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Oksana_A [137]
4 years ago
5

Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of th

e bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity.
Physics
1 answer:
Zolol [24]4 years ago
3 0

Answer:

Point motion will eventually stops due to action of g exactly perpendicular...

Explanation:

If ignoring the air resistance, the magnitude of gravitational acceleration is already strong enough to stops the acceleration. As we know that, the spring constant of a bungee spring cord will be F = -k/x, where x is the stretched length and k is the spring constant of bungee cord. If F = ma = w = mg, the g = -m  k/x. Now we can clearly see that the value of g remains constant due to the fluctuating length of the cord as the motion progresses back and forth in SHM say from x1 to x2 and x2 to x1.

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Attenuation is the correct answer. 
6 0
3 years ago
Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance
xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}

\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}

v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}

v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}

v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}

v₂ = 7.6 x 10⁴ m/s

3 0
3 years ago
Which statement about velocity is true?
LekaFEV [45]

Answer:

Velocity has both speed and direction. Speed is constant. ... Speed is measured over time.

Explanation:

8 0
3 years ago
Jackson throws a football 30 meters at a speed of 15 m/s. How long was the football in the air before Laurence caught it for tou
ch4aika [34]

Answer:

2s

Explanation:

Given parameters:

Distance  = 30m

Speed  = 15m/s

Unknown:

Time before Laurence caught it = ?

Solution:

To solve this problem;

    Speed  = \frac{disance }{time}  

 Time taken  = \frac{distance }{speed }   = \frac{30}{15}  = 2s

The time it takes is 2s

6 0
3 years ago
Read 2 more answers
CONFUSED!!!
otez555 [7]

A. Impulse is simply the product of Force and time. Therefore,

I = F * t                                 ---> 1

where I is impulse, F is force, t is time

 

However another formula for solving impulse is:

I = m vf – m vi                    ---> 2

where m is mass, vf is final velocity and vi is initial velocity

 

Therefore using equation 2 to solve for impulse I:

I = 2000kg (0) – 2000kg (77 m/s)
I = -154,000 kg m/s

 

B. By conservation of momentum, we also know that Impulse is conserved. That means that increasing the time by a factor of 3 would still result in an impuse of -154,000 kg m/s. So,

I = F’ * (3 t) = -154,000 kg m/s

Since t is multiplied by 3, therefore this only means that Force is decreased by a factor of 3 to keep the impulse constant, therefore:

(F/3) (3t) = -154,000 kg m/s

 

 

Summary of Answers:

A. I = -154,000 kg m/s

B. Force is decreased by factor of 3

8 0
4 years ago
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