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pickupchik [31]
2 years ago
13

Because of your knowledge of physics, you have been hired as a consultant for a new James Bond movie, "Oldfinger". In one scene,

Bond jumps horizontally off the top of a cliff to escape a villain. To make the stunt more dramatic, the cliff has a horizontal ledge a distance h beneath the top of the cliff which extends a distance LL from the vertical face of the cliff. The stunt coordinator wants you to determine the minimum horizontal speed, in terms of L and h, with which Bond must jump so that he misses the ledge.
Physics
1 answer:
Elis [28]2 years ago
8 0

Answer:

v = [√(g/2h)]L

Explanation:

Let v be the initial horizontal velocity, t be the time James Bond uses to jump over the ledge of length, L.

So, vt = L and t = L/v

Also, since James Bond has no initial horizontal velocity, he falls freely through the distance, h so we use the equation y - y' = ut - 1/2gt², where y = 0 (at the top of the cliff) and y' = -h, u = 0 (initial vertical velocity), g = acceleration due to gravity = 9.8 m/s² and t = the time it takes to jump off the cliff = L/v.

Substituting these values into the equation, we have

y' - y = ut - 1/2gt²

-h - 0 = 0 × t - 1/2g(L/v)²

-h  = - 1/2gL²/v²

v² = gL²/2h

taking square root of both sides, we have

v = [√(g/2h)]L

So, James Bond's minimum horizontal speed is v = [√(g/2h)]L

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An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined
MrRa [10]

Incomplete question as the car's  speed is missing.I have assumed car's  speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

F_{B}=-5755N

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

\frac{mv^{2} }{R}=W+F_{B}\\  F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N

The minus sign for downward direction

6 0
2 years ago
A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4
HACTEHA [7]

Answer:

7.9 m/s

Explanation:

When both balls collide, they have spent the same time for their motions.

Motion of steel ball

This is purely under gravity. It is vertical.

Initial velocity, <em>u </em>= 0 m/s

Distance, <em>s</em> = 4.0 m - 1.2 m = 2.8 m

Acceleration, <em>a</em> = g

Using the equation of motion

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

2.8 \text{ m} = 0+\dfrac{gt^2}{2}

t = \sqrt{\dfrac{5.6}{g}}

Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, <em>u </em>= ?

Distance, <em>s</em> = 1.2 m

Acceleration, <em>a</em> = -<em>g                   </em> (It is going up)

Using the equation of motion

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

1.2\text{ m} = ut-\frac{1}{2}gt^2

Substituting the value of <em>t</em> from the previous equation,

1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}

u\sqrt{\dfrac{5.6}{g}} = 4.0

Taking <em>g</em> = 9.8 m/s²,

u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}

This is the vertical component of the initial velocity

Considering the horizontal motion which is not accelerated,

horizontal component of the initial velocity is horizontal distance ÷ time.

u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}

The initial velocity is

v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}

4 0
2 years ago
Electromagnetic radiation is emitted when a charged particle moves through a medium faster than the local speed of light. This r
alexandr1967 [171]

Answer:

to create the particle the speed must be greater than 2.25 10⁸ m / s

Explanation:

In this exercise we must use the relation of the index of refraction with the speed of light in a vacuum and a material medium

           n = c / v

where c is the speed of light in the vacuum, v the speed of light in the material medium and n the ratio of rafraccio

in this case they give us that the medium matter water them that has a refractive index of

              n = 1,333

we clear

          v = c / n

let's calculate

           v = 3 10⁸ / 1,333

           v = 2.25 10⁸ m / s

to create the particle the speed must be greater than 2.25 10⁸ m / s

6 0
2 years ago
Nathan accelerates his skateboard uniformly along a straight path from rest to 12.5 m/s in 2.5 s.
kicyunya [14]

Answer:

<h2>a) Nathan's acceleration is 5 m/s² </h2><h2>b) Nathan's displacement during this time interval is 15.625 m</h2><h2>c) Nathan's average velocity during this time interval is 6.25 m/s</h2>

Explanation:

a) We have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = 12.5 m/s    

     Time, t = 2.5 s

     Substituting

                      v = u + at  

                      1.25 = 0 + a x 2.5

                      a = 5 m/s²

     Nathan's acceleration is 5 m/s²

b) We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 5 m/s²  

        Time, t = 2.5 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 2.5 + 0.5 x 5 x 2.5²

                      s = 15.625 m

      Nathan's displacement during this time interval is 15.625 m

c) Displacement = 15.625 m

   Time = 2.5 s

  We have

           Displacement = Time x Average velocity

           15.625 = 2.5 x  Average velocity

           Average velocity = 6.25 m/s

     Nathan's average velocity during this time interval is 6.25 m/s

5 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work
aksik [14]
It is required an infinite work. The additional electron will never reach the origin.

In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
F=k_e  \frac{q_e q_e}{d^2}
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
5 0
2 years ago
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