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Ahat [919]
2 years ago
12

A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is

the spring constant of the bungee cord, assuming it has negligible mass compared to that of the jumper in N/m
Physics
1 answer:
andre [41]2 years ago
8 0

Answer:

2.80N/m

Explanation:

Given data

mass m= 56kg

perios T= 11.2s

The expression for the period is given as

T=2π√m/k

Substitute

11.2= 2*3.142*√56/k

square both sides

11.2^2= 2*3.142*56/k

125.44= 351.904/k

k=351.904/125.44

k= 2.80N/m

Hence the spring constant is 2.80N/m

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A skier is accelerating down a 30.0-degree hill at 3.80 m/s^2.
Bond [772]

Answer:

ax = -3.29[m/s²]

ay = -1.9[m/s²]

Explanation:

We must remember that acceleration is a vector and therefore has magnitude and direction.

In this case, it is accelerating downwards, therefore for a greater understanding we will make a diagram of said vector, this diagram is attached.

a_{x}=-3.8*cos(30) = -3.29 [m/s^{2}]\\ a_{y}=-3.8*sin(30) = -1.9 [m/s^{2}]

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2 years ago
Describe what happens to the particles of a liquid, in terms of movement and kinetic energy, as it is cooled to its freezing poi
nataly862011 [7]
As a system is cooled to its freezing point, the kinetic energy of the particles in the system will lower so the movement are much slower. Cooling means taking out heat from the system. This process is a physical change because it is only the phase of the system is changed and it is still the substance after the process.
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3 years ago
If the speed of light in a substance is 2.26 x 10^8 m/s, what is the index of refraction of that substance?
vladimir1956 [14]

Answer:

1.33

Explanation:

speed of light in vacuum, c = 3 x 10^8 m/s

speed of light in medium, v = 2.26 x 10^8 m/s

The refractive index of the medium is given by

μ = speed of light in vacuum / speed of light in medium

μ = (3 x 10^8) / (2.26 x 10^8)

μ = 1.33

4 0
3 years ago
A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 50.9 J and a maximum d
Llana [10]

(a) 2446 N/m

When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:

E=U=\frac{1}{2}kA^2

where

U is the elastic potential energy

k is the spring constant

A is the maximum displacement (the amplitude)

Here we have

U = E = 50.9 J

A = 0.204 m

Substituting and solving the formula for k,

k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m

(b) 50.9 J

The total mechanical energy of the system at any time during the motion is given by:

E = K + U

where

K is the kinetic energy

U is the elastic potential energy

We know that the total mechanical energy is constant: E = 50.9 J

We also know that at the equilibrium point, the elastic potential energy is zero:

U=\frac{1}{2}kx^2=0 because x (the displacement) is zero

Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:

K=E=50.9 J

(c) 8.55 kg

The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when

K = 50.9 J (at the equilibrium position)

Kinetic energy can be written as

K=\frac{1}{2}mv^2

where m is the mass

Solving the equation for m, we find the mass:

m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg

(d) 2.14 m/s

When the displacement is

x = 0.160 m

The elastic potential energy is

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

So the kinetic energy is

K=E-U=50.9 J-31.3 J=19.6 J

And so we can find the speed through the formula of the kinetic energy:

K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s

(e) 19.6 J

The elastic potential energy when the displacement is x = 0.160 m is given by

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

And since the total mechanical energy E is constant:

E = 50.9 J

the kinetic energy of the block at this point is

K=E-U=50.9 J-31.3 J=19.6 J

(f) 31.3 J

The elastic potential energy stored in the spring at any time is

U=\frac{1}{2}kx^2

where

k = 2446 N/m is the spring constant

x is the displacement

Substituting

x = 0.160 m

we find the elastic potential energy:

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

(g) x = 0

The postion at that instant is x = 0, since it is given that at that instant  the system passes the equilibrium position, which is zero.

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Square feet or square meters?

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