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Savatey [412]
3 years ago
10

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
Physics
1 answer:
Alexus [3.1K]3 years ago
8 0

Answer:

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

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2 years ago
The terrestrial planets are made almost entirely of elements heavier than hydrogen and helium. According to modern science, wher
Vera_Pavlovna [14]

Answer:

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Explanation:

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8 0
2 years ago
What is the first law of thermodynamics?​
jok3333 [9.3K]

Answer:

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4 0
2 years ago
Read 2 more answers
If a conducting loop of radius 10 cm is onboard an instrument on Jupiter at 45 degree latitude, and is rotating with a frequency
Pepsi [2]

Answer:

a)  fem = - 2.1514 10⁻⁴ V,  b) I = - 64.0 10⁻³ A, c)    P = 1.38  10⁻⁶ W

Explanation:

This exercise is about Faraday's law

         fem = - \frac{ d \Phi_B}{dt}

where the magnetic flux is

        Ф = B x A

the bold are vectors

        A = π r²

we assume that the angle between the magnetic field and the normal to the area is zero

         fem = - B π 2r dr/dt = - 2π B r v

linear and angular velocity are related

        v = w r

        w = 2π f

        v = 2π f r

we substitute

        fem = - 2π B r (2π f r)

        fem = -4π² B f r²

For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T

we reduce the magnitudes to the SI system

       f = 2 rev / s (2π rad / 1 rev) = 4π Hz

we calculate

       fem = - 4π² 428 10⁻⁶ 4π 0.10²

       fem = - 16π³ 428 10⁻⁶ 0.010

       fem = - 2.1514 10⁻⁴ V

for the current let's use Ohm's law

        V = I R

        I = V / R

         I = -2.1514 10⁻⁴ / 0.00336

         I = - 64.0 10⁻³ A

Electric power is

        P = V I

        P = 2.1514 10⁻⁴ 64.0 10⁻³

        P = 1.38  10⁻⁶ W

6 0
2 years ago
A proton moving at 8.00 106 m/s through a magnetic field of magnitude 1.72 T experiences a magnetic force of magnitude 7.20 10-1
gladu [14]

Answer:

19.1 deg

Explanation:

v = speed of the proton = 8 x 10⁶ m/s

B = magnitude of the magnetic field = 1.72 T

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N

θ = Angle between proton's velocity and magnetic field

magnitude of magnetic force on the proton is given as

F = q v B Sinθ

7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ

Sinθ = 0.327

θ = 19.1 deg

4 0
3 years ago
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