Physics, 14.04.2020 16:53 brandon1748

The jumper has a mass of 50 kg, and the bridge's height above the river is 150 m. The rope has an unstretched length of 11 m and stretches a distance of 4 m before bringing the jumper to a stop. Determine the maximum speed of the jumper and the spring constant of the rope. In your calculation, use g

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Answer from: julesR9814

The speed of jumper is 54.22 m/s and the spring constant of the rope is 9187.5 N/m.

Given data:

The mass of jumper is, m = 50 kg.

The height of bridge above the river is, h = 150 m.

The length of unstretched rope is, L = 11 m.

The stretched length is, L' = 4 m.

In this problem, when the jumper will jump then the potential energy of jumper will get converted into the spring potential energy of the rope. Then the relation will be,

potential energy of jumper = spring potential energy of rope

$mgh=\dfrac{1}{2}(kL'^{2} )$

Here, k is the spring constant.

Solving as,

$mgh=\dfrac{1}{2}(kL'^{2} )\\\\50 \times 9.8 \times 150=\dfrac{1}{2}(k \times 4^{2} )\\k = 9187.5 \;\rm N/m$

Now, apply the third kinematic equation of motion to calculate the final speed of jumper as,

$v^{2} = u^{2} +2gh\\\\v^{2} = 0^{2} +2 \times 9.8 \times 150\\\\v= \sqrt{2 \times 9.8 \times 150}\\\\v =54.22 \;\rm m/s$

Thus, we can conclude that the speed of jumper is 54.22 m/s and the spring constant of the rope is 9187.5 N/m.

Learn more about the spring constant here:

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Answer from: blue5488

The maximum speed of the jumper is 54.2m/s

The spring constant of the rope is

K=9.19KN/m

Explanation:

Step one

According to hook's law the applied force F is proportional to the extension e provided the elasticity is maintained

Step two

Given that

mass= 50kg

Height of bridge h=150m

Extention e=4m

Step three

To determine the maximum speed of the jumper

Since he is going with gravity we assume g=9.81m/s²

And we apply the equation of motion

V²=U²+2gh

where u= initial velocity of the jumper =0

h=height of bridge

Step four

Substituting we have

V²=0²+2*9.81*150

V²=2943

V=√2943

V=54.2m/s

Step five

To solve for the spring constant

We have to equate to potential energy of the jumper to the energy stored in the spring

Potential energy of the jumper =mgh

Energy stored in the spring =1/2ke²

Hence mgh=1/2ke²

Making k subject of formula we have

K=2mgh/e²

Substituting our data into the expression we have

K=2*50*9.81*150/4²

K=147150/16

K=9196.9N/m

K=9.19KN/m

Answer from: Quest

answer: the correct answer id option(d).

explanation:

snell's law refraction:

$\frac{n_2}{n_1}=\frac{sin \theta_1}{sin\theta_2}$

the angle of refraction = $\theta_2=38 ^o$

the angle of incidence = $\theta_1=?$

refractive index of air = $n_1$ = 1

refractive index of glycerol $n_2$ = 1.47

$\frac{1.47}{1}=\frac{sin\theta_1}{sin 38^o}, (sin38^o=0.61)$

$sin\theta_2=0.8967$

$\theta_2=24.2^o, (sin^{-1}0.8967=63.72^o)$

from the given options, the most closest answer to 63.72° is 65° of option (d).

Answer from: Quest

elctricity and magnet is conversion to it

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