 1/6  9.8 m/s² = 1.63 m/s²

We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

    R= 332 m

We will calculate the time of flight,

   Y = Voy t – ½ g t2  

   Voy = Vo sin θ

When the ball reaches the end point has the same initial  height Y=0

0 = Vo sin  t – ½  g t2

0 = 25 sin (30)  t – ½ 1.63 t2

0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

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In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.
Murrr4er [49]

The central force acting on the electron as it revolves in a circular orbit is 9.52 \times 10^{-8} \ N.

The given parameters;

  • <em>speed of electron, v = 2.2 x 10⁶ m/s</em>
  • <em>radius of the circle, r = 4.63 x 10⁻¹¹ m</em>

<em />

The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

F = \frac{M_e v^2}{r} \\\\

where;

M_e is mass of electron = 9.11 x 10⁻³¹ kg

F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N

Thus, the central force acting on the electron as it revolves in a circular orbit is 9.52 \times 10^{-8} \ N.

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